1.1. Indefinite Integrals
Learning objectives:
To define the indefinite integral of a function.
To list some standard indefinite integrals.
To study some standard arithmetic rules for indefinite integration.
To determine a function from one of its known values and its derivative.
AND
To practice the related problems.
Definition
A function F ( x ) is an anti-derivative of a function f ( x ) if
F ( x ) f ( x )
for all x in the domain of f .
The set of all anti-derivatives of f is the indefinite integral of f
with respect to
x,
denoted by
f ( x)dx
The symbol ∫ is an integral sign. The function
is the variable of integration.
is the integrand of the integral and
x
Once we have found one anti-derivative F of a function f , the other anti-derivatives
of differ from by a constant. We indicate this in the integral notation in the
following way:
f ( x)dx F x C ........(1)
The constant C is the constant of integration or arbitrary constant. Equation (1) is read,
“The indefinite integral of
with respect to
is
F ( x ) C , we say that we have integrated
F ( x ) C .” When we find
and evaluated the integral.
Example 1:
Evaluate
2x dx
Solution
The formula
2x dx x2 C
+
generates all the anti-derivatives of the function 2 . The functions
x 2 1, x 2 , and x 2 2 are all anti-derivatives of the function 2
, as can
be verified by differentiation.
Many of the indefinite integrals needed in scientific work are found by reversing
derivative formulas. The following table lists a number of standard integral forms side
by side with their derivative-formula sources.
Indefinite Integral
Reversed derivative formula
x n1
x dx
C,
n 1
n 1, n rational
n
dx 1 dx x C (special case)
cos kx
sin
kx
dx
C
k
sec
csc
cos kx dx
sin kx
C
k
2
x dx tan x C
2
x dx cot x C
sec x tan x dx sec x C
csc x cot x dx csc x C
d x n1
n
x
dx n 1
d
x 1
dx
d cos kx
sin kx
dx
k
d sin kx
cos kx
dx k
d
tan x sec2 x
dx
d
cot x csc2 x
dx
d
sec x sec x tan x
dx
d
csc x csc x cot x
dx
Example 2:
a)
b)
c)
d)
x6
x dx
C
6
5
1
dx
x
1
1
1
1
1
2
2
x
x
x 2 dx
C
2x 2 C 2 x C
1
1
1
2
2
cos 2 x
C
2
sin x / 2
x
x
cos dx
C 2sin C
2
1/ 2
2
sin 2 x dx
Once an integral formula is identified, it can be easily checked. The derivative should
be the integrand.
Example 3:
Suppose we think
x cos x dx x sin x cos x C
whether this is correct.
. We can easily check
d
x sin x cos x C x cos x sin x sin x 0
dx
x cos x
The derivative of the right-hand side is the integrand, and so it is correct. The standard
arithmetic rules for indefinite integration are
k f ( x) dx k f ( x) dx Constant Multiple Rule
2. f ( x ) dx f ( x ) dx Rule for Negatives
3. f ( x ) g ( x) dx f ( x ) dx g ( x ) dx
1.
Sum and Difference Rule
Example 4:
5sec x tan x dx 5 sec x tan x dx
5 sec x C 5sec x 5C
The constant of integration 5 times C is an arbitrary constant and so it can be written
simply as C itself. Therefore, we can write simply
5sec x tan x dx 5sec x C
The Sum and Difference Rule for integration enables us to integrate expressions term
by term. When we do so, we combine the individual constants of integration into a
single arbitrary constant at the end.
Example 5:
Evaluate
x 2 2 x 5 dx
Solution: We can generate the anti-derivative term by term with the Sum and
Difference Rule.:
x
2
2 x 5 dx x2 dx 2 xdx 5 dx
x3
C1 x2 C2 5 x C3
3
If we combine C1 , C2 , and C3 into a single constant
C C1 C2 C3 , the formula simplifies to
x3
x2 5 x C
3
and still gives all the anti-derivatives there are. For this reason, we can go right to the
final form when you integrate term by term.
2
2
x
2 x 5 dx x dx 2 xdx 5 dx
x3
x2 5 x C
3
We find the simplest anti-derivative for each part and add the constant at the end.
We can sometimes use trigonometric identities such as
sin 2 x
1 cos 2 x
2
,
cos 2 x
1 cos 2 x
2
to transform integrals into integrals that can be evaluated using standard integral
formulas. For example:
sin 2 x dx
1
1
1
1 cos 2 x dx dx cos 2 x dx
2
2
2
1
1 sin 2 x
x
C
2
2 2
x sin 2 x
C
2
4
1 cos 2 x
x sin 2 x
cos2 x dx
dx
C
2
2
4
1 cos 2 x
dx
2
The process of determining a function from one of its known values and its derivative
( ) has two steps. The first is to find a formula that gives us all the functions that
could possibly have
f
the anti-derivatives of
as a derivative. As we have seen earlier, these functions are
f
and the set of all anti-derivatives of
f
is the indefinite
integral of f . The second step is to use the known function value to select the
particular anti-derivative we want from the indefinite integral. This is illustrated
through an example.
Example 6:
Find the function ( ) whose derivative is
and whose graph passes through the
point (0, 2).
Solution:
Since the derivative of
cos x is sin x , the function f x will be
f x cos x C for some constant
. Since
f 0 2 ,
2 cos 0 C C 3
The formula for f is f x cos x 3
PROBLEM SET
IP1: Evaluate
x2 5x 1
dx
x
Solution:
x2 5x 1
dx
x
x2
dx
x
5x
dx
x
1
dx
x
x3 2 dx 5 x1 2 dx x 1 2 dx
x3 2 1
x1 2 1 x 1 21
5
C
3 2 1
1 2 1 1 2 1
where
is an arbitrary constant
2 52
2
x 5. x3 2 2 x1 2 C
5
3
2
10
x5 2 x3 2 2 x1 2 C
5
3
5
P1: Find 4 x3 2
sin x dx
x
Solution:
1
32 5
32
4
x
sin
x
dx
4
x
dx
5
dx sin x dx
x
x
x3 2 1 x 1 21
4
5
cos x C
3
2
1
1
2
1
where
is an arbitrary constant
2 5
4 x 2 10 x cosx C
5
5
8
x 2 10 x cosx C
5
IP2: Evaluate
sin 3 2 x dx
Solution:
3sin 2 x sin 6 x
sin 3 2 x dx
dx
4
1
3sin 2 x sin 6 x dx
4
3
1
sin 2 x dx
sin 6 x dx
4
4
3 cos 2 x 1 cos 6 x
C
4 2
4 6
where
is an arbitrary constant
cos 6 x 3cos 2 x
C
24
8
x
1 sin dx
2
P2: Find
Solution:
1 sin
x
dx
2
cos 2
x
x
x
x
sin 2 2sin .cos dx
4
4
4
4
2
x
x
sin 4 cos 4 dx
x
x
sin cos dx
4
4
x
x
sin dx cos dx
4
4
cos x 4 sin x 4
C
14
14
4 sin x 4 cos x 4 C
IP3: Evaluate
sin 6 x cos6 x
sin2 x.cos2 x
dx
Solution:
To evaluate
sin 6 x cos6 x
sin2 x.cos2 x
The integrand can be written as
dx
6
6
sin x cos x
sin 2 x.cos 2 x
sin
sin 2 x
3
cos 2 x
3
sin 2 x.cos 2 x
3
3 sin2 x.cos2 x sin 2 x cos2 x
sin2 x.cos2 x
13 3 sin 2 x.cos 2 x 1 1 3 sin 2 x.cos 2 x
2
2
sin x.cos x
sin 2 x.cos2 x
2
2
x cos x
1
sin 2 x.cos 2 x
1
2
2
3sin 2 x.cos 2 x
sin 2 x.cos 2 x
3
sin x.cos x
sin6 x cos6 x
1
dx
3
2
dx
2
2
2
sin x.cos x
sin x.cos x
1
dx 3dx
2
2
sin x.cos x
sin2 x cos2 x
sin2 x.cos2 x
dx 3x
sin2 x
cos2 x
2
dx 3x
sin x.cos2 x sin2 x.cos2 x
1
1
2 2 dx 3x
cos x sin x
sec2 xdx csc2 xdx 3x
tan x cot x 3x C
where
P3:
2cos3 x 3sin 3 x
2
2
cos x.sin x
Solution:
is an arbitrary constant
dx f x C f x ?
To evaluate
2cos3 x 3sin 3 x
2
2
dx
cos x.sin x
The integrand can be written as
2cos3 x 3sin 3 x
2
2
cos x.sin x
2
2
2
3sin 3 x
cos x.sin x cos2 x.sin 2 x
2cos x 3sin x
sin 2 x cos 2 x
cos x 1
sin x 1
2
3
sin x sin x
cos x cos x
2cot x.csc x 3tan x.sec x
2cos3 x 3sin 3 x
2
2cos3 x
dx
cos x.sin x
2 csc x.cot x dx 3 sec x.tan x dx
2csc x 3sec x C
3sec x 2csc x C f x C
f x 3sec x 2csc x
where
is an arbitrary constant
IP4: Find the function ( ) whose derivative is
+
through the point ( , ).
Solution:
Given the derivative of ( ) is ( ) =
+
∴ The function ( ) is the anti-derivative of ( ) and
f (t ) dt C
cos t sin t dt C
F (t )
sin t cos t C
where
Given ( ) = 1
⟹
−
+ =1
⟹ 0+1+ = 1 ⟹ =0
∴ ( )=
−
is an arbitrary constant
and whose graph passes
P4: Find the function ( ) whose derivative is
−
passes through the point (− , ).
Solution:
Given the derivative of ( ) is ( ) = 9 − 4 + 5
∴ The function ( ) is the anti-derivative of ( ) and
+
and whose graph
f ( x) dx C
9 x 2 4 x 5 dx C
F ( x)
x3
x2
9 4 5x C
3
2
3 x3 2 x2 5 x C
where
is an arbitrary constant
Given (−1) = 0
⟹ 3(−1) − 2(−1) + 5(−1) + = 0
⟹ −3 − 2 − 5 + = 0 ⟹ = 10
∴ ( ) = 3 − 2 + 5 + 10
EXERCISES:
Find an anti-derivative for each function. Do as many as you can mentally. Check
your answers by differentiation.
1. 2x, x 2 ,
2. 3x 4 ,
3.
4.
5.
6.
7.
8.
9.
x2 , 2x 1
x 4 , x4 2 x 3
1 5
5
,
,
2
x2 x 2
x2
3
1
1
x,
,
x
2
2 x
x
2 1/3
1 2/3
1
x
,
x
, x 4/3
3
3
3
sin x, 3sin x,
sin x 3sin x
2 2x
3x
sec2 x,
sec , sec2
3
3
2
x
x
csc x cot x, csc5 x cot 5 x, csc cot
2
2
2
sin x cos x
Evaluate the integrals. Check your answers by differentiation.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
x 1 dx
t
3
t
2 dt
2 x 5x 7 dx
1
1
x
dx
x
3
x dx
x x dx
2
8
y
y dy
2 x 1 x dx
t t t
t dt
2 cos t dt
7sin
3 d
3csc x dx
csc cot
2 d
4sec x tan x 2sec x dx
sin 2 x csc x dx
4sin y dy
1 cos 4t
2 dt
1 tan d
cot x dx
cos tan sec d
2
3
2
2
1/3
3
1/4
3
2
2
2
2
2
2
2
In problems 30-32, find the function with the given derivative whose graph passes
through the point P.
30.
31.
f x 0
f x 2x 1
32.
r 8 csc2
P( 1,3)
P (0,0)
P( ,0)
4
1.2. Integration by Substitution
Learning objectives:
To study power rule in integral form.
To evaluate indefinite integrals by substitution method.
AND
To practice the related problems.
A change of variable can often turn an unfamiliar integral into one we can evaluate.
The method for doing this is called the substitution method of integration. It is one of
the principal methods for evaluating the integrals.
Example 1:
Evaluate
5
x
2
dx
Solution:
We put the integral in the form
u x 2 ,
Then
by substituting
du d x 2 1 0 dx dx
6
x 2
u6
x
2
dx
u
du
C
C
6
6
Example 2:
2
u n du
5
5
1 y 2 y dy u1/2du ,
u(1/2)1
2
C u3/2 C
3
1/ 2 1
3/2
2
1 y2
C
3
u 1 y 2 , du d 1 y2 2 ydy
Example 3:
1
4t 1 dt u1/2 . du ,
4
1
u 4t 1; du d 4t 1 du 4dt du dt
4
1 1/2
1 u 3/2
1
u du
C u 3/2 C
4
4 3/ 2
6
1
3/2
4 t 1 C
6
The companion formula for the integral of sin u when is a differentiable function is
sin u du cos u C
Example 4:
1
cos 7 5 d cos u du ,
7
1
u 7 5 du d 7 5 du 7d d du
7
1
1
cos u du sin u C
7
7
1
sin 7 5 C
7
Example 5:
x2 sin x3 dx sin x3 x2 dx
u x3 du d x3 du 3 x 2dx dx
sin u x2
1
3x
1
3x 2
du
2
du
1
sin udu
3
1
1
cos u C cos x3 C
3
3
Example :
x
2
2x 3
2
x 1dx
put u x 2 2 x 3; du 2 xdx 2dx 2 x 1 dx
1 2
1 u3
u du
u du C
2
2
2 3
3
1
1
u 3 C x2 2 x 3 C
6
6
2 1
Example 8:
4
sin
t cos t dt u 4 du
, u sin t ; du cos t dt
u5
sin 5 t
C
C
5
5
The success of the substitution method depends on finding a substitution that will
change an integral we cannot evaluate directly into one that we can.
PROBLEM SET
IP1: Evaluate
x
4
14
x
x
dx
5
Solution:
14
4
1
14
x
1
4
x x
x3
dx
5
x
x5
dx
1
14
1
4 4
x
1 3
x
x
5
dx
14
Put 1 −
=
⟹ 0−
1
x 1 3
x
x5
⟹
1−
=
⟹
14
1
1
1
x 4 x3
dx
14
x
1
dx
1
x5 x3
14
1
1
dx
1
x 4 x3
=
=
1
x
⟹
14 x
u
4
=
4
3
du
1 14
u du
3
54
1 u1 4 1
4 54
4
1
C u C 1 3 C ,
3 1 4 1
15
15 x
where C is an arbitrary constant
dx
P1: Evaluate
3x 1
3
dx
3x2 2 x 1
Solution:
To evaluate
3x 1
3
dx
3 x 2 2 x 1
Put 3 x 2 2 x 1 u
6 x 2 dx du 3x 1 dx
3x 1
1
du
2
1 du
dx
3
2 u3
2
3x 2 x 1
1 3
1 u 31
u du
2
2 3 1
1
4u
1
C
2
4 3x2 2 x 1
where
IP2: Evaluate
2
C
is an arbitrary constant
x2
dx , ∈ (−5, ∞)
x5
Solution:
To evaluate
x2
dx
x5
> 0 on (−5, ∞)
=
−5
Put + 5 =
so that
⟹
=2
and
x2
dx
x5
u 5
2
u
2
.2u du 2
u 4 25 10u 2 du
2 u 4 du 25 du 10 u 2 du
u5
u3
2 25u 10 C
3
5
2
20
x 55 2 50 x 5 x 5 3 2 C
5
3
where
x2
P2: Evaluate
is an arbitrary constant
dx
1 x
Solution:
To evaluate
Put 1 +
=
1 x
⟹
dx
=2
2
and
=
−1
u 2 1
2
x
dx
1 x
x2
u .2u du
2 u 4 1 2u 2 du
2 u 4 du du 2 u 2 du
u5
u3
2 u 2. C
3
5
2
4
1 x 5 2 2 1 x 1 x 3 2 C
5
3
where
IP3: Evaluate
,
is an arbitrary constant
sin 2 x.cos5 x dx
Solution:
To evaluate
=
sin 2 x.cos5 x dx sin 2 x.cos 4 x. cos x dx
Put
=
sin 2 x.cos5 x dx
⟹
2
sin 2 x. 1 sin 2 x . cos x dx
2
du u 2 1 u 4 2u 2 du
u 3 u 7 2u 5
2
6
4
u u 2u du
C
3
7
5
2
u 1 u
2
,
P3: Evaluate
sin 3 x sin 7 x 2sin 5 x
C , where is an arbitrary constant
3
7
5
sin 4 x
dx
6
cos x
Solution:
sin 4 x
dx tan 4 x. sec2 x dx
cos6 x
=
Put
=
u5
dx u du
C
5
cos6 x
sin 4 x
⟹
4
, where
is an arbitrary constant
tan 5 x
C
5
tan x.sec2 x
IP4: Evaluate
a b tan x
2
dx
2
Solution:
To evaluate
Put
+
tan x.sec2 x
a b tan2 x
=
tan x.sec2 x
a b tan x
2
2
⟹2
⟹
dx
dx
2
.
=
.
=
1 1
1 1
du C
2b u 2
2b u
1
2
2b a b tan x
tan 4 x .sec2 x
dx
x
To evaluate
tan 4 x .sec2 x
dx
x
Put √ =
⟹
P4: Evaluate
Solution:
√
=
⟹
= 2√
C , where is an arbitrary constant
tan 4 x .sec2 x
dx 2 tan 4 u.sec2 u du
x
tan
=
Again put
4
2
x .sec
x
x
⟹
=
v5
dx 2 v dv 2 C
5
2
tan 5 u C
5
2
tan 5 x C
5
4
where C , is an arbitrary constant
EXERCISES:
Evaluate the integrals
1.
2 t
3t 2 dt , 2.
4. x1 3 dx , 5.
7.
1
1
2 x3 5 x 17 dx , 3. x 2 dx
3
x3
x 3 x dx , 6.
2 x 1 x 3 dx ,8.
2
8
y
dy
1 4
y
t t t
dt , 9.
2
t
2cos t dt,10. 7sin
3
d
Evaluate the indefinite integrals by using the given substitution to reduce the integrals
to standard form.
1.
2.
3.
sin3x dx, u 3x
u 2t
sec2t tan 2t dt ,
5
28
7
x
2
dx,
u 7x 2
9r dr
u 1 r
1 r ,
x sin x 1dx , u x 1
csc 2 cot 2 d
2
4.
3
3
5.
6.
2
3/ 2
2
a. using u cot 2
b. using u csc 2
3/
2
Evaluate the integrals
3 2s ds
7.
8.
4
1 2 d , 3 y 7 3 y 2 dy
x
x
dx , cos 3 z 4 dz , sec 3 x 2 dx , sin 5 cos dx
2
3
3
x
1
1
ds ,
5s 4
,
2
x 1
5
9.
10.
3
2 r
r 1 dr ,
18
x1/2 sin x3/2 1 dx , sec v tan v dv ,
2
2
sin 2t 1
1
1
cot y csc2 y dy ,
cos 1dt ,
dt
t
t2
cos 2 2t 1
1
11.
2
12.
s
3
1
1
sin cos d ,
2
2
3
2 s 5 s 5 3s 4 s 5 ds , t 1 t
4
3
dt
1.3. Approximation by Finite Sums
Learning objectives:
To show how areas, volumes and the distances traveled by objects over time
can be approximated by the finite sums.
AND
To practice the related problems.
One of the great achievements of classical geometry was to obtain formulas for the
areas and volumes of triangles, spheres and cones. We now study a method to
calculate the areas and volumes of these and other more general shapes. The method
is integration and it is a tool for calculating much more than areas and volumes. The
integral has many applications in sciences, engineering, economics and statistics.
The idea behind integration is that we can effectively compute many quantities by
breaking them into small pieces and then summing the contributions from each small
part. We begin with examples involving finite sums. These lead to question of what
happens when more and more terms are summed. Passing to the limit, as the number
of terms goes to infinity, gives an integral.
This module shows how areas, volumes and distance travelled by an object over time
can be approximated by finite sums. Finite sums are the basis for defining the integral.
Distance Traveled
ds
f t m/s of a car moving down a
dt
highway and want to know how far the car will travel in the time interval a t b .
Suppose we know the velocity function v
We can approximate the distance traveled with a sum in the following way. We partition
a, b into short time intervals on each of which v is fairly constant. We approximate the
distance traveled on each time interval with the formula
Distance f t t
and add the results across a, b .
Suppose the partitioned interval looks like this
with the subintervals all of length t . Let t1 be a point in the first subinterval. If the
interval is short enough so the rate is almost constant, the car will move about f t1 t
m during that interval. If t 2 is a point in the second interval, the car will move an
additional f t 2 t m during that interval, and so on. The sum of these products
approximates the total distance D traveled from t
a to t b .
f t1 t f t2 t f t n t
If we use n subintervals, then D
Example 1:
The velocity function of a projectile fired straight into the air is f t 160 9.8 t .
Use the summation technique just described to estimate how far the projectile rises
during the first 3 seconds. How close do the sums come to the exact figure of 435.9 m?
Solution
We consider 3 subintervals of length 1, with evaluated at left endpoints.
D f t1 t f t2 t f t3 t
160 9.8 0 1 160 9.8 1 1 160 9.8 2 1 450.6
We now consider 3 subintervals of length 1, with
evaluated at right endpoints.
D f t1 t f t 2 t f t3 t
160 9.8 1 1 160 9.8 2 1 160 9.8 3 1 421.2
With 6 subintervals of length , we get
Using left endpoints:
D = 443.25
Using right endpoints:
D = 428.55
These six-interval estimates are somewhat closer than three-interval estimates. The
results improve as the subintervals get closer.
As we can see from these figures, the left endpoint sums approach the true value
435.9 from above while the right-endpoint sums approach it from below. The true
value lies between these upper and lower sums.
Area and Cardiac output
The number of liters of blood the heart of a person pumps in a minute is called Cardiac
output. In a clinical test for determining the cardiac output, 5.6 mg of dye is injected in
a main vein near the heart. The concentration (mg/L) of the dye is measured every few
seconds and the data is plotted as shown below.
The formula for the cardiac output is given by
Cardiac output
amount of dye
60 .
area under curve
If we determine the area under the curve, we can find the cardiac output of the
patient.
There are no area formulas for this irregularly shaped region. But we can get a good
estimate of this area by approximating the region between the curve and the −axis
with rectangles and adding the areas of rectangles.
Each rectangle omits some of the area under the curve but includes area from outside
the curve, which compensates. Each rectangle has a base 2 units long and a height that
is equal to the height of the curve above the midpoint of the base. The rectangle’s
height acts as a sort of average value of the function over the time interval on which
the rectangle stands.
Area under curve sum of rectangle areas
f 6 2 f 8 2 f 10 2 f 28 2
1.4 2 6.3 2 7.5 2 0.1 2
28.8 2 57.6 mg s/L
5.6 mg
60 s
Cardiac output
5.8 L / min
57.6 mg.s / L 1 min
The patient’s cardiac output is about 5.8 L/min.
Additional Examples
Example 2:
A solid lies between planes perpendicular to the −axis at = −2 and = 2. The
cross sections of the solid perpendicular to the axis between these planes are vertical
squares whose base edges run from the semicircle
y 9 x2 .
y 9 x2
to the semicircle
2
The height of the square at is 2 9 x . Estimate the volume of the solid.
Solution
We partition the interval [−2, 2] on the −axis into four subintervals of length ∆ = 1.
The solid’s cross section at the left endpoint of each subinterval is a square.
On each of these squares we construct a right cylinder (Square slab) of height 1
extending to the right.
We add the cylinders’ volumes to estimate the volume of the solid.
The area of the solid’s cross section at is
A( x ) 2 9 x 2
2
4 9 x2 .
So, the sum of the volumes of the cylinders is
S 4 A c1 x A c2 x A c3 x A c4 x
4 9 c12 1 4 9 c22 1 4 9 c32 1 4 9 c42 1
2
2
2
2
4 9 2 9 1 9 0 9 1
4 9 4 9 1 9 0 9 1
4 30 120
We will show later that the true volume of solid is V 368 / 3 122.67 . The
difference between S and V is a small percentage of V :
Error percentage
V S4
V
368 / 3 120 8 2.2%
368
368 / 3
With a finer partition (more subintervals) the approximation would be even better.
Example 3:
Estimate the volume of a solid sphere of radius 4.
Solution
We picture the sphere as if its surface were generated by revolving the graph of the
function
f x 16 x 2
about the
−axis.
We partition the interval 4 x 4 into 8 subintervals of length x = 1. We then
approximate the solid with right circular cylinders based on cross sections of the solid
by planes perpendicular to the x -axis at the subintervals’ left-hand endpoints.
The cylinder at = −4 is degenerate because the cross section there is just a point.
We add the cylinders’ volumes t o estimate the volume of a sphere.
The sum of the eight cylinders’ volume is
The sphere’s true volume is V
The difference between
4 3 4
256
r (4)3
3
3
3
S8 and V
is a small percentage of
V:
Error percentage
V S8
V
256 / 3 84
256 / 3
256 253 1
1.6%
256
64
All the examples above describe instances in which sums of function values multiplied
by interval lengths provide approximations that are good enough to solve practical
problems.
PROBLEM SET
IP1: A 5-mg dye is injected in a main vein near the heart. The concentration of the
dye (in milligrams per liter) is measured in the aorta at one-second intervals as
shown in the chart. Estimate the patient’s cardiac output.
Seconds after
Dye concentration
injection (t)
(mg/L)
0
0
1
0.4
2
2.8
3
6.5
4
9.8
5
8.9
6
6.1
7
4.0
8
2.3
9
1.1
10
0
Solution:
From the figure, each rectangle has a base of 1 unit. The rectangle’s height acts as
average value of the function over the time interval on which the rectangle stands.
∴ Area under the curve ≈ sum of the rectangles areas
= (0). 1 + (1). 1 + (2). 1 +……….+ (9) + (10)
0 + 0.4 + 2.8 + 6.5 + 9.8 + 8.9 + 6.1 + 4.0
=
+2.3 + 1.1 + 0
= 41.9 mg.s/L
Now, the cardiac output is =
.
. /
×
= 7.2 L/min
Hence the patient’s cardiac output is . L /min
P1: The table below gives dye concentration for a dye-dilution cardiac output
determination. The amount of dye injected is 5mg. use rectangles to estimate the
area under the dye concentration curve and then estimate the patient’s cardiac
output.
Seconds after
Dye concentration
injection(t)
(mg/L)
2
0
4
0.6
6
1.4
8
2.7
10
3.7
12
4.1
14
3.8
16
2.9
18
1.7
20
1.0
22
0.5
24
0
Solution: From the figure, each rectangle has a base of 2 units. The rectangle’s height
acts as average value of the function over the time interval on which the rectangle
stands.
∴ Area under the curve ≈ sum of the rectangle’s areas
= (2). 2 + (4). 2 + (6). 2 +………+ (24). 2
0 + 0.6 + 1.4 + 2.7 + 3.7 + 4.1 + 3.8
=2
+2.9 + 1.7 + 1.0 + 0.5 + 0
= 2(22.4) = 44.8 mg.s/L
Now, the cardiac output is =
.
. /
×
= 6.6964
≈ 6.7 /
Hence the patient’s cardiac output is about 6.7 /
IP2: Distance traveled upstream
You are sitting on the bank of a tidal river watching the incoming tide carry a bottle
upstream. You record the velocity of the flow every 5 minutes for an hour, with the
results shown in the accompanying table. About how far upstream did the bottle
travel during that hour? Find an estimate using 12 subintervals of length 5 using
a. Left endpoints
b. Right endpoints
Time
Velocity
t (minutes)
v m/sec)
0
1
5
1.2
10
1.7
15
2.0
20
1.8
25
1.6
30
1.4
35
1.2
40
1.0
45
1.8
50
1.5
55
1.2
60
0
Solution:
In upstream the distance travelled by the bottle in an hour can be estimated as
follows:
Using left endpoints
The estimation using 12 subintervals of length 5minutes (300 sec) using left endpoints is
≈ ( )∆ + ( )∆ + ( )∆ +……..+ ( )∆
= [ (0) + (5) + (10 ) +……..+ (55 )](300)
(∵ ∆ = 300
1 + 1.2 + 1.7 + 2.0 + 1.8 + 1.6 + 1.4
=
(300)
+1.2 + 1.0 + 1.8 + 1.5 + 1.2
= 5220
)
Using right endpoints
The estimation using 12 subintervals of length 5minutes (300 sec) using right endpoints is
≈ ( )∆ + ( )∆ + ( )∆ +……..+ ( )∆
= [ (5) + (10 ) + (15) +……..+ (60)](300)
1.2 + 1.7 + 2.0 + 1.8 + 1.6 + 1.4
=
(300)
+1.2 + 1.0 + 1.8 + 1.5 + 1.2 + 0
= 4920
P2. The table below shows the velocity of a model train engine moving along a track
for 10 seconds. Estimate the distance travelled by the engine using 10 subintervals of
length 1 using
a. Left endpoints
b. Right endpoints
Time t (sec)
Velocity
v(t) (m/sec)
0
0
1
12
2
22
3
10
4
5
5
13
6
11
7
6
8
2
9
6
10
0
Solution:
a. Using left endpoints
The approximate distance travelled by the engine using 10 subintervals of length 1
using left endpoints is given by
≈ ( ) ∆ + ( )∆ + ( )∆ +………+ ( )∆
= (0) + (1) + (2) +………+ (9)
( ∵ ∆ = 1)
= [0 + 12 + 22 + 10 + 5 + 13 + 11 + 6 + 2 + 6]
= 87
b. Using right hand endpoints
The approximate distance travelled by the engine using 10 subintervals of length 1
with right endpoints is given by
≈ ( ) ∆ + ( )∆ + ( )∆ +………+ ( )∆
= (1) + (2) + (3) +………+ (10 )
( ∵ ∆ = 1)
= [12 + 22 + 10 + 5 + 13 + 11 + 6 + 2 + 6 + 0](1)
= 87
IP3. Estimate the volume of a solid sphere of radius 4.
Solution
We picture the sphere as if its surface were generated by revolving the graph of the
function
f x 16 x 2
about the
−axis.
We partition the interval 4 x 4 into 4 subintervals of length x 2 . We then
approximate the solid with right circular cylinders based on cross sections of the solid
by planes perpendicular to the x -axis at the subintervals’ left-hand endpoints.
The cylinder at = −4 is degenerate because the cross section there is just a point.
We add the cylinders’ volumes to estimate the volume of a sphere.
The sum of the four cylinders’ volume is
2
2
S4 f c1 x ....... f c4 x
2
2
16 c12 x 16 c4 2 x
The
2
2
2
2
2 16 4 16 2 16 0 16 2
2 0 12 16 12 80
4 3 4
256
3
sphere’s true volume is V r (4)
3
3
3
The difference between S 4 and V is a small percentage of V :
Error percentage
V S 4 256 / 3 80 256 240 1
6.2%
V
256
16
256 / 3
P3. A solid lies between planes perpendicular to the −axis at = − and = .
The cross sections of the solid perpendicular to the axis between these planes are
vertical squares whose base edges run from the semicircle y 9 x2 to the
semicircle y 9 x 2 .
2
The height of the square at is 2 9 x . Estimate the volume of the solid.
Solution:
We partition the interval [−2, 2] on the −axis into eight subintervals of length ∆ =
0.5. The solid’s cross section at the left endpoint of each subinterval is a square.
On each of these squares we construct a right cylinder (Square slab) of height 1
extending to the right.
We add the cylinders’ volumes to estimate the volume of the solid.
The area of the solid’s cross section at is
A( x ) 2 9 x 2
2
4 9 x2 .
So, the sum of the volumes of the cylinders is
S8 A c1 x A c2 x ...... A c8 x
4 9 c12 0.5 4 9 c22 0.5 ...... 4 9 c82 0.5
2
2
2
We will show later
2 9 2 9 1.5 ....... 9 1.5
2 9 4 9 0 2 9 2.25 9 1 9 0.25
2 61 122
that the true volume of solid is V 368 / 3 122.67 . The difference between
and V is a small percentage of V :
Error percentage
S
V S8 368 / 3 122
0.67
0.5461%
V
122.67
368 / 3
IP4. The nose “cone” of a rocket is a paraboloid obtained by revolving the curve
x , 0 x 5 , about the x -axis, where x is measured in meters. To
estimate the volume V of the nose cone, we partition [ , ] into five subintervals of
equal length, slice the cone with planes perpendicular to the x -axis at the
y
subintervals’ left-hand endpoints, and construct cylinders of height 1 based on cross
sections at these points.
S5 of the volumes of the cylinders. Do you expect to S5 to
overestimate V , or to underestimate V ? Give reasons for your answer.
The true volume of the nose cone is V = 25 π/2 m3 . Express V S5 as a
percentage of V to the nearest percent.
a. Find the sum
b.
Solution:
a. We partition the interval [0, 5] into 5 subintervals of length ∆
= 1 and by slicing
the cone with planes perpendicular to the −axis at the subintervals’ left hand
endpoints, we construct cylinders of height 1 based on cross sections at these
points.
Given that ( ) = √
Now, we add the cylinders’ volumes to estimate the
volume of the nose cone
2
2
2
S5 f c1 x f c2 x ..... f c5 x
2
2
2
c1 c2 ..... c5 x
0 1 2 3 4 (1) 10
Given true volume of the nose cone is
By comparing the estimated volume
underestimated V, because
>
V
25
12.5m3
2
S5 and true volume V of the nose cone, S5 is
S5
b. Error percentage is
=
S5
=
|
|
.
.
= 0.2 ≈ 20%
P4: A solid lies between planes perpendicular to the x -axis at x 0 and x 4 .
The cross sections of the solid perpendicular to the axis between these planes are
vertical squares whose base edges run from the parabolic curve
parabolic curve
y
c. Find the sum
y x
to the
x.
S4 of the volumes of the cylinders obtained by partitioning
0 x 4 into four subintervals of length 1 based on the cross sections
at the subinterval’s right-hand endpoints.
d. The true volume is V
the nearest percent.
Solution:
32 . Express |V S 4 | as a percentage of V
a. We partition the interval [0, 4] into 4 subintervals of length ∆
to
= 1. The cross
sections of the solid perpendicular to the axis between these planes are vertical
squares whose base edges runs from the parabolic curve = −√ to the parabolic
curve = √ at the left hand endpoints is a square.
The area of the solid’s cross section at is
( )= √ +√
=4
Now, the sum of the volumes of the cylinders is
= ( )∆ + ( )∆ + ( )∆ + ( )∆
= [4(1) + 4(2) + 4(3) + 4(4)](1) = 40
b.Given that true volume is
The difference between
Error percentage is
|
|
|
=
=
= 32
and V is a small percentage of V
|
=
≈ 25%
EXERCISES
1. The table below gives dye concentrations for a dye-dilution cardiac-output
determination. The amount of dye injected is 5 mg. Use rectangles to estimate the
area under the dye concentration curve and then estimate the patient’s cardiac
output.
Seconds after
Dye concentration
Injection (t)
(c)
2
0
4
0.6
6
1.4
8
3.7
10
4.7
12
5.1
14
4.2
16
3.0
18
2.7
20
1.5
22
0.6
24
0
2. The table below shows the velocity of a model train engine moving
along a track for 12 seconds. Estimate the distance traveled by the
engine using 10 subintervals of length 2 with (a) left-endpoint values
and (b) right-endpoint values.
Time (seconds)
0
2
4
6
8
10
12
Velocity (m/sec)
0
14
24
11
6
3
0
3. Read example 2. Suppose we use only two square cylinders to
estimate the volume V of the solid.
a. Find the sum S2 of the volumes of the cylinders.
4.
5.
b. Express V S 2 as a percentage of V to the nearest percent.
Read example 3. Suppose we approximate the volume V of the
sphere by partitioning the interval 4 x 4 into four subintervals
of length 2 and using cylinders based on the cross sections at the
subintervals’ right-hand endpoints.
a. Find the sum S4 of the volume of the cylinders.
b. Express V S 4 as a percentage of V to the nearest percent.
To estimate the volume V of a solid hemisphere of radius 4, imagine
its axis of symmetry to be the interval [0, 4] on the x -axis. Partition [0,
4] into eight subintervals of equal length and approximate the solid
with cylinders based on the circular cross sections of the hemisphere
perpendicular to the x -axis at the subintervals’ right-hand endpoints.
S8 of the volumes of cylinders. Do you expect
S8 to overestimate V , or to underestimate V ? Give reasons
a. Find the sum
for your answer.
b. Express V S8 as a percentage of V to the nearest percent.
6. A solid lies between planes perpendicular to the x -axis at x = 0 and
x = 4. The cross sections of the solid perpendicular to the axis
between these planes are vertical squares whose base edges run from
the parabolic curve
a. Find the sum
y x
to the parabolic curve
y
x.
S 4 of the volumes of the cylinders obtained by
partitioning 0 x 4 into four subintervals of length 1
based on the cross sections at the subinterval’s left-hand
endpoints.
b. The true volume is V = 32. Express |V
S4 | as a percentage
of V to the nearest percent.
c. Repeat parts (a) and (b) for the sum
S8
7. A reservoir shaped like a hemispherical bowl of radius 8 m is filled with
8.
water to a depth of 4 m.
a. Find an estimate S of the water’s volume by approximating the
water with eight circumscribed solid cylinders.
b. The true water’s volume is V = 320 π/3 m3. Find the error |
V S | as a percentage of V to the nearest percent.
The nose “cone” of a rocket is a paraboloid obtained by revolving the
curve y
x , 0 x 5 , about the
x -axis, where x is measured in meters. To estimate the volume V
of the nose cone, we partition [0,5] into five subintervals of equal
length, slice the cone with planes perpendicular to the x -axis at the
subintervals’ right-hand endpoints, and construct cylinders of height 1
based on cross sections at these points.
S5 of the volumes of the cylinders. Do you expect
to S 5 to overestimate V , or to underestimate V ? Give
a. Find the sum
reasons for your answer.
b. The true volume of the nose cone is V = 25 π/2 m3 . Express
V S5
as a percentage of V to the nearest percent.
1.4. Average value of a Non-Negative function
Learning objectives:
To estimate the average value of a non-negative continuous function in the given
interval.
AND
To practice the related problems.
The average value of a collection of numbers , , … is obtained by adding them
together and dividing by . But what is the average value of a continuous function
on an interval [ , ] ? Such a function assumes infinitely many values.
The average value of a non-negative continuous function
The average value of a non-negative continuous function over an interval [ , ] can
be approximated by finite sums.
We first subdivide the interval [ , ] into subintervals of equal width ( . . ,
ℎ)
∆ =
. If ( ) is the value of at the chosen point in the
subinterval then
Average value of over [ , ]
f c1 f c2 ..... f ck ..... f cn
n
f c1
n
f c2
n
...
f ck
n
...
f cn
n
1
1
1
1
f c1 f c2 ... f ck ... f cn
n
n
n
n
1
1
1
1
1
b
a
f
c
f
c
...
f
c
...
f
c
1
2
k
n
b a
n
n
n
n
1
ba
ba
ba
b a
f
c
f
c
...
f
c
...
f
c
1
2
k
n
b a
n
n
n
n
1
f c1 .x f c2 .x ... f ck .x ... f cn .x
ba
1
.A
length of a, b
where A is the approximate area that lies above −axis, below the graph of the
funtion ( ) and between vertical lines = and = .
By taking more and more subintervals (i.e., n sufficiently large), we get better and better
approximation for A and thereby we get better and better average values of over
[ , ].
We define the average value of a non-negative function on an interval [ , ] as the area
under its graph divided by − .
Example 1:
Estimate the average value of the function
f x x 2 on the interval [−1, 1].
Solution:
2
The graph of y x and the partition of the interval [−1, 1] into 6 subintervals of
length x 1/ 3 are shown below.
The six subintervals are
−1, − , − , − , − , 0 , 0, ,
, ,
,1
We estimate the average value with a finite sum.
It appears that a good estimate for the average of square on each subinterval is the
square of the midpoint of the subinterval.
We have = − ,
=− ,
=− ,
= ,
= ,
=
are the midpoints of the above six subintervals respectively.
The average value of the function on [−1, 1] is
1
f c1 .x f c2 .x ... f c6 .x
length 1, 1
2
2
2
2
2
2
1 5 3 1 1 3 5 1
1 (1) 6 6 6 6 6 6 3
1 25 9 1 1 9 25 70
216 0.324
6
36
Note: We can calculate the average value in the above example using anti-derivative,
as we see in a module on definite integrals. Later, we will show that the true average
value of the function is 1/3, and this approximation compares well with the true value.
PROBLEM SET
IP1: Use a finite sum to estimate the average value of a function ( ) =
on the
interval [ , ] by partitioning the interval into 4 subintervals of equal length and
evaluating at the subinterval midpoints.
Solution:
We partition the interval [0, 8] into 4 subintervals of length ∆ = 2 which are as
follows.
[0, 2] , [2, 4] , [4, 6] , [6, 8]
We estimate the average value with a finite sum by using midpoint rule.
We have = 1,
= 3,
= 5,
= 7 are the midpoints of the above four
subintervals respectively.
Average value of the function on [0, 8] is
1
f c1 .x f c2 .x f c3 .x f c4 .x
length 0, 8
= [ (1). 2 + (3). 2 + (5). 2 + (7). 2]
= [2(1) + 2(3) + 2(5) + 2(7) ] ≈ 42
P1: Use a finite sum to estimate the average value of a function ( ) =
on the
interval [ , ] by partitioning the interval into 4 subintervals of equal length and
evaluating at the subinterval midpoints.
Solution:
We partition the interval [0, 2] into 4 subintervals of length ∆ =
= 0.5 which are
[0, 0.5] , [0.5, 1] , [1, 1.5] , [1.5, 2]
We estimate the average value with a finite sum by using midpoint rule.
We have = 0.25,
= 0.75,
= 1.25,
= 1.75 are the midpoints of the above
four subintervals respectively.
Average value of the function on [0, 2] is
1
f c1 .x f c2 .x f c3 .x f c4 .x
length 0, 2
= [ (0.25)(0.5) + (0.75)(0.5) + (1.25)(0.5) + (1.75)(0.5)]
= (0.5)(0.5)[(0.25) + (0.75) + (1.25) + (1.75) ]
= 0.25[0.01563 + 0.42187 + 1.95312 + 5.35937] ≈ 1.9375
IP2: Estimate the average value of the function ( ) = +
on the
interval [ , ].
Solution:
We partition the interval [0, 2] into 4 subintervals of lengths ∆ = 0.5 = which are
as follows.
[0, 0.5] , [0.5, 1] , [1, 1.5] , [1.5, 2]
We estimate the average value of finite sum by using midpoint rule.
We have = ,
= ,
= ,
= are the midpoints of the above four
subintervals respectively.
Average value of the function on [0, 2] is
1
f c1 .x f c2 .x f c3 .x f c4 .x
length 0, 2
(0.5) +
=
where,
(0.5) +
(0.5) +
= +
= + =1
= +
= + =1
= +
= + −
= +
Average value=
1
= + −
+1
√
√
+1
(0.5)
=1
=1
+1
≈1
P2: Estimate the average value of the function ( ) =
on the interval [ , ].
Solution:
We partition the interval [0, ] into 4 subintervals of lengths ∆ = which are as
follows.
0,
,
,
,
,
,
,
We estimate the average value of finite sum by using midpoint rule.
We have = ,
= ,
= ,
=
are the midpoints of the above four
subintervals respectively.
Average value of the function on [0, ] is
1
f c1 .x f c2 .x f c3 .x f c4 .x
length 0,
=
=
+
.
+ .
+
+ .
+
+ .
= [0.3825 + 0.9247 + 0.9237 + 0.3825] ≈ 0.65335
IP3: Estimate the average value of the function ( ) =
+ on the interval
[ , ] using rectangles of equal width.
Solution:
In general the average value can be approximated by using the formula
ℎ
Average value=
ℎ
Here ∆ = and =
Using Right hand endpoint approximation, we have
Average value
1
2
=
.
+
.
+ ⋯+
=
+
=
+⋯+
+2 +
=
+
.
+2 +⋯+
+
+⋯+
+2
+2
P3: Estimate the average value of the function ( ) =
interval [ , ].
−
on the
Solution:
We partition the interval [0, 4] into 4 subintervals of lengths ∆ = 1 which are as
follows.
[0, 1] , [1, 2] , [2, 3] , [3, 4]
We estimate the average value with a finite sum by using midpoint rule.
We have = ,
= ,
= ,
= are the midpoints of the above four
subintervals respectively.
Average value of the function on [0, 4] is
1
f c1 .x f c2 .x f c3 .x f c4 .x
length 0, 4
=
(1 ) +
=
+
where,
(1) +
+
=1−
=1−
(1 ) +
+
= 0.2712
= 0.97842
(1)
=1−
= 0.97876
=1−
= 0.27313
Average value
1
= [0.2712 + 0.97842 + 0.97876 + 0.27313] ≈ 0.62538
4
IP4: What is the average value of the of the function ( ) =
on the interval
[ , ]?
Solution:
The average value equals the area under the graph divided by the width of the
interval. In this case we don’t need finite approximation to estimate the area of the
region under the graph: a triangle of height 6 and base 2 has area 6 (from figure)
The width of the interval is − = 2 − 0 = 2
The average value of the function is = = 3
Hence the average value of the function ( ) = 3 over the interval [0, 2] is 3.
P4: Use a finite sum to estimate the average value of a function ( ) = on the
interval [ , ] by partitioning the interval into 4 subintervals of equal length and
evaluating at the subinterval midpoints.
Solution:
We partition the interval [1, 9] into 4 subintervals of lengths ∆ = 2 which are as
follows.
[1, 3] , [3, 5] , [5, 7] , [7, 9]
We estimate the average value with a finite sum by using midpoint rule.
We have = 2,
= 4,
= 6,
= 8 are the midpoints of the above four
subintervals respectively.
Average value of the function on [1, 9] is
1
f c1 .x f c2 .x f c3 .x f c4 .x
length 1, 9
= [ (2). 2 + (4). 2 + (6). 2 + (8). 2]
=
(2) + (2) + (2) + (2) ≈ 0.26
EXERCISES:
Use a finite sum to estimate the average value of the function on the given interval by
partitioning the interval into 4 subintervals of equal length and evaluating at the
subinterval midpoints.
[0, 4]
1. ( ) = 3
,
[1, 9]
2. ( ) =
,
[0, ]
3. ( ) =
,
[−2, 2]
4. ( ) = 4 −
,
[0, 2]
5. ( ) =
,
1.5. Algebra of Finite sums
Learning objectives:
To introduce sigma notation to write a sum with many terms in the compact form.
To study the rules for finite sums.
To state the formula for the sum of the first natural numbers and formulas for
the sums of the squares and cubes of the first natural numbers.
AND
To practice the related problems.
We introduce a compact notation for sums that contain large number of terms.
Sigma Notation for Finite Sums
We use the capital Greek letter
n
The symbol
ak
(“sigma”) to write an abbreviation for the sum
denotes the sum a1 a2
an . The a ’s are terms of the
k 1
sum: a1 is the first term, a2 is the second term, ak is the k th term, and an is the
nth and last term. The variable is the index of summation. The values of runs
through the integers from 1 to n . The number 1 is the lower limit of summation; the
number n is the upper limit of summation.
Example 1:
5
k 1 2 3 4 5 1
k 1
3
1
k 1
2
k
k
1
k 1 1 1
1
2
1
2
2
2 13 3 1 2 3 2
7
k 1 11 2 1 2 3 6
k 1
The lower limit of summation does not have to be 1; it can be any integer.
Example 2:
Express the sum 1 + 3 + 5 + 7 + 9 in sigma notation.
Solution:
6
Starting with
k 2:
1 + 3 + 5 + 7 + 9 =
(2k 3)
k2
1
Starting with
k 3 :
1 + 3 + 5 + 7 + 9 =
(2k 7)
k 3
The formula generating the terms changes with the lower limit of summation, but the
terms generated remain the same. It is often simplest to start with k 0 or k 1.
6
Starting with
k 0:
1 + 3 + 5 + 7 + 9 =
(2k 1)
k 0
5
Starting with
k 1:
1 + 3 + 5 + 7 + 9 =
(2k 1)
k 1
Algebra with Finite Sums
We can use the following rules whenever we work with finite sums.
n
n
n
ak bk ak bk
1. Sum Rule:
k 1
n
k 1
n
k 1
n
ak bk ak bk
2. Difference Rule:
k 1
n
k 1
k 1
n
cak c ak
3. Constant Multiple Rule:
k 1
k 1
(any number
c)
n
c n c ( c is any constant value)
4. Constant Value Rule:
k 1
The formal proofs of these rules can be done by mathematical induction.
Example 3:
n
n
k 1
k 1
3k k 3 k k 2
a)
k 1
n
n
n
n
ak 1 ak 1 ak ak
b)
k 1
3
c)
n
2
k 1
3
k 1
k 1
3
k 4 k 4 1 2 3 3 4 6 12 18
k 1
k 1
k 1
Sum Formulas for Positive Integers
Gauss discovered the formula for the sum of first n integers. The formulas for the
sums of the squares and cubes of the first n integers are also famous.
n
n integers:
The first
k
k 1
n
n squares:
The first
k2
k 1
n
n n 1 2 n 1
6
n n 1
k
2
k 1
n cubes:
The first
n n 1
2
2
3
Example 4:
Evaluate
4
k
4
2
4
k 3 k
k 1
k 1
2
3k
k 1
4 4 1 8 1 4 4 1
3
30 30 0
6
2
PROBLEM SET
IP1
5
1. sin kπ
k 1
. −1
Answer: B
Solution:
. 0
. 1
. 4
5
sinkπ
(1. ) +
(2 ) +
(3 ) +
(4 ) +
(5 )
k 1
= 0+0+0+0+0 = 0
4
k
2. 1 cos kπ
k 1
. −1
Answer: D
Solution:
4
1
k
. 0
cos kπ = (−1)
. 1
+ (−1)
k 1
=−
+
2 −
3 +
= −(−1) + 1 − (−1) + 1 = 4
4
. 4
2 + (−1)
3 + (−1)
4
P1:
4
1.
3k
k 2
k 1
.
Answer: C
Solution:
4
.
3k
k 2
( )
+
( )
+
.
( )
( )
+
=
.
+ + +
=
k1
3
2.
k 1
k
k 1
.
.
.
.
Answer: B
Solution:
3
k 1
k
k 1
+
+
IP2: Which of the following express
6
A.
2
5
K 1
B.
k 1
1
k 0
K
2k
= 0+ + =
−
+
−
+
3
C.
1
−
in sigma notation?
K 1 k 2
2
k 2
Solution:
6
A. 2
K 1
(2)11 (2)21 (2)31 (2)4 1 (2)51 (2)61
k 1
= 1 − 2 + 4 − 8 + 16 − 32
5
K
B. 1 2k = (−1)
(2) + (−1) (2) + (−1) (2)
k 0
+(−1) (2) + (−1) (2)
= 1 − 2 + 4 − 8 + 16 − 32
3
C.
1
K 1 k 2
2
k 2
= (−1)
2
+ (−1)
2
+ (−1) 2
+(−1) 2
+ (−1) 2
+ (−1) 2
= −1 + 2 − 4 + 8 − 16 + 32
and represents 1 − 2 + 4 − 8 + 16 − 32, but does not represents the given
pattern.
P2.
1. Express
+
+
+
in sigma notation
4
4
4
2k
A.
B.
k 1
k
C.
k 1
k
4
2
D.
k 1
2k 2
k 1
Answer: C
Solution:
1 + 4 + 9 + 16 = (1) + (2) + (3) + (4)
4
k2
=
k 1
2. Express
+
5
A.
k
+
+
+
in sigma notation
5
2
B.
k 1
4
2K
C.
k 1
2K
5
D.
k 1
2K 2
k 1
Answer: B
Solution:
2 + 4 + 6 + 8 + 10 = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)
5
2K
=
k 1
n
IP3: Suppose that
n
ak 5
and
k 1
bk 2
. Find the values of
k 1
n
a)
2ak 3bk
k 1
n
b)
5ak 3bk 1
k 1
n
c)
5ak 13bk 5
k 1
Solution:
n
n
n
2ak 3bk 2ak 3bk
a)
k 1
k 1
k 1
n
n
2 ak 3 bk = 2(5) + 3(−2) = 4
k 1
n
b)
k 1
n
n
n
5ak 3bk 1 5ak 3bk 1
k 1
k 1
k 1
k 1
n
n
5 ak 3 bk n.1
k 1
k 1
= 5(5) − 3(−2) +
n
c)
= 31 +
n
n
n
6ak 13bk 5 6ak 13bk 5
k 1
k 1
k 1
n
k 1
n
6 ak 13 bk 5n
k 1
k 1
6(5) 13(2) 5n 4 5n
P3:
n
Suppose that
n
ak 0
and
k 1
n
a)
8ak
bk 1 then find the values of
k 1
n
n
250bk
b)
k 1
c)
k 1
ak 1
k 1
Solution:
n
a)
n
8ak 8 ak 8 0 0
k 1
k 1
n
b)
n
250bk 250 bk 250(1) 250
k 1
n
c)
d)
n
n
ak 1 ak 1 0 n n
k 1
k 1
k 1
n
n
n
bk 1 bk 1 1 n
k 1
10
IP4:
k 1
k 1
k 1
k k 2 2k 1 2
k 1
Solution:
n
d)
bk 1
k 1
10
k k2 2k 1 2
k 1
10
3
10
2
(k 2k k) 2
k 1
10
3
10
k 1
10
k 1
10
k 1
10
k 2k 2 k 2(10)
k 1
10
k 3 2 k2 k 20
k 1
k 1
k 1
2
10 10 1 10(10 1)(2(10) 1) 10(10 1)
20
2
6
2
3870
7
P4:
k(2k 1)
k 1
Solution:
7
7
k(2k 1)
k 1
(2k2 k)
k 1
7
2
7
2k2
k 1
7
k
k 1
7
k k
2
k 1
k 1
7(7 1) 2(7) 1 7(7 1)
2
6
2
7.8.15 7.8
280 28 308
3
2
EXERCISES:
1. Write the sums without sigma notation. Then evaluate them.
2
a.
k 1
6k
k 1
4
cos k
b.
k 1
3
1
c.
k 1
sin
k 1
k
2. Which of the following express 1 2 4 8 16 32 in sigma notation?
6
a.
2k 1
k 1
5
b.
2k
k 0
4
c.
2k 1
k 1
3. Which formula is not equivalent to the other two?
4
a.
k 2
2
b.
k 0
1
c.
(1) k 1
k 1
(1)k
k 1
k 1
(1) k
k 2
4. Express the sums in problems 6-8 in sigma notation.
1 2 3 4 5 6
1 1 1 1
b.
2 4 8 16
1 1 1 1
c. 1
2 3 4 5
a.
n
5. Suppose that
n
ak 5 bk 6
and
k 1
. Find the values of
k 1
n
3ak
a.
k 1
n
b.
k 1
n
bk
6
(ak bk )
c.
k 1
n
d.
(ak bk )
k 1
n
e.
(bk 2ak )
k 1
6. Evaluate the sums
10
(a)
10
k
(b)
k 1
10
k2
k 1
(c)
k 1
7
d.
(2k )
k 1
6
e.
3 k2
5
,
f.
k 1
5
k 1
5
k
g.
k
225
k 1
k 1
k 3k 5
3
3
k3
1.6. Limit of Riemann Sums
Learning objectives:
To define a partition of an interval and its norm.
To define the definite integral of a given function on a given interval as a limit of
Riemann sums.
AND
To practice the related problems.
To define Riemann sum of a given function on an interval for a partition of the
interval.
In the preceding module, we estimated distances, areas, volumes, and average values
with finite sums. The terms in the sums were obtained by multiplying selected function
values by the lengths of intervals. We now inquire what happens to the sums like
these as the intervals involved become more numerous and shorter.
Riemann Sums
The approximating sums in the modules 3.3 and 3.4 are examples of a more general kind
of sum called a Riemann sum. The functions in the examples had nonnegative values, but
the more general notion has no such restriction.
Given an arbitrary function y f x on an interval
into
a, b , we partition the interval
n subintervals by choosing n 1 points, say x1 , x2 ,, xn 1 , between
subject to the only condition that,
a x1 x2 xn 1 b
and
To make the notation consistent, we usually denote
by
x0 and
by
xn . The set
P { x0 , x1 , , xn } is called a partition of a, b .
The partition
P defines n closed subintervals
x0 , x1 , x1, x2 , , xn1, xn .
th
The typical closed subinterval xk 1 , xk is called the k subinterval of P .
The width (length) of the
k th subinterval is xk xk xk 1 .
xk 1, xk , we select a point ck and construct a vertical
rectangle on the subinterval xk 1 , xk to touch the curve y f x at the point
ck , f ck . The choice of ck does not matter as long as it lies in xk 1, xk .
In each subinterval
f ck is positive, the number f ck xk height base is the area of the
rectangle. If f ck is negative, the number f ck xk is the negative of the area.
If
In any case, we add the
n products f ck xk to form the sum
n
SP
f ck xk
k 1
This sum, which depends on
Riemann sum for
f
P and the choice of the numbers ck , is called a
on the interval
who studied the limits of such sums.
a, b , after German mathematician Riemann,
As the partitions of
a, b become finer, the rectangles defined by the partition
approximate the region between the x -axis and the graph of
accuracy.
Finer partitions create more rectangles with shorter bases
f
with increasing
So, we expect the associated Riemann sums to have a limiting value. The norm of a
partition
P is the partition’s widest (longest) subinterval length. It is denoted by P
(read “the norm of P”).
As the norms of the successive partitions approaches zero, the subintervals become
shorter and their number approaches infinity.
Example 1:
The set = {0, 0.2, 0.6, 1, 1.5, 2} is a partition of [0, 2]. There are five subintervals of
: [0, 0.2], [0.2, 0.6], [0.6, 1], [1, 1.5] and [1.5, 2].
The lengths of the subintervals are
x1 0.2, x2 0.4, x3 0.4, x4 0.5 and x5 0.5 . The longest
subinterval length is 0.5, so the norm of the partition is P 0.5 . In this example,
there are two subintervals of this length.
Definition
Let
f x be a function defined on a closed interval a, b We say that the limit of
n
the Riemann sums
f ck xk on a, b as
k 1
following condition is satisfied:
P 0 is the number I
if the
Given any number
every partition
0 , there exists a corresponding number 0 such that for
P of a, b
n
P
f ck xk I
k 1
for any choice of the numbers ck in the subinterval
xk 1, xk .
n
If the limit exists, we write
f ck xk I
lim
P 0 k 1
a, b , we say that f is integrable over
a, b , and we say that the Riemann sums of f on a, b converge to the number
We call
I
the definite integral of
f
over
I.
b
We usually write
I
as
f x dx
, which is read “integral of
f
from
a to b .” Thus
a
if the limit exists,
n
lim
P 0 k 1
When each partition has
write
n
lim
n k 1
b
f ck xk f ( x) dx
a
equal subintervals, each of width ∆ =
, we will also
b
f ck xk I f ( x) dx
a
Despite the variety in the Riemann sums
f ck xk
as the partitions change
and the arbitrary choice of ck ’s in the interval of each new partition, the sums always
have the same limit as
P 0 as long as f
is continuous.
Theorem 1:
All continuous functions are integrable. That is, if a function
interval
a, b , then its definite integral over a, b exists.
The theorem is proved in advanced calculus.
f
is continuous on an
Example 2:
Express the limit of Riemann sums
n
2
3
c
2ck 5 xk
k
P 0
lim
k 1
as an integral if
Solution:
P denotes a partition of the interval [−1, 3].
The function being evaluated at ck in each term of the sum is
f x 3x 2 2 x 5 a polynomial and hence a continuous function. The
interval being partitioned is [− , ].
The limit is therefore the integral of
n
lim
P 0
k 1
f
2
from −1 to 3:
3
3ck 2ck 5 xk
3 x2 2 x 5 dx
1
Constant Functions
Suppose that
f
has the constant value
f x c over a, b . Then, no matter
how the ck ’s are chosen,
n
n
n
f ck xk c xk c xk c b a
k 1
k 1
Since the sums all have the value
k 1
c b a , their limit, the integral, does too. We
have the following result.
If
f x has the constant value c on a, b , then
b
b
a f x dx a c dx c b a
Example 3:
4
3 dx
a)
1
b)
4
3 4 1 3 5 15
1 3 dx 3 4 1 3 5 15
PROBLEM SET
IP1: Let ( ) =
be a function defined over the interval [− , ]. Partition the
interval into 4 subintervals of equal length. Then compute the Riemann sum
4
f ck xk . Given that
is the
k 1
a. Left hand endpoint
b. Right hand endpoint
c. Midpoint of the
subinterval.
Solution:
The given function is ( ) =
, ∈ [− , ]
By partitioning the interval [0, 2] into 4 subintervals with ∆ =
We have – , −
,
− ,0 ,
0,
,
(
)
=−
,
= .
,
Left hand endpoints:
The left hand endpoints of the subintervals are = − ,
We now compute the corresponding Riemann sum.
= 0,
=
4
f ck xk f (c1)x1 f (c2 )x2 f (c3 )x3 f (c4 ) x4
k 1
sin sin sin 0 sin
2
2
2
0 1 0 1 0
2
Right hand endpoints:
The right hand endpoints of the subintervals are = −
We now compute the corresponding Riemann sum.
= 0,
,
=
,
4
f ck xk f (c1)x1 f (c2 )x2 f (c3 )x3 f (c4 ) x4
k 1
sin sin 0 sin sin
2 2
2
1 0 1 0 0
2
C. Midpoints of the subintervals:
The midpoints of the subintervals are = − ,
We now compute the corresponding Riemann sum.
=−
,
=
,
=
=
4
f ck xk f (c1)x1 f (c2 )x2 f (c3 )x3 f (c4 ) x4
k 1
3
3
sin
sin sin sin
2 4
4
4
4
1
1
1
1
0
2
2
2
2 2
P1: Let ( ) =
− be a function defined over the interval [ , ]. Partition the
interval into 4 subintervals of equal length. Then compute the Riemann sum
4
f ck xk
. Given that
is the
k 1
d. Left hand endpoint
e. Right hand endpoint
f. Midpoint of the
subinterval
Solution:
The given function is ( ) =
− 1, ∈ [0, 2]
By partitioning the interval [0, 2] into 4 subintervals with ∆ =
= 0.5
We have, [0, 0.5], [0.5, 1], [1,1.5], [1.5, 2]
Left hand endpoints:
The left hand endpoints of the subintervals are = 0,
= 0.5
=1 ,
We now compute the corresponding Riemann sum.
= 1.5
4
f ck xk f (c1 )x1 f (c2 )x2 f (c3 )x3 f (c4 )x4
k 1
2
2
0.5 0 2 1 0.5 1 12 1 1.5 1
0.5 1 0.75 0 1.25 0.25
Right hand endpoints:
The right hand endpoints of the subintervals are = 0.5 ,
=1
We now compute the corresponding Riemann sum.
= 1.5,
=2
4
f ck xk f (c1 )x1 f (c2 )x2 f (c3 )x3 f (c4 ) x4
k 1
2
2
0.5 0.5 1 12 1 1.5 1 2 2 1
0.5 0.75 0 1.25 3 1.75
C. Midpoints of the subintervals:
The midpoints of the subintervals are
= 0.25, = 0.75,
= 1.25,
= 1.75
We now compute the corresponding Riemann sum.
4
f ck xk f (c1 )x1 f (c2 ) x2 f (c3 )x3 f (c4 )x4
k 1
2
2
2
2
0.5 0.25 1 0.75 1 1.25 1 1.75 1
0.5 0.9375 0.4375 0.5625 2.0625 0.625
IP2: Find the norm of the partition = {− , − . , − . , , . , }
of the interval [− , ]?
Solution:
The set = {−2, −1.6, −0.5, 0, 0.8, 1} is a partition of the interval [−2, 1], since
−2 < −1.6 < −0.5 < 0 < 0.8 < 1
There are 5 subintervals of :
[−2, −1.6], [−1.6, −0.5], [−0.5, 0], [0, 0.8 ], [0.8, 1]
The lengths of the subintervals are
∆ = −1.6 − (−2) = 0.4 ; ∆ = −0.5 − (−1.6) = 1.1
∆ = 0 − (−0.5) = 0.5 ; ∆ = 0.8 − 0 = 0.8 ; ∆ = 1 − 0.8 = 0.2
The longest subinterval length is 1.1. So, the norm of the partition is ‖ ‖ = 1.1
P2. Find the norm of the partition = {− , − . , − . , , . , . , }
of the interval [− , ]?
Solution:
The set = {−1, − 0.5, − 0.1, 1, 1.05, 1.5, 2} is a partition of the
interval [−1, 2], since
−1 < −0.5 < −0.1 < 1 < 1.05 < 1.5 < 2
There are 6 subintervals of :
[−1, −0.5], [−0.5, −0.1], [−0.1, 1], [1, 1.05 ], [1.05, 1.5], [1.5, 2]
The lengths of the subintervals are
∆ = −0.5 − (−1) = 0.5 ; ∆ = −0.1 − (−0.5) = 0.4
∆ = 1 − (−0.1) = 1.1 ; ∆ = 1.05 − 1 = 0.05
∆ = 1.5 − 1.05 = 0.45 ; ∆ = 2 − 1.5 = 0.5
The longest subinterval length is 1.1. So, the norm of the partition is ‖ ‖ = 1.1
IP3: Express the limits in the problems and as definite integrals.
n
4 ck2 . xk ,where P is a partition of [ , ]
a. lim
P 0 k 1
n
b.
tan ck . xk , where P is a partition of
P 0
lim
k 1
,
Solution:
a. The function evaluated at
in each term of the sum is ( ) = √4 −
interval partitioned is [0, 1].
Notice that ( ) is a continuous function on [0, 1].
The limit is therefore the integral of from 0 to 1.
n
1
lim
4 ck2 . xk
4 x2
0
P 0 k 1
b. The function being evaluated at
. The
dx
in each term of the sum is ( ) = tan . The
interval partitioned is 0, .
Notice that ( ) is a continuous function on 0,
The limit is therefore the integral of
n
lim
tan ck . xk
.
from 0 to .
4
tan x dx
0
P 0 k 1
P3. Express the limits in the problems a, b as definite integrals.
n
c. lim
2ck3 . xk ,where P is a partition of – ,
P 0 k 1
n
d. lim
2ck3 . xk , where P is a partition of [ , ]
P 0 k 1
Solution:
c. The function evaluated at
in each term of the sum is ( ) = 2 and it is a
continuous function on – 1, 0 . The interval partitioned is – 1, 0 .
The limit is therefore the integral of from −1 to 0.
n
0
lim
2ck3 . xk 2 x3 dx
1
P 0 k 1
d. The function being evaluated at
in each term of the sum is ( ) = 2
continuous function on [1, 4] . The interval partitioned is [1, 4].
The limit is therefore the integral of from 1 to 4.
n
4
lim
2ck3 . xk 2 x3 dx
1
P 0 k 1
and it is a
IP4. If the function ( ) is defined on the interval – . , . as ( ) = . then
1.5
f ( x) dx
find
1.05
Solution:
The function defined on the interval – 1.05, 1.5 is ( ) = 3.5 , which is a constant
function and therefore,
1.5
1.5
1.05 f ( x) dx 1.05
3.5 dx 3.51.5 (1.05) 8.925
P4: If the function ( ) is defined on the interval [ , ] as ( ) =
5
then find
3 f ( x) dx
Solution:
The function ( ) defined on the interval [3, 5] is ( ) = 3, which is a constant
function and therefore,
5
5
3 f ( x) dx 3 3 dx 35 3 6
EXERCISES
1. Partition the interval into four subintervals of equal length. Then compute the
4
Riemann sum
f ck xk
, given that ck is the (a) left-hand endpoint, (b)
k 1
right-hand endpoint, (c) midpoint of the
a. f x x 2 1 ,
b. f x sin x ,
0,2
,
2. Find the norm of the partition
a.
b.
P 0,1.2,1.5,2.3, 2.6,3
P 0, 0.2, 0.5,0.7, 0.9, 1
3. Express the limits as definite integrals.
k th subinterval.
n
a.
lim
P 0
b.
lim
P 0
c.
lim
P 0
d.
k 1
n
k 1
n
ck2 xk , where P is a partition of 0,2
1
xk , where P is a partition of 2,3
1 ck
k 1
n
sec ck xk , where P is a partition of /4,0
P 0
lim
k 1
4. Evaluate the integrals
1
2 5 dx
3
b. 160 dt
0
3.4
c. 0.5 ds
2.1
a.
ck2 3ck xk , where P is a partition of 7,5
2.1. Definite integrals
Learning objectives:
To define the area under a curve as a definite integral.
To evaluate definite integrals
b
b
b
2
c
dx
,
x
dx
,
x dx a b
a
a
a
AND to practice the related problems.
b
The terminology associated with symbol
a f x dx is illustrated below.
The value of the definite integral of a function over any particular interval depends on
the function and not on the letter we choose to represent its independent variable. If
we decide to use t or instead of , we simply write the integral as
b
a
f t dt or
b
a
f u du
instead of
b
a f x dx
No matter how we write the integral, it is still the same number defined as a limit of
Riemann sums. Since it does not matter what letter we use, the variable of integration
is called a dummy variable.
The Area under the Graph of a Non-negative Function
The sums, we used to estimate the height of the projectile in the projectile example in
Module 3.3 (example 1), were Riemann sums for the projectile’s velocity function
v f t 160 9.8t on the interval [0, 3].
From the above figure, we see how the associated rectangles approximate the
trapezoid between the t -axis and the curve v 160 9.8t . As the norm of the
partition goes to zero, the rectangles fit the trapezoid with increasing accuracy and the
sum of the areas they enclose approaches the trapezoid’s area, which is
160 130.6
Trapezoid area 3
435.9
2
Thus the sums we constructed in the projectile example approached a limit of 435.9.
Since the limit of these sums is also the integral of f from 0 to 3, we know the value
of the integral as well:
3
0 160 9.8t dt trapezoid area 435.9
Definition
Let f x 0 be continuous on a, b . The area of the region between the graph of
the x -axis and the vertical lines = , = is
b
A
a f x dx
b
Example 1: Evaluate
a x dx
0 a b.
Solution:
We sketch the region under the curve
y x, a x b .
The region is a trapezoid with height b a and bases
and .
The area of the trapezoid = ( − ).
= −
The value of the integral is the area of the trapezoid:
b
a
b2 a 2
xdx
2
2
2
For example,
1
5
5
x dx
2
1
2
2
2
We notice that x 2 / 2 is an anti-derivative of
between anti-derivatives and summation.
x , indicating a possible connection
f
,
Example 2: Find the area of the region between the parabola y x 2 and the
on the interval a, b , < < .
Solution:
We evaluate the integral for the area as a limit of Riemann sums.
We sketch the region and partition
x
x -axis
a, b into n subintervals of length
b a
n
The points of the partition are
x0 a, x1 a x, x2 a 2x, , xn 1 a n 1 x, xn a n x b
We are free to choose ck ’s any way we please. We choose each ck to be the righthand endpoint of its subinterval. Thus = , = , …, = . The rectangles
defined by these choices have areas
2
f c1 x f a x x a x x
2
f c2 x f a 2 x x a 2x x
2
f cn x f a nx x a nx x
n
n
k 1
k 1
S n f ck x ( a k x) 2 x
n
The sum of these areas
a2 k 2 (x)2 2ak x x
k 1
n
k 1
a 2 x k 2 ( x)3 2 ak (x) 2
2
a x.n ( x)
3
n
k
2
2a ( x)
2
k 1
n
k
k 1
3
2ba
2
b a n ( n 1)(2n 1)
b a n( n 1)
a
.
n
2
a
n
n . 2
6
n
1
1
1
3 1
2
a 2 (b a ) b a 1 2 a b a 1
6 n
n
n
We now use the definition of definite integral
n
b
f x dx lim
f ck x
a
P 0 k 1
to find the area under the parabola from x
b 2
x
0
a to x b a
dx lim S n
n
2
a (b a )
2
a (b a )
b a 3
6
b a 3
3
1
1
1
2
lim 1 2 a b a lim 1
n
n
n
n
n
ab a
2
b3 a 3
3
3
Again, we notice that
is an anti-derivative of
between anti-derivatives and definite integrals.
With different values of
1 2
x
0
, indicating a possible connection
b , we get
13 1 1.5 2
(1.5)3 3.375
dx , x dx
1.125,
3 3 0
3
3
and so on.
PROBLEM SET
IP1. i). Evaluate
5
3 x dx ?
b2 a 2
, ab
Solution: We have x dx
a
2
2
5
52 32
x dx
8
3
2
2
b
2
ii). Evaluate
Solution:
0
2 d ?
b 2
b3 a3
We have
x dx
, ab
a
3
3
3
2
2
3
2
0
d
P1: i). Evaluate
a
3
0
24
3a
x dx ?
b2 a 2
Solution: We have x dx
, ab
a
2
2
2
3
a
3a
a2
x dx
a2
a
2
2
3b 2
b
ii ). Evaluate
x dx ?
0
b 2
b3 a3
x dx
, ab
Solution: We have
a
3
3
3
3b
3b 2
x dx
0 9b3
0
3
IP2: Graph the integrand and use areas to evaluate the integral
Solution: We have to compute the integral
integrand is ( ) = √16 −
4.
=
( )
16 x 2 dx by using areas. Here the
=
0
= 4 Square units.
4
, where is the radius of the circle.
16 x2 dx 4
P2: Graph the integrand and use areas to evaluate the integral
32
12
2 x 4 dx
16 x 2 dx
, its graph is a circle with center at the origin and radius
We know that area of the quarter circle is
∴
0
4
0
4
Solution: We have to compute the integral
32
12
2 x 4 dx by using the areas. Here
integrand is ( ) = (−2 + 4) and its graph is a straight line with
= 3,
= 1.
We know the area of the trapezoid is
From the graph ℎ = − = 1,
∴
32
12
= ℎ[ + ]
=
= 3,
=
=1
= (1)[3 + 1] = 2 Square units
2 x 4 dx 2
IP3: Use a definite integral to find the area of the region between the curves =
and the −axis on the interval [ , ].
Solution:
We have to compute the area of under the graph of the function =
and the
−axis on the interval [ , ].
Now, we sketch the region and partition [ , ] into subintervals of length ∆ =
The points of the partition are
= ,
= +∆ ,
= + 2∆ , ……..
=
Choose each to be the right hand endpoint of its subintervals.
Thus = ,
= ,
= , …,
= .
The rectangles defined by these choices have areas
3
f c1 x f a x x a x x
3
f c2 x f a 2x x a 2x x
3
f c3 x f a 3x x a 3x x
3
f cn x f a nx x a nx x
Then the sum of these areas is
n
n
k 1
n
k 1
3
S n f ck x a k x x
3
a 3 k 3 x 3a 2 k x 3ak 2 x
2
k 1
3
a x x
4
n
k
3
3a
2
x
n
2
n
3
x k 3a x k 2
k 1
k 1
k 1
2
4
3
2
b a n n 1
b a n (n 1)(2n 1)
2 b a n (n 1)
a
3a
3a
n
4
2
6
n
n
n
n
3 b a
3
a ( b a)
b a 4
4
2
3a
1
1
n
2
2
b a 2
3
1 3a b a 1
1
1
1 2
6
n
n
n
2
We now use the definition of definite integral,
to find the area under the graph from
b 3
x
a
a
3
=
n
b
a f x dx
to
=
lim
as
dx lim Sn
n
4
b a
(b a )
4
2
3a
1
lim 1
n
n
2
b a 2
2
1
lim 1
n
n
3
3a b a
1
1
lim 1 2
n
6
n
n
a
3
4
2
b a
3a 2 b a
(b a )
a
b4 a 4
4
4
4
2
b
a
b a 3
b4 a 4
x dx
4
4
3
f ck x
P 0 k 1
P3: Use a definite integral to find the area of the region between the curves =
and the −axis on the interval [ , ].
Solution:
We have to compute the area of under the graph of the function =
and the
−axis on the interval [0, ].
Now, we sketch the region and partition [0, ] into subintervals of length
∆ =
= .
The points of the partition are
= 0,
=∆ ,
= 2∆ , ……..
= ∆ =
Choose each to be the right hand endpoint of its subintervals.
Thus,
= ,
= ,
=
and so on.
The rectangles defined by these choices have areas
2
f c1 x f x x x x .12 x
3
2
3
2
3
2
3
f c2 x f 2x x 2x x .22 x
f c3 x f 3x x 3x x .32 x
f cn x f n x x nx x .n 2 x
Then the sum of these areas is
n
n
2
3
3
S n f ck x k ( x) (x)
k 1
3
k 1
n
k2
k 1
n n 1 2n 1 b3 n 1 2 n 1
3
6
6
n
n2
b
b3
3 1
2
6
n n2
We now use the definition of definite integral
b
n
f ck x
a f x dx Plim0
k 1
to find the area under the graph from
=
to
=
as
b
0 x
2
dx lim S n lim
n
b3
6
n
b3
3 1
2
6
n n2
2 0 0
b
0 x
2
dx
b3
3
b3
3
IP4: Use a definite integral to find the area of the region between the curves
= + and the −axis on the interval [ , ].
Solution:
We have to compute the area of under the graph of the function = + 1 and the
−axis on the interval [0, ].
Now, we sketch the region and partition [0, ] into subintervals of length
∆ =
= .
The points of the partition are
= 0,
=∆ ,
= 2∆ , ……..
= ∆ =
Choose each to be the right hand endpoint of its subintervals.
Thus = ,
= ,
= and so on. The rectangles defined by these choices
have areas
1
x
2
f c1 x f x x 1 x x x
2
2
1
2x
2
f c2 x f 2x x
1 x .2 x x
2
2
1
3x
2
f c3 x f 3x x
1 x .3 x x
2
2
1
2
nx
f cn x f nx x
1 x .n x x
2
2
Then the sum of these areas is
n
n
1
2
S n f ck x k x x
k 1
k 1 2
2
x
2
n
n
k x 1
k 1
k 1
b 2 n ( n 1) b
b2 1
.n 1 b
2
n
4 n
2n2
We now use the definition of definite integral
n
b
a f x dx
lim
f ck x
P 0 k 1
to find the area under the graph from
= 0 to
= 3 as
x
b2 1
b2
S n lim
b
1 b
0 2 1 dx nlim
n
4
n 4
b
P4: Use a definite integral to find the area of the region between the curves
=
+ and the −axis on the interval [ , ].
Solution:
We have to compute the area of under the graph of the function =
+ and the
−axis on the interval [1, 4].
Now, we sketch the region and partition [1, 4] into subintervals of length
∆ =
= .
The points of the partition are
= 1,
= 1+ ,
= 1+2
,…
=1+
=4
Choose each to be the right hand endpoint of its subintervals.
Thus = ,
= ,
=
and so on.
The rectangles defined by these choices have areas
3 2 3
3
f c1 x f 1 x 1 1 x
n
n n
2
3
3
3
f c2 x f 1 2 x 1 2 1 2 x
n
n
n
2
3
3
3
f c3 x f 1 3 x 1 3 1 3 x
n
n
n
2
3
3
f cn x f 1 n x 1 n
n
n
3
1 n x
n
Then the sum of these areas is
2
3
3
S n f ck x x 1 k 1 k
n
n
k 1
k 1
2
n
3
3
2 3
( x) 1 k 2k 1 k
n
n
n
k 1
2
2 n
n
n
3
3 3
3
23
2 9
2 k 3k 2n k k
n k 1
n
n n
n k 1
n k 1
n
n
9 n( n 1)(2n 1) 9 n ( n 1)
3
2n 2
n
6
n 2
n
3 1 2 9 1
3 2 1 1 1
2 n n 2 n
We now use the definition of definite integral
n
b
a f x dx
lim
f ck x
P 0 k 1
to find the area under the graph from
4
1 x
2
= 1 to
= 4 as
x dx lim S n
n
3 1 2 9 1 57
lim 3 2 1 1 1
2 n n 2 n 2
n
EXERCISES
1. Graph the integrands and use areas to evaluate the integrals.
a.
4
x
3 dx
2 2
3
b.
c.
d.
3
1
2 x dx
1
1 2 x dx
b
e.
f.
9 x 2 dx
0
b
a
x dx , b 0
2 s ds , 0 a b
2. Evaluate the integrals in problems 7-12.
2
a.
b.
1 x dx
2
d
7
2
x
0 dx
1/2 2
0 t dt
2a
a x dx
b
2
x
0 dx
3
c.
d.
e.
3
f.
3. In problems 13 and 14, use a definite integral to find the area of the region
between the given curve and the x -axis on the interval [0, b ] .
a.
b.
y 3 x2
y 2x
2.2. Properties of Definite Integrals
Learning objectives:
To study the properties of the definite integrals.
AND
To practice the related problems.
In this module, we describe working rules for integrals.
The following rules hold for definite integrals.
These rules will enable us to add and subtract definite integrals, multiply their
integrands by constants, and compare them with other definite integrals.
All the rules except the first two follow from the way the integrals are defined
with Riemann sums. We omit the proofs. As the sums have these properties so
their limits should have them too. Although this is not that simple, for the
present we contend with this argument. The proofs will be covered in advanced
calculus.
Rules 1 and 2 are definitions. We want every integral over an interval of zero
length to be zero. Rule 1 extends the definition of definite integral to allow for
the case a b .
Rules 3 and 4 are like the analogous rules for limits and indefinite integrals. Once
we know the integrals of two functions, we automatically know the integrals of all
constant multiples of these functions and their sums and differences. We can also
use Rules 3 and 4 repeatedly to evaluate integrals of arbitrary finite linear
combinations of integrable functions term by term. For any constants
c1, , cn , regardless of sign, and functions f1 x , ,
f n x , integrable on
a, b
b
b
b
c
f
x
c
f
x
dx
c
f
x
dx
c
n n
1 1
n f n x dx
a 1 1
a
a
The figure below illustrates Rule 5 with a positive function, but the rule applies to
any integrable function.
The Additivity for definite integrals:
b
a
c
b
f x dx
f x dx
c
b
c
a
c
f x dx
f x dx
a f x dx
b
a f x dx
Example 1:
Suppose that
1
1
f x dx 5 ,
1
1
1
h x dx 7 ,
4
1
f x dx 2 .Then
4
4 f x dx 1 f x dx (2) 2
1
1
1
(ii ) [2 f x 3h( x)] dx 2 f x dx 3 h x dx
1
1
1
(i )
2(5) 3 7 31
4
(iii )
1
f x dx
1
1
f x dx
4
1
f x dx
5 2 3
In the previous module and in this module, we learned to evaluate three general
integrals:
b
a c dx c b a
Any constant c
b2 a 2
x dx
a
2
2
b
b3
2
x dx
0
3
b
0 a b
b 0
The rules stated earlier for the definite integrals will enable us to build on these
results.
Example 2:
2 t2
2
2
1 2 2
7
t
5
dt
t
dt
7
t
dt
0 4
0
0
0 5 dt
4
1 23
22
10
7 5 2 0
4 3
2
3
Example 3:
Evaluate
3 2
x
2
dx
Solution:
b3
We cannot apply the equation
directly because the lower limit of
x dx
0
3
b
2
integration is different from 0. We can, however, use the Additivity Rule to express
3 2
x
2
dx as a difference of two integrals that can be evaluated with this equation.
2 2
x
0
3 2
x
2
dx
dx
3 2
x
2
3 2
x
0
dx
dx
3 2
x
0
dx
2 2
x
0
33 23 19
dx
3
3
3
The Max-Min inequality for definite integrals says that m(b a) is a lower bound
b
a f x dx and M b a is an upper bound.
1
Example 4: Show that the value of
0 1 cos x dx cannot possibly be 2.
for the value of
Solution:
The maximum value of 1 cos x on [0, 1] is 1 1 2 , so
1
0
1 cos x dx max. 1 cos x . (1 0) 2 1= 2
The integral cannot exceed
2 , so it cannot possibly equal 2.
Example 5: Use the inequality cos x 1 x 2 / 2 , which holds for all
x , to find a
1
0 cos x dx .
lower bound for the value of
Solution:
1
x2
1 1 2
cos
x
dx
1
dx
1
dx
0
0 2 0
0 x dx
2
1
1 1 0 13 / 3 5 / 6 0.83
2
1
1
The value of the integral is at least
PROBLEM SET
1
is integrable and that h r dr 0 and
1
IP1: Suppose that
find a .
3
1
.
h r dr
b.
3
1 h r dr 6 then
1
3 h r dr
Solution:
3
a
1
h r dr
3
1
1
1 h r dr Additivity
3
3
1
h r dr h r dr h r dr 6 0 6
1
1
1
1
3
b h r dr 1 h u du order of int egration
3
1
h r dr
3
6
P1: Suppose that
u is
1 h r dr
and
by a
dummy
above
are integrable and that
9
1
f x 1
9
7 f x dx 5
9
7 h x dx 4 , then find
9
7
7
i ). 2 f x 3h x dx, ( ii ). h x f x dx, ( iii ). 2 f x dx
7
9
1
Solution:
i.
9
7
2 f x 3h x dx
9
7
2
2 f x dx
9
7
9
7 3h x dx
f x dx 3
9
7 h x dx 2 5 3 4 2
By Difference rule and constant rule
and
ii.
7
h x f x dx
9
7
9
7
h x dx
9
7
9 f x dx Difference rule
h x dx
9
7 f x dx order of integration
4 5 1
9
1
iii.
7
1
7
2 f x dx
2 f x dx
1
9
2 f x dx
1
2 f x dx
2
9
1
9
7 2 f x dx Additivity
9
7 2 f x dx
f x dx 2
9
7 f x dx Constant multiple rule
2 1 2 5 12
IP2. I). Evaluate
1
u
1 du
2
2
Solution:
1
2
u
1 2 du
2
1
2
1
u
1 2 du
1 2
du
u du
2 1
2
1 u2
1 22 12
7
u 2 1
2 2
2 2 2
4
1
2
1
II). Evaluate
2
0 t 2 dt
Solution:
2
0 t 2 dt 0
0
2
t dt
0
2
t dt 2
2
2 dt
0
2
dt
2
t2
2
2 t 0
2 0
2
2
2
0 2 2 0 1
P2: I. Evaluate
4
2
2
x
1 x 5 dx
Solution:
4
1 2 x
2
4
1 2 x
x 5 dx
2
4 2
x
1
dx
4
1
2
dx
4
4
1 x dx 1 5 dx
4
1 dx
x dx 5
4
4
x3 x2
4
2 5 x 1
3 1 2 1
2
1
15
99
64 1 16 1 5 4 1 42 15
3
2
2
2
II. Evaluate
0
2
3
x
1 x 5 dx
Solution:
0
1
3 x 2 x 5 dx
1
2
3
x
0 x 5 dx
1 2
0
3
x dx
1
1
0
1
0
x dx 5 dx
1
x3 x 2
1
3 5 x 0
3 0 2 0
1 0
IP3: Show that the value
Solution:
Given
1
0
1
0
1
1
7
1 0 51 0 1 5
2
2
2
x 8 dx lies between √ and .
x 8 dx
The integrand ( ) = √ + 8 is increasing on [0, 1]
⟹
= (1) = √1 + 8 = 3 and
= (0) = √0 + 8 = 2√2
From the Max-Min inequality, we have
If has maximum value
and minimum value
b
min f b a f x dx max f b a
a
2 2 1 0
1
0
1
2 2
0
x 8 dx 3 1 0
x 8 dx 3
on [ , ], then
1
2
0 sin x dx cannot possibly be 2.
P3: Show that the value of
( ) ≤ 1, ∀
Solution: We have −1 ≤
≤ 1 ⟹ −1 ≤
From Max-Min inequality, we have
If has maximum value
and minimum value
on [ , ], then
b
min f b a f x dx max f b a
a
1
11 0 sin x 2 dx 11 0
0
1
1
0 sin x
1 sin x 2 dx 1 or
0
2
dx 1
1
2
0 sin x dx cannot possibly be 2.
IP4: Use Max-Min inequality to find upper and lower bounds for the integrals
0.5
0
1
1 x2
dx and
1
1
0.5 1 x 2 dx
Add these two integrals to arrive at an improved estimate of
1
1
0 1 x 2 dx
Solution:
We have to compute the upper and lower bounds of the integral
0.5
0
1
1 x
2
dx .
Here integrand is ( ) =
,
⟹ is decreasing function on [0, 0.5]
∴ Maximum value of occurs at = 0
⟹
= (0) =
=1
∴ Minimum value of occurs at = 0.5
⟹
= (0) = ( . ) = 0.8
By Max-Min inequality, we have
( 0.5 0 ) min f
0 .5
0
1
dx ( 0.5 0 )max f
1 x2
0 .5
1
0.5 0.8
dx 0.5 1
0 1 x2
0 .5
1
0.4
dx 0.5 ............ 1
0 1 x2
⟹ Upper bound is 0.5 and Lower bound is 0.4
Again, we have to compute the upper and lower bounds of the integral
Here integrand is ( ) =
,
1
1
0.5 1 x 2 dx .
⟹ is decreasing function on [0.5, 1]
∴ Maximum value of occurs at = 0.5
⟹ max = (0.5) = ( . ) = 0.8
∴ Minimum value of
⟹ min = (0) =
occurs at = 1
= 0.5
( )
By Max-Min inequality, we have
( 1 0.5 )min f
0.5 0.5
1
0.5 1 x 2 dx ( 1 0.5 )max f
1
1
0.5 1 x 2 dx 0.5 0.8
1
0.25
1
1
0.5 1 x2 dx 0.4 ............ 2
⟹ Upper bound is 0.4 and Lower bound is 0.25
By adding (1) and (2), we get
0.25 0.4
0.65
0.5
0
1
1 x2
1
dx
1
1
0.5 1 x2 dx 0.5 0.4
1
0 1 x2 dx 0.9
⟹ Upper bound is 0.9 and Lower bound is 0.65
P4: Use Max-Min inequality to find upper and lower bounds for the interval of
1
1
0 1 x 2 dx
Solution: We have to compute the upper and lower bounds of the integral
1
1
0 1 x 2 dx . Here integrand is
( )=
⟹ is decreasing function on [0, 1]
∴ Maximum value of occurs at = 0
⟹
= (0) =
=1
∴ Minimum value of occurs at = 1
⟹
= (0) =
= 0.5
By max-min inequality, we have
1 0 min f
0. 5
1
1
1
0 1 x2 dx 1 0 max f
1
0 1 x 2 dx 1
⟹ Upper bound is 1 and Lower bound is 0.5
EXERCISES:
f
1. Suppose
2
1
and
g are integrable and that
f x dx 4,
5
5
f x dx 6,
1
g x dx 8 .
1
Use properties of definite integrals and find the following integrals
2
2
A).
E ).
5
1
g x dx , B ).
2
1
g x dx , C ).
5
f x g x dx , F ).
2. Suppose that
A).
1
2
1
f
1
3 f x dx , D ).
4
3
2
1
3 f z dz , C ).
is integrable and that
f z dz
4. Evaluate the integrals.
a.
1
3 7 dx
2
b. 5x dx
0
2
c. 2t 3 dt
0
1
z
d. 1 dz
2
2
e.
f.
2
2
f x dx
4 f x g x dx
1
1
2
3
0
f t dt , D ).
1 3u 2 du
2
0 3x 2 x 5 dx
b.
3
4
2
1 [ f x ] dx
f z dz 3 and
Find
a.
5
f x dx 5 . Find
f u du , B ).
3. Suppose that
5
2
f t dt
4
0
f z dz 7 .
2.3. Area and Integrals
Learning objectives:
To find the area of the region between a curve
−axis.
AND
To practice the related problems.
= ( ) over the interval [ , ] and
If an integrable function y f x has both positive and negative values on an interval
a, b , then the Riemann sums for f on a, b is obtained by adding the areas of the
rectangles that lie above the x -axis to the negatives of the areas of the rectangles that
lie below it.
The Riemann sums are algebraic sums of areas and so is the integral to which they
converge.
The value of the integral of
b
a
f x dx
x1
a
f x dx
f
from
x2
x
1
to
f x dx
is
b
x
f x dx
2
A1 A2 A3
The resulting cancellation reduces the sums, so their limiting value is a number whose
magnitude is less than the total area between the curve and the x -axis. The value of
the integral is the area above the axis minus the area below the axis.
This means that we must take special care in finding areas by integration.
Example 1:
Find the area of the region between the curve y 4 x 2 , 0 x 3 , and the
Solution:
x -axis.
The x -intercept of the curve partitions [0, 3] into subintervals on which f x 4 x 2
has the same sign. The curve crosses the −axis at 4 −
= 0 i.e., = ±2 and
−2 ∉ [−2,2].
Now [0, 3] is partitioned into subintervals [0, 2] , [2, 3]
To find the area of the region between the graph of
f
f
and the
x -axis, we integrate
over each subinterval and add the absolute values of the results.
23 16
Integral Over [0, 2]:
4 x dx 4 dx x dx 4 2 0
0
0
0
3
3
33 23
3
3
3
7
2
2
Integral Over [2, 3]:
4 x dx 4 dx x dx 4 3 2
3
2
2
2
3
3
16
7 23
The region’s area: Area
3
3
3
2
2
2
2
2
Procedure
The following is a step-by-step procedure of how to find the area of the region between
a curve = ( ), ≤ ≤ , and the -axis.
1. Partition [ , ] with the zeros of .
2. Integrate over each subinterval.
3. Add the absolute values of the integrals.
PROBLEM SET
IP1. Find the area under the graph =
over the interval [ ,
Solution:
Given = ( ) =
. It is a continuous function on [0, 2 ].
(I). Partition [ , ] with zeros of :
The zeros of ( ) on [0, 2 ] are
= 0 ⟹ = 0, ,
,2
Partition [0, 2 ] into subintervals 0,
(II). Integrate over each subinterval:
Integral over 0,
A1
2
0
A2
2
3 2
,
,
,
:
cos x dx sin x 0
Integral over
,
1 0 1
:
3 2
cos x dx sin x 2 1 1 2
2
,2 .
]?
,2
Integral over
A3
2
3 2
:
2
cos x dx sin x 3
2
0 1 1
(III). Add absolute values of the integrals:
Area of the curve ( ) =
on [0, 2 ] is
= | | + | | + | | = |1| + |−2| + |1| = 4
P1. Find the area under the graph =
over the interval [ ,
Solution:
Given = ( ) =
. It is a continuous function on [0, 2 ].
(I). Partition [ , ] with zeros of :
The zeros of ( ) on [0, 2 ] are
= 0 ⟹ = 0, , 2
Partition[0, 2 ] into subintervals [0, ] , [ , 2 ].
(II). Integrate over each subinterval:
Integral over [0, ]:
A1
0
]?
sin x dx cos x 0 1 ( 1) 2
Integral over [ , 2 ]:
A2
2
2
sin x dx cos x 1 1 2
(III). Add absolute values of the integrals:
Area of the curve ( ) =
on [0, 2 ] is
= | | + | | = |2| + |−2| = 2 + 2 = 4
IP2: Find the area bounded by the curve
− =
and the −axis?
Solution: The given curve is = ( ) = ( − 1).
Notice that ( ) is continuous since it is a polynomial. It crosses −axis at
(I). Partition [− , ] with zeros of :
The zeros of ( ) are ( − 1) = 0 ⟹ = ±1
The Partition of [−1, 1] with zeros of is itself.
(II). Integrate over each subinterval:
Integral over [−1, 1]:
y2 1
1
A1
dy
1
2
2
1
1
1 y3
2 3
1
1
1
1
y dy
2
2
1
1 dy
1 1
1 1 1 1
2
y 1 1 (1)
2
2 3 3 2
3
(III). Add absolute value of the integral:
The area bounded by the curve − 1 = 2 and the
=| |= − =
−axis is
= ±1
P2: Find the area between the −axis and the curve = ( − ) −
interval – , ?
Solution:
Given = ( ) = ( − 1) − 25. It is a continuous function on [−4, 6].
(I). Partition [− , ] with zeros of :
The zeros of ( ) are
( − 1) − 25 = 0 ⟹ = −4, 6
The Partition of [−4, 6] with zeros of is itself.
(II). Integrate over each subinterval:
Integral over [−4, 6]:
over the
6
x 13
2
A1
25 x
x 1 25 dx
4
3
4
6
6 13
4 13
500
25(6)
25( 4)
3
3
3
(III). Add absolute values of the integrals:
The area between the −axis and the curve
= ( − 1) − 25 over the interval – 4, 6 is
∴
=|
|= −
=
IP3. Find the area bounded by the parabola
=− , = .
Solution:
=
, the
−axis and the lines
Given = ( ) = . It is a continuous function on [−2, 4].
(I). Partition [− , ] with zeros of :
The zeros of ( ) are
=0⟹ =0
Partition[−2, 4] into subintervals [−2, 0] , [0, 4].
(II). Integrate over each subinterval:
Integral over [−2, 0]:
Integral over [0, 4]:
x2
1 x3
A1
dx
8 3
2 8
0
2
1 8 1
0
8 3 3
4
1 x3
1 64
8
A2
dx 0
0 8
8 3
8 3
3
0
4 x2
0
(III). Add absolute values of the integrals:
The area bounded by the parabola
= 8 , the
= 4 is,
= | |+| |=
+
=3
−axis and the lines
= −2,
P3: Find the area bounded by the parabola = , the −axis and the lines
=− , = .
Solution:
Given = ( ) = . It is a continuous function on [−1, 2].
(I). Partition [− , ] with zeros of :
The zeros of ( ) are
=0⟹ =0
Partition [−1, 2] into subintervals [−1, 0] , [0, 2].
(II). Integrate over each subinterval:
3 0
13 1
x
2
x dx 0
Integral over [−1, 0]: A1
1
3
3
3
1
0
Integral over [0, 2]: A2
2 2
x dx
0
2
2 3
8
x3
0
3
3
3
0
(III). Add absolute values of the integrals:
The area bounded by the parabola = , the
the lines = −1, = 2 is
= | |+| |=
+
=3
−axis and
IP4: Using the integration find the area of region bounded by the triangle whose
vertices are ( , ), ( , ), ( , ).
Solution:
(2, 2), and (3, 1) are the vertices of a triangle ABC which is as
Let (1, 0),
shown in figure.
From the graph,
Area of ∆
=
+
ℎ
−
∆
∆
Now, the equations of the sides AB, BC, and CA are given by
= 2( − 1), = 4 − ,
= ( − 1) respectively.
Hence,
Area of ABC
2
1
2 x 1 dx
3
2
3
x 1
dx
1 2
4 x dx
2
3
3
x2
x2
1 x2
2 x 4x x
2
2 2
2
1
1
2
2
22
1
1
32
22 1 32
2
2 1 4(3) 4(2) 3 1
2
2
2
2 2 2
2
1
3
2
∴ Required area is =
P4: Find the area bounded by =
+ , the −axis and the ordinates
=− , = .
Solution:
Given = ( ) = 3 + 2. It is a continuous function on [−1, 1].
(I). Partition [− , ] with zeros of :
The zeros of ( ) are 3 + 2 = 0 ⟹ = −
Partition [−1, 1] into subintervals −1, −
(II). Integrate over each subinterval:
, − ,1 .
Integral over −1, − :
2 3
A1
1
3x 2 dx 3
2 3
1
2 3
x dx 2
1
dx
2 3
2 3 2 1 2
x2
2
1
2 3
2 ( 1)
3
2 x 1 3
2
6
2
3
2 1
Integral over − , 1 :
1
A2
2 3
3x 2 dx 3
1
2 3
1
x dx 2
2 3 dx
1
12 2 3 2
x2
25
1
2 1 ( 2 3)
3
2 x2 3 3
2
6
2
2 2 3
(III). Add absolute values of the integrals:
The area bounded by the parabola = 3 + 2, the
= −1, = 1 is
= | |+| |= − +
=
−axis and the ordinates
EXERCISES:
In problems 1, 2 and 3, find the total shaded area.
1.
2.
3.
Find the total area between the region and
2
1. y x 6 x 8
2. y 2 x x 2
3. = − − 2 ,
4. = 3 − 3,
5. =
−3 +2 ,
6. =
−4 ,
⁄
7. =
,
[0,3]
[0,3]
−3≤
−2 ≤
0≤
−2≤
−1≤
≤2
≤2
≤2
≤2
≤8
−axis.
2.4. The Mean Value Theorem for Definite Integrals
Learning objectives:
To define the average value of an integrable function on an interval.
To state and prove the Mean value theorem for definite integrals.
AND
To practice the related problems.
In an earlier module, we discussed the average value of a nonnegative continuous
function. Now, we define the average value without requiring f being nonnegative,
and also showing that every continuous function assumes its average value at least
once.
The Average value
In arithmetic, we know that the average of n numbers is the sum of the numbers
divided by . For a continuous function on a closed interval a,b there may be
infinitely many values to consider, but we can sample them in an orderly way. We
partition a,b into
evaluate
subintervals of equal length (the length is Δx
at a point ck in each subinterval.
The average of the
sampled values is
f c1 f c2 f cn 1 n
f ck
n
n k 1
Δx n
f ck
b a k 1
n
1
f ck Δx
b a k 1
ba
) and
n
Thus, the average of the sampled values is always 1 / ( b a ) times a Riemann sum for
on a,b . As we increase the size of the sample and let the norm of the partition
1 b
approach zero, the average must approach
f x dx .
b a a
Definition
If is integrable on a,b , then its average (mean) value on a,b is denoted
by ̅ or ( ) and
f
1
ba
b
a f x dx
Example 1:
Find the average value of f x 4 x 2 on [0, 3]. Does
some point in the given domain?
Solution:
1
b a
1
3 0
f
actually take on this value at
b
a f x dx
3
1 3
4 x2 dx 4 dx x 2 dx
0
0
3 0
3
1
33 1
4 3 0 12 9 1
3
3 3
The average value of f x 4 x 2 over the interval [0, 3] is 1. The function assumes this
value when 4 x2 1 or x 3 . Since one of these points, x 3 , lies in [0, 3], the
function does assume its average value in the given domain.
The Mean Value Theorem for Definite Integrals
The statement that a continuous function on a closed interval assumes its average
value at least once in the interval is known as the Mean Value theorem for Definite
Integrals.
Theorem 1:
If f is continuous on a,b , then at some point in a,b ,
f c
1 b
f x dx
b a a
Proof
If we divide both sides of the Max-Min inequality by b a , we obtain
b
1
f x dx M
b a a
is continuous, the Intermediate Value Theorem for Continuous Functions says
m
f
Since
must assume every value between m and
b
1
value
f x dx at some point c in a,b .
ba a
Hence the theorem.
that
f
M . It must therefore assume the
Note 1:
The following figure shows a positive continuous function
interval [ , ].
= ( ) defined over the
Geometrically, the Mean value theorem says that there is a number in [ , ]
such that the rectangle with height equal to the average value ( ) of the function
and base width − has exactly the same area as the region beneath the graph of
from to .
Note 2:
The continuity of is important in the Mean value theorem of definite integrals. A
discontinuous function can step over its average value.
The above figure shows that a discontinuous function need not assume its average
value.
Example 2:
is continuous on a,b ,
Show that if
b
a b , and if a f x dx 0 then f x 0 at
least once in a,b .
Solution:
The average value of
f
b
1
1
f
x
dx
0 0
b a a
ba
assumes this value at some point c in a,b .
on a,b is
By the Mean value theorem,
f
f
PROBLEM SET
IP1: Find the average value of the function ( ) = −| | on
a. [− , ]
b. [ , ]
c. [− , ]
Solution:
Notice that ℎ( ) = −| | is a continuous function and so
its definite integral
b
a h( x ) dx exists.
The average value of a function ℎ( ) on [ , ] is
av h h
1
b a
b
a h( x) dx
a. The average value of the function ℎ( ) = −| | on [−1, 0] is
0
0
1
av h h
x dx x dx
1
1 0 1
0
12
x2
1
x dx 0
1
2
2
2 1
b. The average value of the function ℎ( ) = −| | on [0, 1] is
0
1
12
x2
1
x dx x dx
0
0
0
2
2
2 0
c. The average value of the function ℎ( ) = −| | on [−1, 1] is
1
av h h
1 0
av h h
1
1 1
0
1
1
1 x dx
1
x dx
1
0
1 1 1
1
from parts a and b
2 2 2
2
1
2
1
x dx
P1: Find the average value of the function ( ) = | | − on
b. [− , ]
b. [ , ]
c. [− , ]
Solution:
Notice that ( ) = | | − 1 is a continuous function and so its definite integral
b
a g ( x) dx exists.
The average value of a function ( ) on [ , ] is
av g g
1
ba
b
a g ( x) dx
a. The average value of the function ( ) = | | − 1 on
[−1, 1] is
av g g
1
2
1
1 1
1
1 x 1 dx
0
1
x
1
dx
x
1
dx
1
0
1
1 0
x 1 dx x 1 dx
0
2 1
0
1
x2
1 x2
x x
2 2
1 2
0
1 1 1
1
0 1 1 0
2 2 2
2
b. The average value of the function ( ) = | | − 1 on [1, 3] is
av g g
1
3 1
3
1
x 1 dx
1
2
3
1 x 1 dx
3
1 x2
1 9 1
x 3 1 1
2 2
1 2 2 2
c. The average value of the function ( ) = | | − 1 on [−1, 3] is
1
av g g
3 1
3
1
x 1 dx
1
4
1 1
1
x 1 dx
4 1
4
1 1 1
1
1
4 2 4
4
3
1 x 1 dx
3
1 x 1 dx
from parts a and b
IP2. Find the average value of ( ) = − over the interval [ , ]. At what point or
points in the given interval does the function assume its average value.
Solution:
Notice that ( ) = −
is a continuous function and so its
3
f ( x) dx exists.
definite integral
0
The average value of a function ( ) on [ , ] is
1
av f f
ba
b
f ( x) dx
a
∴ The average value of the function ( ) = −
1
av f f
3 0
3
3
on [0, 3] is
x2
1 2
x dx
dx
2
6
0
0
3
1 x3
1 27
3
0
6 3
6 3
2
0
Notice that −
=− ⟹
= ±√3.
By the mean value theorem, the function assumes the value when = √3 ∈ [0, 3].
P2. Find the average value of ( ) =
− over the interval [ , ]. At what point
or points in the given interval does the function assume its average value.
Solution:
Notice that ( ) = 3 − 3 is a continuous function and so its definite integral
1
f ( x) dx exists.
0
b
1
The average value of a function ( ) on [ , ] is av f f
f ( x) dx
ba
a
∴ The average value of the function ( ) = 3
1
av f f
1 0
− 3 on [0, 1] is
1
3x2 3 dx
0
1
1
1
x3
1
1
2
3 x dx 3 dx 3 3 x0 3 0 31 0 2
3
3 0
0
0
Notice that 3
− 3 = −2 ⟹
=±
By the mean theorem, the function
√
assumes this value when
=
√
∈ [0, 1]
IP3. Find the average value of ( ) =
−
+
over the interval [ , ]. At what
point or points in the given interval does the function assume its average value.
Solution:
Notice that ( ) =
− 3 + 2 is a continuous function and so its definite integral
2
f ( x) dx exists.
0
1
The average value of a function ( ) on [ , ] is av f f
ba
∴ The average value of the function ( ) =
av f f
1
20
−3
b
f ( x) dx
a
+ 2 on [0, 2] is
2
x3 3x2 2 x dx
0
2
2
2
1 3
2
x dx 3 x dx 2 xdx
2
0
0
0
2
1 x4 3 2 2 2 1
x
x
4 8 4 0
2 4 0 0 2
0
Notice that − 3 + 2 = 0 ⟹ = 0, 1, 2
By the mean theorem, the function assumes this value when = 0, 1, 2 ∈ [0, 2]
P3. Find the average value of ( ) = √ − over the interval [− , ]. At what
point or points in the given interval does the function assume its average value.
Solution:
The graph of the function ( ) = √4 − over the interval is plotted over the
interval [−2, 2], which is shown below.
We recognize ( ) = √4 − as a function whose graph is the upper semi-circle of
radius centered at the origin.
The area between the semi-circle and the −axis from -2 to 2 can be computed using
the geometric formula.
Area of a semi-circle=
=
( )
=2
Because is non-negative, the area is also the value of the integral of
from -2 to 2
2
4 x 2 dx 2
2
Therefore, the average value of the function ( ) = √4 −
1
av f
2 2
Notice that √4 −
2
4 x 2 dx
1
2
4
2
⟹4−
=
2
=
⟹
⟹
=± 4−
By the mean value theorem, the function
is
= 4−
∈ [−2, 2]
assumes the value when = ± 4 −
.
IP4. Find the average value of ( ) = + over the interval [− , ]. At what point
or points in the given interval does the function assume its average value.
Solution:
Notice that ( ) = 1 +
is a continuous function and so its definite integral
2
f ( x) dx exists.
1
b
1
The average value of a function ( ) on [ , ] is av f f
f ( x) dx
ba
a
The average value of the function ( ) = 1 +
2
on [−1, 2] is
2
2
1
2
2
av f f
1 x dx
dx x dx
2 1
3
1
1
1
3 2
1 2
x
1
8 1
x 1 2 1
3
3 3
3 1 3
6
2
3
1
Notice that 1 +
= 2 ⟹ = ±1 and = ±1 ∈ [−1, 2]
By the mean value theorem, the function assumes this value when = ±1.
P4. Find the average value of ( ) = − over the interval [ , ]. At what point or
points in the given interval does the function assume its average value.
Solution:
3
Notice that ( ) = 4 −
is a continuous function and so its definite integral
f ( x) dx
0
exists.
b
1
f ( x) dx
The average value of a function ( ) on [ , ] is av f f
ba
a
∴ The average value of the function ( ) = 4 − on [0, 3] is
3
3
3
1
1
av f f
4 x dx 4 dx x dx
30
3
0
0
0
2 3
32
1
3 x 1
4 x 0 4(3) 0 0
3
2 0 3
2
3 5
4
2 2
Notice that 4 − = ⟹ = ∈ [0, 3]
By the mean value theorem, the function
assumes this value when
=
EXERCISES
1. Find the average value over the given interval. At what point or points in the given
interval does the function assume its average value?
8. f x x 2 1 on 0 , 3
9. f t ( t 1 )2
f x 3 x2 1 on
;
on 0 ,3
0,1
10. f t t 2 t , on 2, 1
2. Find the average value of the function over the given interval from the graph of
(without integrating).
x 4 4 x 1
1 x 2
x 2
a) f x
b)
f t sint
on
on
0,2π
4,2
f
2.5. The Fundamental Theorem
Learning objectives:
To state and prove the first part of the Fundamental theorem of calculus.
AND
To practice the related problems.
The Fundamental Theorem of Integral Calculus gives the connection between
integration and differentiation. It is independently discovered by Leibniz and Newton.
The Fundamental Theorem of Calculus, Part 1
If f t is an integral function over a finite interval then the integral from any fixed
number ∈ to another number ∈ defines a function whose value at is
F x
x
a f t dt
........... 1
Equation (1) gives an important way to define new functions, but its importance now is the
connection it makes between integrals and derivatives.
If f is any continuous function, then F is a differentiable function of
derivative is
f
itself. At every value of
x
whose
xI ,
d
d x
F x f t dt f x
dx
dx a
This idea is the first part of the Fundamental Theorem of Calculus.
Theorem 1:
The Fundamental Theorem of Calculus, Part 1
If
f
is continuous on a, b , then F x
every point of a, b and
x
a
f t dt has a derivative at
dF d x
f t dt f x ,
dx dx a
a x b .......... 2
Proof
We prove the theorem by showing that the limit of the difference quotient
F x h F x
h
as h 0 is the number f x , where and ( + ℎ) are in ( , ).
We replace F x h and F x by their defining integrals. Then
F x h F x
x h
a
f t dt
x
a f t dt
xh
The Additivity Rule for integrals simplifies the right-hand side to
x
f t dt
Now,
F x h F x 1
1 x h
F x h F x
f t dt
h
h
h x
According to the Mean Value Theorem for Definite Integrals, the value of the last
expression in the previous equation is one of the values taken on by f in the interval
joining
x
and
x h . That is, for some number c in this interval,
1 x h
x f t dt f c
h
As h 0 , the endpoint x h approaches to
x , forcing c approach
also. Since
f
is continuous at x , f c approaches f x .
lim f c f x
h 0
Going back to the beginning, then, we have
F x h F x
dF
lim
dx h 0
h
1 x h
lim
f t dt lim f c f x
h 0 h x
h 0
If = or then the limit is interpreted as one sided limit with ℎ → 0 or ℎ →
0 respectively. Then
dF
f ( x) , a x b
dx
Hence the theorem
Note1: The equation (2) says that every continuous function
some other function, namely
f
is the derivative of
x
a f t dt . It says that every continuous function has an
anti-derivative. And it says that the process of integration and differentiation are
inverses of one another.
Note2: If the values of f are positive, the equation
d x
f t dt f x
dx a
has a geometric interpretation.
The integral of f from to is the area ( ) of the region between the graph
of and the x -axis from to . Imagine covering this region from left to right
by unrolling a carpet of variable width f t .
x , the rate at which the floor is being covered is f x .
As the carpet rolls past
Example 1:
d
dx
d
dx
x
cos t dt cos x
x
1
0 1 t
dt
2
1
1 x2
Example 2:
Find dy / dx if y
Solution
We treat
x2
1
cos t dt
y as the composite of the two functions,
y
u
u x2
1 cos t dt ,
and apply the Chain Rule.
dy dy du d
dx du dx du
u
1
cos t dt
du
du
cos u
dx
dx
cos x 2 2 x 2 x cos x 2
Example 3:
Find a function y F x whose derivative is tan x and which passes through the
point (1, 5).
Solution:
The Fundamental theorem makes it easy to construct a function with derivative
that equals 0 at = 1
y
Since y (1)
x
1 tan t dt
1
1 tan t dt 0 , we have to add 5 to this function to construct the function
with derivative is
and whose value is 5 when
=1
F x
x
1 tan t dt 5
Therefore, the required function is
y F x
x
1 tan t dt 5
PROBLEM SET
x
2
0
IP1. If y
cos t dt , x
Solution: Given y
Put
=
⟹
Then y
x2
0
2
then find
cos t dt , x
2
=2 .
u
0 cos
t dt F (u )
By the Fundamental Theorem of calculus (part1), we have
dy
cos u
du
By the chain rule, we have
dy dy du
.
cos u . 2 x 2 x cos x
dx du dx
P1. If y
x
0
1 t 2 dt then find
Solution: Given
y
x
0
1 t 2 dt
Notice that ( ) = √1 +
is a continuous function on [0, ],
Then by the Fundamental Theorem of calculus (part1),
y F ( x) y
x
0
f (t ) dt
is differentiable on (0, ) and its derivative
3
t 2 t dt then find
6x x
Solution: Given y
Put
=6 √ ⟹
)= ( )
dy
1 x2
dx
6x x
IP2. If y
ʹ(
3
= 9√
t 2 t dt
u
then y
3
t 2 t dt F (u )
> 0.
By the fundamental theorem of calculus (part1),
dy
u2 u
du
By the chain rule, we have
2
dy dy du 2
32
32
. u u .9 x 6 x
6 x .9 x
dx du dx
36 x3 6 x3 2 .9 x 9 x 36 x 2 6 x1 2
5
P2. If y
2x
3t sin t dt then find
3
5
Solution: Given y
Put
=2
⟹
2x
=6
u
then
u
5
y
3t sin t dt
3
3t sin t dt
5 3t sin t dt F (u)
By the Fundamental Theorem of calculus (part1), we have
dy
3u sin u
du
By the chain rule, we have
dy dy du
. 3u sin u . 6 x 2 36 x5 sin 2 x3
dx du dx
IP3. If y
sin x
1
3t 2 dt
Solution: Given y
Put
=
⟹
then find
sin x
1
=
3t 2 dt
. Then y
u
1
3t 2 dt
By the Fundamental Theorem of calculus (part 1), we have
By the Chain rule, we have
dy dy du
. 3u 2 .cos x 3sin 2 x.cos x
dx du dx
P3. If y
tan
0
sec 2 t dt
Solution: Given y
Put =
⟹
=
tan
0
then find
sec 2 t dt
. Then y
u
0
sec 2 t dt F (u )
dy
3u 2
du
By the Fundamental Theorem of calculus (part1), we have
dy
sec2 u
du
By the Chain rule, we have
dy dy du
.
sec2 u.sec2 sec2 tan .sec2
d du d
IP4. Find the function = ( ) whose derivative is ( ) and which passes
)?
through ( ,
Solution: The Fundamental theorem makes it easy to construct a function with
derivative ( ) that equals 0 at =
, v (t )
t
t g ( x) dx C
0
Since v(t0 )
t0
t
g ( x) dx C , we have to add
to this function to construct the function
0
with derivative is ( ) and whose value is
v(t )
when =
0
t
t g ( x) dx v0
0
Therefore, the required function is v(t )
t
t g ( x) dx v0
0
P4. Find the function = ( ) whose derivative is √ +
and which passes
through (− , )?
Solution:
The Fundamental Theorem makes it easy to construct a function with derivative
√ +
that equals 0 at = − .
y
Since y
1
1
x
1 t 2 dt
1
1 t 2 dt 0 , we have to add 2 to this function to construct the function
with derivative is √1 +
F ( x)
x
1
2
and whose value is 2 when
= −1
1 t 2 dt 2
Therefore, the required function is F ( x)
x
1
1 t 2 dt 2
EXERCISES:
dy
if
dx
x
d
1. y
cos t dt ;
dx 0
Find
4. y
0
x
dt ;
sin t
2
d
2. y
dt
5. y
sin x
0
t4
0
u du ;
dt
1 t2
,
3. y
x
2
x
0
1
dt
t
0
; 6. y
tan x
dt
1 t2
2.6. Evaluation of Definite Integrals
(The Fundamental Theorem of Calculus part2)
Learning objectives:
To state and prove the Fundamental Theorem of Calculus part2.
AND
To practice the related problems on the evaluation of definite integrals.
The second part of the Fundamental Theorem of Calculus describes how to evaluate
definite integrals.
Theorem
The Fundamental Theorem of Calculus, Part 2:
If f is continuous at every point of a , b and F is any anti-derivative of f on a , b ,
then
b
a f x dx F b F a
Proof:
To prove the theorem, we use the fact that the functions with identical derivatives
differ only by a constant. We know one function whose derivative equals f , name
G x
x
a f t dt
Therefore, if F is any other such function, then F x G x C throughout a , b
for some constant C . We use this equation to calculate F b F a . Thus,
F b F a G b C G a C
G b G a
b
a
b
a
f t dt
a
a f t dt
f t dt 0
b
a f t dt
This concludes the proof.
b
a
The usual notation for the number F b F a is F x when F x has a single
b
a
term or F x when F x has more than one term.
The Fundamental Theorem of Calculus (part2) says that to evaluate the definite integral
of a continuous function f from a to b , we find an anti-derivative F of f and
calculate the number F b F a . The existence of an anti-derivative is assured by
the first part of the Fundamental Theorem.
Example 1:
0
a)
b)
cos x dx sin x0 sin sin 0 0 0 0
0
0
/4 sec x tan x dx sec x /4 sec0 sec / 4 1
2
c)
4
4 3
4
4
x 2 dx x3/2
1 2
x 1
x
4
4
43/2 13/2
4
1
8 1 5 4
The above theorem explains the formulas we derived for the integrals of
derived earlier. We can now see
2 b
x
b2 a2
because
x dx
a
2
2
2
a
b
b
x 2 /2 is an anti-derivative of x .
x3
b3 a 3
because
x dx
a
3
3
3
b
2
x and x 2
x 3 /3 is an anti-derivative of x 2 .
a
Example 2:
We model the voltage in our home wiring with the sine function
V Vmax sin120 t
which expresses the voltage V in volts as a function of time t in seconds. The function
runs through 60 cycles each second (its frequency is 60 hertz or 60 Hz). The positive
constant Vmax is the peak voltage.
The average value of V over a half-cycle (duration 1/120 second) is
Vav
1
(1/120) 0
1/120
0
Vmax sin120 t dt
1/120
120Vmax
1
cos120
t
120
0
Vmax
cos cos 0
2Vmax
The average value of the voltage over a full cycle is zero. If we measured the voltage
with a standard moving-coil galvanometer, the meter would read zero.
To measure the voltage effectively, we use an instrument that measures the square
root of the average value of the square of the voltage, namely
Vrms
V 2 av
The subscript “rms” stands for “root mean square.” Since the average value of
2
V 2 Vmax sin 2 120 t over a cycle is
V av
2
1
(1/ 60) 0
1/60
0
Vrms
(Vmax )2
(Vmax ) sin 120 t dt
, the rms voltage is
2
2
Vmax 2
2
2
V
max
2
The values given for household currents and voltages are always rms values. Thus, “115
volts ac” means that the rms voltage is 115. The peak voltage,
Vmax 2Vrms 2 115 163 volts
is considerably higher.
PROBLEM SET
IP1:
4
2
4
sec
t
dt ?
3
t2
Solution:
4
2
4sec t 2 dt
3
t
4
4
3
sec2 t dt
4
4
t 2 dt 4 tan t 3
3
1
1
4 tan tan
4
3
3
4
4 1 3 1 4 3 3
4
t 21
2 1 3
3 1 cos 2t
3
P1:
2
Solution:
3 1 cos 2t
3
2
dt ?
dt
3 1
1
cos 2t dt
3 2 2
1 3
1 3
1 3
1 sin 2t 3
dt
cos 2t dt t 3
2 3
2 3
2
2 2 3
1 1
sin 2 sin 2
2 3 3 4
3
3
1 3
3
3
3 4 2
2 3 4
4
IP2:
4
x dx ?
Solution:
0
4
x2
x2
x dx
x dx
x dx
( x ) dx
x dx
4
4
0
4
0
2 4 2 0
4
0
4
0
4
4 2 4 2
0
0 16
2 2
1
P2:
cos x cos x dx ?
0 2
Solution:
1
cos x cos x dx
0 2
1 2
1
cos x cos x dx 2 cos x cos x dx
2 0
2
1 2
1
cos x cos x dx
cos x cos x dx
2 0
2 2
2
0
2
cos x dx 0
2
sin x0
0
cos x dx
sin sin 0 (1 0) 1
2
IP3: Evaluate
1 7
u
1
du
2 u5
2
Solution:
1 7
u
1
du
2 u5
2
1
√
=
√
=
− (1 )
√2
=−
−1 −
2
1
u 5 du
2
−
− (1)
√2
[16 − 1]
=−
4
P3: Evaluate
2
√
−
−
1
du
2 u5
1
√
−
=−
=
2 7
u
9
1 u
u
du
Solution:
4
9
1 u
du
u
=
9
4
−
1 u
du
u
=
9
4
1
1 u du
−2
= 9 − 2(9) − 4 − 2(4)
= [9 − 6 ] − [4 − 4 ] = 3
IP4:
Find the area of the shaded region in the given graph.
1
u 7 du
Solution:
The area of the shaded region in the given graph can be calculated as follows.
On − ,
:
The area of the rectangle bounded by the lines
= 0 and = − is = √2 × =
The area between the curve =
= √2,
√
.
and
0
= 0,
= 0 is
0
4
sec . tan d sec
4
sec 0 sec 4 1
2
2 1
Therefore, the area of the shaded region on − , 0 is
√
=
+ √2 − 1
,
On
:
The area of the rectangle bounded by the lines
= 0,
=
= √2
√
= √2 × =
The area between the curve
=
.
and
= 0 is
4
s e c . ta n d
0
s e c
4
4
s e c 0
s e c 0
Therefore, the area of the shaded region on 0,
=
√
− √2 − 1
Thus, the area of the total shaded region on − ,
=
is
√
− √2 − 1 +
√
− √2 − 1 =
√
is
2 1
= 0 is
P4: Find the area of the shaded region in the given graph
Solution:
The area of the shaded region in the given graph can be computed as follows:
The area of the rectangle bounded by the lines
=−
and = 1 is = 1 + × 2 = 2 +
Now, the area under the curve
0
=
= 2,
=0,
on − , 0 is
0
se c 2 t d t ta n t 4
4
ta n 0 ta n 0 1 1
4
The area under the curve
on [0, 1] is
=1−
1
3
13 2
t
1 t 2 dt t 1 0
3 3
0
3 0
1
Thus, the total area under curves on − , 0 is 1 + =
Therefore, the area of the shaded region is
= 2+ − = +
EXERCISES:
Evaluate the following definite integrals:
0
a.
(2 x 5) dx
2
4
b.
x3
3 x dx
4
0
2
c.
x3 2 x 3 dx
x 3 x dx
2
2
1
d.
0
1
e.
x
2
dx
2
f.
0
sin x dx
2
g.
sin x dx
2
0
h.
1 cos 2t
dt
2
2
i.
(1 cos x ) dx
r 12 dr
0
1
j.
1
2
k.
8 y 2 sin y dy
2
1
l.
1
1
3
v2
v
dv
12
3 2
m.
4
csc .cot d
2.7. Substitution in Definite Integrals
Learning objectives:
To evaluate the Definite integrals by the method of substitution.
AND
To practice the related problems.
There are two methods for evaluating a definite integral by substitution. One is to find
the corresponding indefinite integral by substitution and use one of the resulting antiderivatives to evaluate the definite integral by the Fundamental Theorem. The other is
to use the following formula.
g b
b
a f g x g x d x g a
f u d u
We make the same u -substitution = ( ) and
= ( )
that we use to
evaluate the corresponding indefinite integral. We then integrate with respect to
from the value u at x a to the value u at x b .
Example 1:
1
Evaluate
1
u
3x 2 x3 1 dx
Solution
Method 1: Transform the integral as an indefinite integral, integrate, change back to x ,
and use the original x -limits.
3x 2 x3 1 dx
u x3 1,
udu
du 3x 2 dx
3/2
2 3/2
2 3
u C x 1
C
3
3
1
3/2
2
3x 2 x3 1 dx x 3 1
1
3
1
3/2
3/2
2 3
3
1 1
1 1
3
1
2 3/2
2
4 2
2 0 2 2
3
3
3
Method 2: Transform the integral and evaluate the transformed integral with the
transformed limits.
1
1
3x
2
3
x 1 dx
2
0
u x 3 1,
u du
du 3 x 2 dx
3
when x 1, u 1 1 0
3
when x 1, u 1 1 2
2
2
2
2
4 2
u 3/2 23/2 0 2 2
3
3
3
0 3
Example 2:
/2
/4
cot csc2 d
put u cot du csc 2 d
when / 4, u cot / 4 1
when / 2, u cot / 2 0
/2
/4 cot csc
2
d
0
0
1 u du 1 u du
0
u2
02 12 1
2 1
2 2 2
PROBLEM SET
IP1: Evaluate the integrals
1
a.
10 v
0 1 v3 2 2
Solution:
4
dv
b.
1
10 v
32 2
1 v
dv
⁄
=1+
10 v
⟹
= √
0 1 v3 2 2 dv
a. To evaluate
Put
1
⟹√
Limits:
=0⟹
1
=
= 0 + 1 = 1 and
=1⟹
2 1 20
20 2 2
dv 2
du
u du
2
1 u 3
3 1
32
10 v
0 1 v
20
3
4
b. To evaluate
Put
=1+1= 2
=1+
1
⁄
4
1
32 2
1 v
⟹
1 10
2 1 3
dv
= √
=
= 1+1= 2
= 1+8= 9
10 v
1 v
2
20
1
u
3
1
10 v
⟹√
Limits:
=1⟹
=4⟹
3 2 2
dv
9 1 20
d
u
2 u2 3
20
3
9
2
u 2 du
9
20 1
20 1 1 70
3 u 2
3 9 2
27
P1: Evaluate the integrals
7
a.
0
13
0 t t 1
2
dt
b.
13
7 t t 1
2
dt
Solution:
7
0 t t 1
a. To evaluate
Put = + 1 ⟹
=2
Limits:
= 0⟹ = 0+1= 1
= √7 ⟹ = 7 + 1 = 8
7
13
2
0 t t
13
2
1
⟹
8
dt
1 u
dt
=
1 8 13
.
u du
2
2 1
1 3 du
1 u 1 3 1
2 1 3 1
0
b. To evaluate
7 t t 1
13
7 t t 1
2
8
1
3 4 3
4
8
1
8
13
2
Put = + 1 ⟹
=2
Limits:
= −√7 ⟹ = 7 + 1 = 8
= 0⟹ = 0+1= 1
0
⟹
45
8
dt
=
1
dt
3
1 8 13
.
u du
2
2 1
1 3 du
8 u
8
1
3
1
1 u
1 31
2
1
3 4 3
45
4 3
8 1
8
8
IP2: Evaluate the integrals
1
a.
0
1
2
r 1 r dr
1
b.
r 1 r 2 dr
Solution:
1
a. To evaluate
0
r 1 r 2 dr
Put = 1 − ⟹
= −2
Limits:
= 0⟹ = 1−0= 1
= 1⟹ = 1−1= 0
1
0
r 1 r
2
⟹
0
dr
1
=−
1
u du
2
1 1
u du
2 0
1
1 u1 2 1
1
1
1 0
2 1 2 1
3
3
0
b. To evaluate
1
1
r 1 r 2 dr
Put = 1 − ⟹
= −2
Limits:
= −1 ⟹ = 1 − 1 = 0
= 1⟹ = 1−1= 0
1
1
r 1 r
2
⟹
0
dr
0
=−
1
u
du 0
2
P2:
Evaluate the integrals
a.
0
Solution:
2
3cos x .sin x dx
3
b.
2
3cos 2 x .sin x dx
a. To evaluate
0
3cos 2 x .sin x dx
Put =
⟹
=−
Limits:
= 0 ⟹ = 1 and =
0
⟹
⟹
=−
= −1
1
2
3 cos x .sin x dx
1
1
2
3u du 3
1
u 2 du
1
u3
3
3
3 1 1 2
3 1
b. To evaluate
3
2
Put =
Limits:
⟹
=2 ⟹
=3 ⟹
3
2
3 cos 2 x .sin x dx
=−
=
=
⟹
=−
2 =1
3 = −1
1
2
3 cos x .sin x dx
1
3u
2
1
du 3
1
1
u 2 du
u3
3
3
3 1 1 2
3 1
IP3: Evaluate the integrals
0
t
a.
2 tan
2
2
2
t
b.
2 t
dt
sec
2
2 t
2
tan
sec
dt
2
2
2
Solution:
a. To evaluate
0
t
2 t
2 ta n 2 se c 2 d t
2
Put = 2 + tan ⟹
= sec
⟹ sec
Limits:
= − ⟹ = 2+
− =1
=0⟹
=2+
0⟹
=2
=2
0
2
2
t
2 t
2
tan
sec
dt
u
.(2
du
)
2
u du
2
2
2
1
1
2
u2
2 4 1 3
2 1
b.To evaluate
2
t
2 t
dt
2 ta n s e c
2
2
2
Put = 2 + tan ⟹
= sec
⟹ sec
Limits:
= − ⟹ = 2+
− =1
=
⟹
=2+
2
t
2 tan 2
2
=3
3
3
2 t
sec
dt
u
.(2
du
)
2
u du
2
1
1
u2
2
2
P3:
a.
0
E v alu ate
Solution:
To evaluate
0
2
Put = 3 + 2
Limits:
sin w
2
3 2 co s w
sin w
3 2 cos w 2
⟹
3
9 1 8
1
dw
2
dw
= −2
⟹
=− ⟹ =3+2
− =3
= 0 ⟹ = 3 + 2 cos(0) = 5
0
sin w
dw
2 3 2 cos w 2
5 1 1
1 5 2
u
du
du
2 3
3 u2 2
5
=2
1 1
1 1 1
1
2 u 3 2 5 3
15
=−
b.
2
0
E valuate
Solution:
To evaluate
sin w
dw
3 2 cos w 2
2
0
sin w
3 2 co s w
2
dw
Put = 3 + 2
⟹
= −2
Limits:
= 0 ⟹ = 3 + 2 cos(0) = 5
= ⟹ = 3+2
=3
2
sin w
0
3 2 c os w
IP4:
4
0
Solution:
To evaluate
Put
4
0
=1−
Limits:
=0⟹
= ⟹
4
5
1
1
u
2
3
1 sin 2 t 3
0
=1−
=1−
1
du
1
1 1
5 3 1 5
cos 2 t dt
dt
2(0) = 1
2
=0
3 2
sin 2 t
1
1
2
2
= −2 2
2
=−
0
2
5
3 u
1 sin 2 t 3 2 cos 2t
2 ⟹
⟹
=−
dw
2
3 1 1
1
du
2
2
2
5 u
1
2
⟹
u
cos 2t dt
1
1
d
u
2
2
3 2
0
1
1
u3 2 du
1 u 3 2 1
1 5 2 1
3 2
u
du
u
0
2 3 2 1
5
0
0
1
1
1
1 0
5
5
P4:
3 2
cot 5 .sec 2
6
6
Solution:
To evaluate
3 2
= tan
Put
d
3 2
1
cot 5 .sec 2 d
.sec 2 d
6
6
6
tan 5
6
⟹
= sec
⟹ sec
=6
Limits:
=
⟹
= tan =
=
⟹
= tan
3
2
√
=1
1
.sec 2
d
6
5
tan
6
1
1
u 5
3
u 51
6
5 1
6
du
1
1
3
1
u 5 du
1
3
3 4
u
2
1
6
1
3
3
1 9 1 2
2
3 2
cot 5 .sec 2 d 12
6
6
EXERCISES:
Evaluate the integrals in problems 1-12.
1. a)
2. a)
3. a)
4. a)
3
0
y 1 dy
/4
tan x sec 2 x d x
0
1 3
4 3
t
1
t
dt
0
1
5r
1 4 r 2 2 d r
0
1 y 1 dy
0
2
b)
tan
x
sec
x dx
/ 4
b)
t 1 t d t
b)
1
1
b) 1
0
4 3
3
5r
4 r 2
2
dr
5. a)
0
11.
12.
x
x
0
3
4
b)
dx
1 cos 3t sin 3 t dt
14.
x
x
4
x
2
dx
1
3
dx
9
/ 6 1 c o s 3 t s in 3 t d t
b)
cos z
dz
4 3sin z
t 5 2t 5t 4 2 dt
2/3
2
3
2
4
y
y
4
y
1
12
y
2 y 4 dy
0
1/2
1 3
2
2
y
6
y
12
y
9
y
4 y 4 dy
0
1
t 2 sin 2 1 dt
t
1
3
2
16.
3
0
/6
3
cos
2 sin 2 d
0
1/4
5
5
4cos
t
sin t dt
0
1 2
15.
b)
4x
/3
2
1
cos z
a)
dz
0
4 3sin z
4
dy
dx
2
1
2 y 1 y
3
0
9
1
13.
b)
dx
x 1
/6
1
10.
2
0
7. a)
9.
4x
1
6. a)
8.
3
0
cos 2 3/2 d
2.8. Areas between Curves
Learning objectives:
In this module, we study
To determine the areas of the shaded regions between the given curves and lines.
AND
To practice the related problems.
Suppose we want to find the area of a region that is bounded above by the curve = ( ),
below by the curve = ( ), and on the left and right by the lines = and = .
We first approximate the region with
vertical rectangles based on a partition
P x0 ,x1 , ,xn of a,b
The area of the
rectangle is
ΔAk f ck g ck Δxk
We then approximate the area of the region by adding the areas of the
n
n
A ΔAk f ck g ck Δxk
k 1
k 1
Riemann sum
n rectangles:
P 0 the sums on the right approach the limit
As
because
f
and
g
b
a f x g x dx
are continuous. The area of the region is the value of this integral.
n
b
f ck g ck Δxk f x g x dx
a
P 0
A lim
k 1
Thus, this basic formula for finding the area between curves is a limit of Riemann sums.
g are continuous with f x g x throughout a,b , then the area
of the region between the curves y f x and y g x from a to b is the
If
f
and
integral of
f g from a to b :
b
A f x g x dx
a
.............1
We will graph the curves and draw a representative rectangle. This reveals which curve is
f (upper curve) and which is g (lower curve). It also helps identify the limits of
integration. Write a formula for
f x g x from
f x g x and simplify it. Integrate
to ; this gives the area.
Example 1:
2
Find the area between y sec x and y sin x from 0 to π/4.
Solution:
The curves and a vertical rectangle are sketched.
The upper curve is the graph of
.
f x sec 2 x ; the lower is the graph of
( )=
A
π/4
π/4
2
sec
x
sin
x
dx
tan
x
cos
x
0
0
2
2
1
0 1
2
2
When a region is determined by curves that intersect, the intersection points give the
limits of integration.
Example 2:
Solution Find the area of the region enclosed by the parabola
line y x .
We sketch the curves and a vertical rectangle.
The upper and lower curves are
y 2 x 2 and the
f x 2 x 2 and g x x . The x -
coordinates of the intersection points are the limits of integration.
2 x2 x x 2 x 2 0 x 1 x 2 0
x 1,
x2
The region runs from x 1 to x 2 .
f x g x 2 x2 x 2 x2 x
A
2
2
2
x
x
1
2
x2 x3
dx 2 x
2
3
1
4 8
1 1 9
4 2
2 3
2 3 2
If the formula for a bounding curve changes at one or more points, we partition the
region into subregions that correspond to the formula changes, calculate the area of
each subregion and then add.
Example 3:
Find the area of the region in the first quadrant that is bounded above by
y x
and below by the x -axis and the line y x 2
Solution:
The graphs of the two functions are shown below.
The region’s upper boundary is the graph of
changes from g
f x x . The lower boundary
x 0 for 0 x 2 to g x x 2 for 2 x 4 .
There is agreement at x = 2.
The limits of integration for region
region B is a = 2.
are
a 0 and b 2 . The left-hand limit for
To find the right-hand limit, we solve the equations
simultaneously for x :
y x
and
y x2
2
x x 2 x x 2 x2 4 x 4
x 2 5 x 4 0 x 1 x 4 0 x 1, x 4
x = 4 satisfies the equation x x 2 . The value x = 1 is an
extraneous root introduced by squaring. The right-hand limit is b 4 .
For 0 x 2 : f x g x x 0 x
Only the value
For 2 x 4 :
f x g x x x 2 x x 2
the areas of subregions A and B to find the total area:
2
Total area =
0
4
x dx
2
x x 2 dx
We add
4
2
2
2 3/2 2 3/2 x
x x
2x
2
3
0 3
2
2 3/2
2 3/2
2 3/2
2
0 4 8 8 2 2 4
3
3
3
2
10
8 2
3
3
If a region’s bounding curves are described by functions of y , the approximating
rectangles are horizontal instead of vertical and the basic formula has y in place of
d
A
c f y g y dy
In the equation (2) above,
hand curve, so
f
always denotes the right-hand curve and
g
the left-
f y g y is nonnegative.
Example 4:
Find the area of the region in the previous example by integrating with respect to y .
Solution:
We sketch the region and a typical horizontal rectangle based on a partition of an
interval of y -values.
The region’s right-hand boundary is the line x
left-hand boundary is the curve x
y 2, so f y y 2 . The
y 2 , so g y y 2 . The lower limit of
x.
integration is y 0 . We find the upper limit by solving
simultaneously for y .
x y 2 and x y 2
y 2 y2 y2 y 2 0
y 1 y 2 0 y 1, y 2
The upper limit of integration is b 2
f y g y y 2 y2
A
b
2
f y g y dy 2 y y 2 dy
a
0
2
3 2
y
y
4 8 10
2 y
4
2
3
2 3 3
0
Sometimes, we can combine integrals with formulas from geometry.
The previous example can also be solved the following way.
The area we want is the area between the curve y x , 0
axis, minus the area of a triangle with base 2 and height 2.
4
Area
0
x dx
x 4 , and the x -
1
2 2
2
4
2
2
10
x3/2 2 8 0 2
3
3
3
0
We thus see that, it is sometimes easier to find the area between two curves by
integrating with respect to y instead of x . Also, it may help to combine geometry
and calculus.
PROBLEM SET
IP1. Find the total area of the shaded region of the graph
Solution:
To obtain the total area
of the shaded region in the given graph, we have to
calculate the areas of the three shaded regions , , and add the absolute the
: −2 ≤
values of the areas, where
curve ( ) =
A1
−
and lower
,
( )− ( )=
0
≤ 0, upper curve ( ) =
−
1
f ( x ) g ( x ) dx 3
2
− = (
−4 )
0
x3 4 x dx
2
0
1 x4 4x2
1
4
0 4 8
3 4
2
3
3
2
: For the sketch lower limit is 0 and we find the upper limit by solving the equations
=
−
⟹
and
=
simultaneously for .
4
− = ⟹
−
= 0 ⟹ ( − 2)( + 2) = 0
3
3
3 3
3
= −2, 0 ,2 ⟹Upper limit is 2
Upper curve ( ) =
( )− ( )= −
2
A2
( )=
and lower curve
=− (
−
1
f (x) g (x) dx
3
0
−
−4 )
2
x3 4 x dx
0
2
x4 4 x2
1
4
4 8 0
2
3
3
4
0
≤ 3, upper curve ( ) = − and lower curve ( ) =
1
3
:2 ≤
,
( )− ( )=
3
A3
−
− = (
1
f ( x ) g ( x ) dx
3
2
−4 )
3
x
3
4 x dx
2
3
1 x 4 4 x2
81
16
25
18
8
3 4
2
4
12
4
2
Therefore, the total area of the shaded region is
=| |+| |+| |=
+
+
=
P1. Find the total area of the shaded region of the graph
Solution:
To obtain the total area
of the shaded region in the given graph, we have to
calculate the areas of the two shaded regions , and add the absolute the values of
the areas, where
: −2 ≤ ≤ 0, upper curve ( ) = 2 −
− 5 and lower curve
0
( )=−
+3
A1
f ( x) g ( x) dx
2
0
2 x 3 x 2 5 x x 2 3 x dx
2
0
2 x
2
0
3
8x
x4
dx
4x2
2
2
: 0 ≤ ≤ 2, upper curve ( ) = −
( )=2 −
−5
0
(8 1 6 ) 8
+ 3 and lower curve
2
f (x ) g ( x) dx
A2
0
2
x2 3 x 2 x3 x2 5 x dx
0
2
8x2
2 x4
8 x 2 x3 dx
4
2
0
2
(1 6 8 ) 0 8
0
Therefore, the total area of the shaded region is
= | | + | | = |8| + |8| = 16
IP2. Find the areas of the regions enclosed by the curves
= ⁄ and =
,− ≤ ≤ .
Solution:
Given curves are = ⁄ and = sec
where− ≤
=
The graph is plotted between the curves
−
⁄
and
≤ .
= sec
over the interval
,
From the graph, upper curve is ( ) = sec
and lower curve ( ) =
Here = −1 and = 1
∴ Area
of the shaded region in the given graph is
1
A
f ( x) g ( x) dx
1
1
1
1
2 x
1 3
s ec 3 x
dx
sec
1
2
1
x
dx
3
x1 3 d x
1
1
1
x1 3 1
3
x
ta n
3 1 1 3 1
1
3
3
4
6 3
3 1 1
3
⁄
P2. Find the areas of the regions enclosed by the curves
=
and =
,− ≤ ≤ .
Solution:
Given curves are = 8cos and = sec where− ≤
The graph is plotted between the curves = 8cos and
= sec
over the interval −
≤ .
,
From the graph, upper curve is ( ) = 8cos and lower curve ( ) = sec
Here = − and =
∴ Area
of the shaded region in the given graph is
3
A
f ( x ) g ( x ) dx
3
3
8 cos x sec 2 x dx
3
3
8 sin x tan x 3
3
3
8.
3 8.
3 6 3
2
2
IP3. Find the area of the region bounded between the curves
= −
and =
+
Solution:
Given curves are = 7 − 2 and =
+4
Now, we have to calculate the limits of integration.
7−2 =
+ 4 ⟹ 3 − 3 = 0 ⟹ 3( − 1)( + 1) = 0
⟹ = −1, 1 ⟹ = −1, = 1
The graph is plotted between the curves = 7 − 2 and
=
+ 4 over the interval
[−1 , 1]
From the graph, upper curve is ( ) = 7 − 2 and lower curve ( ) =
∴ Area
of the shaded region in the given graph is
1
A
f (x) g (x) dx
1
2
2
2
2
2
7 2 x x 4 d x 3 3 x d x
1
1
3
1
1
dx 3
1
x3
1
x dx 3 x 1 3
3
2
1
1
1
1 1
3 1 ( 1) 3 4
3 3
P3. Find the area of the region bounded between the curves
=
and = + .
Solution:
Given curves are =
and = + 2
Now, we have to calculate the limits of integration.
= +2⟹
− − 2 = 0 ⟹ ( − 2)( + 1) = 0
⟹ = −1, 2 ⟹ = −1,
=2
The graph is plotted between the curves =
and
= + 2 over the
interval [−1 , 2]
From the graph, upper curve is ( ) = + 2 and lower curve ( ) =
∴ Area
of the shaded region in the given graph is
+4
2
A
f ( y ) g ( y ) dy
1
2
1
2
2
y2 y
dy
1
2
y dy 2
1
2
dy
y 2 dy
1
2
2
y2
y3
2
2 y 1
2
1
3 1
8 1 9
4 1
2 2 ( 1)
2 2
3 3 2
IP4. Find the area of the propeller-shaped region enclosed by the curves
− ⁄ = and − ⁄ =
Solution:
Given curves are
− ⁄ =0⟹ =
and
− ⁄ =0⟹ =
Now, we have to calculate the limits of integration.
=
⟹
−
= 0 ⟹ ( − 1) = 0
⟹ = −1, 0, 1 ⟹ = −1 and = 1
The graph is plotted between the curves =
and =
over the interval [−1, 1],
which is a propelled- shaped region.
To obtain the total area
of the shaded region in the graph, we have to calculate the
areas of the two shaded regions , and add the absolute the values of the areas,
where
: −1 ≤ ≤ 0, ( ) − ( ) =
− and
: 0 ≤ ≤ 1, ( ) − ( ) =
−
By observing the graph, it is symmetry about the origin.
A A1 A2 2 A2
1
2
x 3 x 5 dx
0
1
x4 x6
1 1
2
6
4 6
4
0
1
0
6
Hence, the total area of the shaded region on [−1, 1] is =
P4. Find the area of the region in the first quadrant bounded on the left by the
−axis, below by the curve =
, above left the curve = ( − ) and above
right by the line
= − .
Solution:
Given curves are = 2
, = ( − 1) and lines are −axis that is = 0,
=3− .
Now, we have to calculate the limits of integration.
( − 1) = 3 − ⟹
− −2= 0
⟹ ( − 2)( + 1) = 0 ⟹ = 2 and = −1 is not a solution because > 0
Again, 2
= 3− ⟹ 4 = 9−6 +
⟹
− 10 + 9 = 0 ⟹ ( − 9)( − 1) = 0
⟹ = 1 and = 9 is not a solution because it does not satisfy the equation.
Hence, = 0 and = 2
The graph is plotted between the curves = 2
, = ( − 1) and lines are −axis
that is = 0, = 3 − over [0, 2]
To obtain the total area
of the shaded region in the graph, we have to calculate the
areas , and add the absolute the values of the areas, where
: 0 ≤ ≤ 1, ( ) − ( ) = 2
−0 =2
1
1
A1
2 y d y 2 y1 2 d y
0
0
1
y1 2 1
4
4
2
1 0
3
3
1 2 1 0
:1 ≤
( )−
≤ 2,
( ) = (3 − ) − ( − 1)
2
A2
3 y y 1 2 dy
1
2
3
y 2 y 1
3 y
2
3
1
1
1
7
6 2 3 0
3
2
6
Therefore, the total area of the shaded region on [0, 2] is
=| |+| |=
+
=
EXERCISES:
I. Find the area of the shaded regions in problems 1-4.
II. Find the areas of the regions enclosed by the lines and curves in problems 5-11.
a)
y x 2 2 and y 2
b)
y x 4 and y 8 x
c)
y x 2 and y x 2 4 x
d)
y x 4 4 x 2 4 and y x 2
e)
y
x and 5y x 6
III. Find the areas of the regions enclosed by the lines and curves in problems 12-15.
a.
x 2 y 2 , x 0,
b.
y 2 4 x 4, 4 x y 16
c.
x y 2 0,
x 3 y2 2
d.
x y 2 1,
x y 1 y2
y3
IV. Find the areas of the regions enclosed by the curves in problems 16 and 17.
a.
4 x 2 y 4, x 4 y 1
2
4
b. x 4 y 4, x y 1,
for x 0
V. Find the areas of the regions enclosed by the lines and curves in problems 18-21.
a)
y 2sin x
b)
y cos x / 2
c)
and y sin 2 x,
0 x
and y 1 x 2
y sec2 x, y tan 2 x , x / 4, x / 4
x 3sin y cos y , x 0 , 0 y / 2
d)
e) Find the area of the propeller-shaped region enclosed by the curve
x y 3 0 and the line x y 0 .
f) Find the area of the region in the first quadrant bounded by the line
line
x = 2, the curve y 1/ x 2 , and the x -axis.
y x , the
3.1. Volume of Solids by Slicing
Learning objectives:
To find the volume of a solid by the method of slicing.
o
AND
To practice the related problems.
A cross section of a solid S is the plane region formed by intersecting S with a plane
(see the figure below).
Suppose we want to find the volume of a solid like the one shown below.
At each point
x
in the closed interval
a, b the cross section of the solid is a region
R( x) whose area is A( x) . Then A is a real-valued function of x . If it is also a
continuous function of x , we can calculate the volume of the solid as an integral in the
following way.
We partition the interval
a, b along the x -axis in the usual manner and slice the
x -axis at the partition points. The k th slice, the
one between the planes at xk 1 and xk , has approximately the same volume as
the cylinder between these two planes based on the region R( xk ) .
solid by planes perpendicular to the
The volume of the solid is therefore approximated by the cylinder volume sum
n
A xk xk
k 1
This is a Riemann sum for the function
A( x) on a, b . We expect the
approximations from these sums to improve as the norm of the partition of
a, b
goes to zero, so we define their limiting integral to be the volume of the solid.
Volume of a solid
The volume of a solid of known integrable cross-section area A( x) from x
a to
x b is the integral of A from a to b :
b
V A x dx
.......... (1)
a
Steps involved to calculate the volume of a solid
1.
Sketch the solid and a typical cross section.
2.
Find a formula for A( x) , the area of a typical cross-section
3.
Find the limits of integration.
4.
Integrate A( x) using the fundamental theorem to find the volume of the solid.
Example 1:
A pyramid 3 m high has a square base that is 3 m on a side. The cross section of the
pyramid perpendicular to the altitude x m down from the vertex is a square x m on a
side. Find the volume of the pyramid.
Solution
We draw the pyramid with its altitude along the x -axis and its vertex at the origin.
A x x 2
The squares go from x 0 to x 3 . The volume is
The area of cross section is
V
b
a
A x dx
3 2
0
x dx
3 3
x
3
9 m
3
0
PROBLEM SET
IP1. Find a formula for the area ( ) of the cross sections of the solid perpendicular
to the −axis.
The solid lies between planes at = and = . In each case, the cross sections
are perpendicular to the − axis between these planes run from the parabola
= −√ to the parabola = √
a.
The cross sections are circular disks with diameters in the
−plane.
=√
b.
The cross sections are squares with bases in the
−plane.
c.
The cross sections are squares with diagonals in the
−plane.
d.
The cross sections are equilateral triangles with bases in the
−plane.
Solution:
a.
Since the cross sections are circular disks with diameters in the
− plane,
)
Area of the cross section ( ) = (
From the figure,
=√
∴ ( )= √
=
b.
Since the cross sections are squares with bases in the
−plane ,
Area of the cross section ( ) =
ℎ×ℎ ℎ
From the figure,
ℎ = ℎ ℎ = 2√
∴ ( ) = 2√ × 2√ = 4
c.
Since The cross sections are squares with diagonals in the
−plane,
Area of the cross section A(x) = (side) and
diagonal = √2 side. Thus,
=
√
= 2√
From the figure,
∴
( )=
√
√
=2
d.
Since the cross sections are equilateral triangles with bases in the
√
Area of the cross section ( ) = (
( )=
√
) and
= 2√
From the figure,
∴
−plane,
2√
= √3
P1. Find a formula for the area ( ) of the cross sections of the solid perpendicular
to the −axis.
The solid lies between planes perpendicular to the −axis at = − and = . In
each case, the cross sections are perpendicular to the − axis between these
planes run from the semicircle = −√ − to the semicircle = √ −
e.
The cross sections are circular disks with diameters in the
−plane.
f.
The cross sections are squares with bases in the
g.
The cross sections are squares with diagonals in the
−plane.
−plane.
h.
−plane.
The cross sections are equilateral triangles with bases in the
Solution:
e.
Since the cross sections are circular disks with diameters in the
)
Area of the cross section ( ) = (
From the figure,
= √1 −
∴
( )=
√1 −
= (1 −
)
f.
Since the cross sections are squares with bases in the
Area of the cross section ( ) =
ℎ×ℎ ℎ
From the figure,
ℎ = ℎ ℎ = 2√1 −
∴ ( ) = 2√1 − × 2√1 −
= 4 (1 − )
g.
− plane,
−plane ,
Since The cross sections are squares with diagonals in the
) and
Area of the cross section ( ) = (
= √2
. Thus,
=
−plane,
√
= 2√1 −
From the figure,
∴
h.
( )=
√
√
= 2 (1 −
)
Since the cross sections are equilateral triangles with bases in the
√
Area of the cross section ( ) = (
From the figure,
∴
( )=√
−plane,
) and
= 2√1 −
2√1 −
= √3(1 −
)
IP2. The solid lies between planes perpendicular to the −axis at = − and
= . The cross sections perpendicular to the − axis between these planes are
squares whose diagonals run from the semicircle = −√ − to the semicircle
= √ − . Find the volume of the solid.
Solution: Given the solid lies between planes perpendicular to the −axis at
= −1 and = 1.
The cross sections perpendicular to the − axis between these planes, are squares
whose diagonals run from the semicircle = −√1 − to the semicircle
= √1 −
) and
Area of the cross section is ( ) = (
= √2
Now, diagonal = √1 −
− −√1 −
= 2√1 −
∴
=
⟹ ( )=
√
√
=
√
√
√
= 2(1 −
)
= −1 and
Limits of integration:
=1
∴ Volume of the solid is
b
1
a
1
V A( x ) dx
1
1
x
1
1
2 dx 2
2 1 x 2 dx
2
dx
1
x3
1
2 x 1 2
3 1
1 1 8
2 1 (1) 2
3 3 3
P2. The solid lies between the planes perpendicular to the −axis at
= − and = . The cross sections perpendicular to the −axis, are circular
disks whose diameters run from the parabola =
to the parabola = − .
Find the volume of the solid.
Solution:
Given the solid lies between the planes perpendicular to the
= −1 and = 1.
−axis at
The cross sections perpendicular to the −axis are circular disks whose diameters
run from the parabola =
to the parabola = 2 −
∴ Area of the cross section ( ) = (radius) =
where diameter = (2 − ) −
= 2(1 − )
(
)
⟹ ( )=
Limits of integration:
= (1 − 2 +
= −1 and = 1
(
)
)
∴ Volume of the solid is
b
1
V A( x) dx
a
1 2 x 2 x4 dx
1
1
1
1
2
4
dx 2 x dx x dx
1
1
1
1
1
x3
x5
1
x1 2
3 1 5 1
2
1
1 (1) 1 (1) 1 (1)
3
5
2 1 16
2 1
3 5 15
IP3. The solid lies between the planes perpendicular to the
−axis at
= − and
= . Find the volume of the solid, if the cross sections perpendicular to the
−axis are
a.
Circular disks with diameters running from the curve
=
to the
curve =
.
b.
Squares with bases running from the curve =
to the curve =
.
Solution:
Given the solid lies between the planes perpendicular to the −axis at = − and
= .
a.
Since the cross sections perpendicular to the −axis are circular disks with
diameters running from the curve = tan to the curve = sec
∴ Area of the cross section
(
( ) = (radius) =
diameter = sec − tan
⟹ ( ) = (sec − tan )
)
and
= [sec
+ tan
− 2 sec . tan ]
= [sec
+ sec
− 1 −2 sec . tan ]
= [2 sec
− 1 −2 sec . tan ]
= − and
Limits of integration:
=
Volume of the solid is
3
b
V
A( x) dx
3
a
2 sec2 x 1 2 sec x.tan x dx
4
3
2
sec2 x dx
3
3
4
dx
3
2
3
sec x.tan x dx
3
3
3
3
tan x 3 x 3 sec x 3
2
4
2
3
3
4 3
2
3
3 2
2 2
2
6
b.
Since the cross sections perpendicular to the −axis are squares with bases
running from the curve = tan to the curve = sec
∴ Area of the cross section
( ) = (side) = (sec − tan )
= sec + tan − 2 sec . tan
= sec + sec − 1 −2 sec . tan
= 2 sec − 1 −2 sec . tan
Limits of integration: = − and =
∴ Volume of the solid is
3
b
V
A( x) dx
2sec2 x 1 2sec x.tan x dx
3
a
3
2
3
sec2 x dx
3
3
3
3
3
dx 2
sec x.tan x dx
3
3
2 tan x 3 x 3 2 sec x 3
2 3 3 2
3 3
2
4 3
3
2 2
P3. The base of a solid is the region between the curve = √
and the interval
[ , ] on the −axis. Find the volume of the solid, if the cross sections
perpendicular to the −axis, are
c.
Equilateral triangles with bases running from the −axis to the curve as
shown in figure
d.
Squares with bases running from the −axis to the curve
Solution:
Given the base of a solid is the region between the curve = 2√sin and the interval
[0, ] on the −axis.
c.
Since the cross sections perpendicular to the −axis are equilateral triangles
with bases running from the −axis to the curve as shown in figure.
∴ Area of the cross section
( )= ×(
)× (
)×
= × 2√
× 2√
Limits of integration: = 0 and =
∴ Volume of the solid is
b
V
A( x) dx
a
×
√
= √3
3 sin x dx 3 cos x 0 3 1 1 2 3
0
d.
Since the cross sections perpendicular to the
running from the −axis to the curve.
∴ Area of the cross section
−axis are squares with bases
( ) = (side) = 2√sin
= 4 sin
Limits of integration: = 0 and =
∴ Volume of the solid is
b
V
a
A( x) dx 4sin x dx 4 cos x 0 4 1 1 8
0
IP4.
A curved wedge is cut from a cylinder of radius 3 by two planes. One plane is
perpendicular to the axis of the cylinder. The second plane crosses the first plane at
a 45° angle at the center of the cylinder. Find the volume of the wedge.
Solution:
We draw the wedge and sketch a typical cross section perpendicular to the x -axis.
+
The base of the cylinder is the circle
The cross section at
x
= 9.
is a rectangle of area, A x x 2 9 x2 2 x 9 x 2
The rectangles runs from x 0 to x 3 .
The volume of the wedge is
b
V
3
9 x2 dx
a A x dx 0 2 x
put 9 x 2 u 2 x dx du 2 x dx du
Limits :
x0u 9
3
V
0
and
x 3u 0
2 x 9 x 2 dx
0
9
u1 2 du
9
0 u
12
du
9
u3 2
2
9 9 0 18
3 2 0 3
P4.
A wedge is cut out of a circular cylinder of radius by two planes; one of the planes
is perpendicular to the axis of the cylinder and the other plane intersects the first at
an angle of °along a diameter of the cylinders. Find the volume of the wedge?
Solution:
If we place the −axis along the diameter where the planes meet, then the base of
the solid is semicircle with equation = √16 −
, −4 ≤ ≤ 4.
A cross section perpendicular to the −axis at a distance from the origin is
triangle
, as shown in figure, whose base is = √16 −
and whose height is
| | = tan 30° = √16 − ⁄√3
Thus, the cross section area is
( ) = ×base×height
= × √16 −
=
⁄√3
× √16 −
√
∴ Volume of the wedge is
b
V
4
A( x) dx
a
1
3
4
4
4
16 x
2
0
16 x 2
dx
2 3
16
1
dx
dx
3
3
4
0
0
x 2 dx
4
16
1 x3
64 64
128
4
x 0
3
3 3
3 3 3 3 3
0
EXERCISES
1. The solid lies between planes perpendicular to the x -axis at x = 0 and x = 4. The
cross sections perpendicular to the axis on the interval 0 x 4 are squares whose
diagonal run from the parabola y x to the parabola y x . Find the volume of
the solid.
2. The solid lies between planes perpendicular to the x -axis at x = −1 and x = 1.
The cross sections perpendicular to the axis between these planes are vertical squares
whose base edges run from the semicircle y 1 x2 to the semicircle
y 1 x 2 . Find the volume of the solid.
3. The solid lies between the planes perpendicular to the y -axis at y = 0 and y = 2.
The cross sections perpendicular to the y -axis are circular disks with diameters
running from the
y -axis to the parabola
x 5 y 2 . Find the volume of the solid.
3.2. Volume of Solids of Revolution- Disks
Learning objectives:
To find the volume of solids of revolution by disk method.
AND
To practice the related problems.
The most common application of the method of slicing is to solids of revolution. Solids
of revolution are solids whose shapes can be generated by revolving plane regions
about axes.
If the region can be arranged between the graph of a continuous function
y R x , a x b , and the x -axis, we can find the solid’s volume in the
following way.
The typical cross section of the solid perpendicular to the axis of revolution is a disk of
2
radius R x and area ,
A x R x
The solid’s volume, being the integral of A from x a to x b , is the integral of
2
R x from a to b .
The volume of the solid generated by revolving about the x -axis the region between
the x -axis and the graph of the continuous function y R x , a x b , is
b
2
V R x dx ......... (1)
a
Example 1:
A solid of revolution (Rotation about
−axis)
The region between the curve y x , 0 x 4 , and the x -axis is revolved
about the x -axis to generate a solid. Find its volume.
Solution
We draw figures showing the region, a typical radius, and the generated solid.
The volume is
b
4
2
4
2
V R x dx x dx x dx
a
0
0
4
2
4 8
x2
2
2
0
The axis of revolution need not necessarily be the x -axis, but the rule for calculating
volume is the same.
Example 2:
A solid of revolution (Rotation about the line = )
Find the volume of the solid generated by revolving the region bounded by y
and the lines y = 1, x = 4 about the line y = 1.
Solution
We draw figures showing the region, a typical radius, and the generated solid.
x
The volume is
4
2
4
2
V R x dx x 1 dx
1
1
4
[ x 2 x 1] dx
1
4
x2
2
7
2 x3/2 x
3
6
2
1
To find the volume of a solid generated by revolving a region between the
a curve x R y ,
replaced by
≤
≤
about the
y -axis, we use equation (1) with
y -axis and
x
y.
d
V
c R y
2
dy
Example 3:
A solid of revolution (Rotation about −axis)
Find the volume of the solid generated by revolving the region between the
y -axis
and the curve x 2 / y, 1 y 4 , about the y -axis.
Solution
We draw figures showing the region, a typical radius, and the generated solid.
The volume is
2
4 4
2
V R y dy dy 2 dy
1
1 y
1 y
4
2
4
4
1
3
4 4 3
4
y 1
Example 4:
A solid of revolution (Rotation about the line = )
Find the volume of the solid generated by revolving the region between the parabola
x y 2 1 and the line x 3 about the line x
= 3.
Solution
We draw figures showing the region, a typical radius, and the generated solid.
The volume is
2
2
2
2
V
R y dy 2 y 2 dy
2
2
5 2
4
y
4 4 y 2 y 4 dy 4 y y 3
2
3
5
2
2
64 2
15
PROBLEM SET
IP1. Find the volume of the solid generated by revolving the regions bounded by the
curve =
and lines = , = − , = about the −axis.
Solution: Given curve is = sec and lines are = 0, = − , = . A typical
cross section of the solid perpendicular to the axis of revolution is shown in figure,
which is a disk of radius ( ) = sec .
Area of the region ( ) = [ ( )] = [sec ] = sec
Limits of integration:
= − and =
∴ Volume of the solid generated by revolving the regions bounded by the curve
= sec and lines = 0, = − , = about the −axis is
b
4
4
sec2 x dx tan x 4
4
2
V R( x) dx
a
tan 4 tan 4 2
P1.
Find the volume of the solid generated by revolving the regions bounded by the
curve =
and lines = , = about the −axis.
Solution: Given curve is =
and lines are = 2, = 0.
A typical cross section of the solid perpendicular to the axis of revolution is shown in
figure, which is a disk of radius ( ) =
Area of the region ( ) = [ ( )] = [ ] =
Limits of integration:
= 0 ⟹ = 0 and = 2
⟹ = 0 and = 2
∴ Volume of the solid generated by revolving the regions bounded by the curve
=
and lines = 2, = 0 about the −axis is
b
2
2
x7
2 6
x dx
0
7 0
V R( x) dx
a
27
128
0
7
7
IP2.
Find the volume of the solid generated by revolving the region in the first quadrant
bounded above by the line = , below by the curve =
,
≤ ≤ ⁄ and on the left by the −axis about the line = ?
Solution: Given curve is = 2 sin , 0 ≤ ≤ ⁄2 and line is = 2. A typical cross
section of the solid perpendicular to the axis of revolution is shown in figure, which is
a disk of radius
( ) = (2 − 2 sin ) = 2(1 − sin )
Area of the region
( ) = [ ( )] = [2(1 − sin )]
= [4(1 + sin
=4
− 2 sin )]
1+
− 2 sin
=4
= 0 and
Limits of integration:
−
− 2 sin
=
∴ Volume of the solid generated by revolving the region in the first quadrant
bounded above by the line = 2, below by the curve = 2sin , 0 ≤ ≤ ⁄2 and
on the left by the −axis about the line = 2 is
b
2
V R( x) dx
a
4
2 3
0
2
cos 2 x
2sin x dx
2
2
1 2
3 2
4 dx cos 2 x dx 2 sin x dx
0
2 0
2 0
3 2 1 sin 2 x 2
2
4 x0
2
cos
x
0
2
2
2
0
3
1
4 0 0 0 2 0 1
4
2 2
3
4 2 3 8
4
P2.
Find the volume of the solid generated by revolving the region in the first quadrant
bounded above by the line = √ , below by the curve =
and on the
left by the −axis about the line = √ ?
Solution: Given curve is = sec . tan and line is = √2
A typical cross section of the solid perpendicular to the axis of revolution is shown in figure,
which is a disk of radius
( ) = √2 − sec . tan
Area of the region
( ) = [ ( )] =
=
2 + sec
√2 − sec . tan
. tan
− 2√2 sec . tan
= 0 and sec . tan
Limits of integration:
= 0 and
= √2 ⟹
=
=
∴ Volume of the solid generated by revolving the region in the first quadrant
bounded above by the line = √2, below by the curve = sec . tan and on the
left by the −axis about the line = √2 is
b
2
V R( x) dx
a
4
2
2 tan x sec2 x 2 2 sec x tan x dx
0
2
4
0
dx
4
0
4
tan x 2 sec2 x dx 2 2 0
sec x tan x dx
4
tan x 3
4
2 x 0
3
0
1
0 2 2
2 3
11
2 2
3
2
4
2 2 sec x0
2 1
IP3.
Find the volume of the solid generated by revolving the regions enclosed by the
curve
=
,− ≤
Solution: Given curve is
=
≤
and lines
cos
, −2 ≤
=
about the
≤ 0 and line
−axis.
= 0.
A typical cross section of the solid perpendicular to the axis of revolution is shown in
figure, which is a disk of radius
( )=
cos
Area of the region ( ) = [ ( )]
=
cos
=
cos
Limits of integration:
= −2 and = 0
∴ Volume of the solid generated by revolving the regions enclosed by the curve
=
cos
, −2 ≤
≤ 0 and lines
= 0 about the
−axis is
2
y
2
R( y) dy cos
dy
c
0
4
V
d
0
4 y
. sin
4
0
sin
4
4 2
2
P3.
Find the volume of the solid obtained by rotating the region bounded by the
curve =
, = and = about the −axis.
Solution: The region between the curve =
and lines = 8 and = 0 is shown
in figure and the resulting solid is shown in figure.
The region is rotated about the −axis, it makes sense to slice the solid
perpendicular to the −axis and therefore to integrate with respect to .
If we slice at height , we get a circular disk with radius ( ) = ,
where =
. So the area of a cross section through is
( ) = [ ( )] =
Limits of integration: = 0 and
∴ Volume of the solid is
d
c
2
=
=8
8
0
V R( x) dy y 2 3 dy
8
y 2 31
3 5 3 8
y
0
2
3
1
5
0
3
96
32 0
5
5
⁄
IP4.
Find the volume of the solid generated by revolving the regions enclosed by the
curve = ⁄ and bounded by the lines = , = about the −axis.
Solution: Given curve is = ⁄ , and lines are
=0, =2
A typical cross section of the solid perpendicular to the axis of revolution is shown in
figure, which is a disk of radius
( )= ⁄
⁄
Area of the region ( ) = [ ( )] =
=
Limits of integration:
⁄
= 0 ⟹ = 0 and = 2
= 0 and = 2
∴ Volume of the solid generated by revolving the regions enclosed by the curve
= ⁄ and bounded by the lines = 0, = 2 about the −axis is
d
2
2
V R ( y ) dy y3 dy
c
0
2
y4
24
0 4
4 0
4
P4.
Find the volume of the solid obtained by rotating about the −axis the region
under the curve = √ from 0 to 1.
Solution:
The region under the curve = √ from 0 to 1 is shown in figure. If we rotate about
the −axis, we get the solid which is shown in figure.
When we slice through the point , we get a disk with radius
of this cross section is
( ) = [ ( )] = √
=
Limits of integration: = 0 and = 1
∴ Volume of the solid is
( ) = √ . The area
1
x2
b
1
1
2
V R( x) dx x dx 0
a
0
2 2
2 0
EXERCISES
1.
Find the volumes of solids generated by revolving the regions bounded by the
lines and curves in problems about the x -axis.
a.
b.
c.
y x 2 , y 0,
x2
y 9 x2 , y 0
y cos x , 0 x / 2
2.
Find the volumes of the solids generated by revolving the regions bounded by
the lines and curves in problems about the y -axis.
a)
b)
c)
x 5 y2,
x 0, y 1, y 1
x 2sin 2 y ,
0 y / 2, x 0
x 2 / y 1 ,
x 0,
y 0, y 3
3.3. Volumes of solids of Revolution-Washers
Learning objectives:
To find the volume of a solid of revolution by Washer method.
AND
To practice the related problems.
If the region we revolve to generate a solid does not border on or cross the axis of
revolution, the solid has a hole in it.
The cross sections which are perpendicular to the axis of revolution are washers
(rings) instead of disks. We denote the outer radius by R x and the inner radius by
r x . The washer’s (ring’s) area is
2
2
2
A x R x r x R x r x
2
The volume is
b
2
V R x r x
a
2
dx
This method for calculating volume of a solid of revolution is called the Washer
method, because a slab is a circular washer of outer radius R(x) and inner radius r(x).
Example 1:
2
The region bounded by the curve y x 1 and the line y x 3 is
revolved about the x -axis to generate a solid. Find the volume of the solid.
Solution:
Draw the region and sketch a line segment across it perpendicular to the axis of
revolution – −axis.
When the region is revolved, this segment will generate a typical ring cross section of
the generated solid.
We find the limits of integration by finding the x -coordinates of the intersection
points.
x 2 1 x 3 x 2 x 2 0 x 2 x 1 0
x 2, x 1
We find the outer and inner radii of the ring that would be swept out by the line
segment if it were revolved about the x -axis along with the region. These radii are
the distances of the ends of the line segment from the axis of revolution.
R x x 3 , r x x2 1
The volume is
b
2
2
V R x r x dx
a
1
2
x 3 x2 1
2
2
dx
1
3
1
x5
117
2
4
2 x
8 6 x x x dx 8 x 3 x
2
3
5
5
2
To find the volume of a solid generated by revolving a region about the
use the same procedure but integrate with respect to y instead of x .
Example 2:
2
y -axis, we
The region bounded by the parabola y x and the line y 2 x in the first
quadrant is revolved about the y -axis to generate a solid. Find the volume of the
solid.
Solution
We draw the region and sketch a line segment across it perpendicular to the axis of
revolution - y -axis.
The line and parabola intersect at
y = 0 and y = 4, so the limits of integration are
c = 0 and d
= 4.
The radii of the ring swept out by the line segment are R y y , r y y / 2 .
The volume is
2
2 y
d
4
2
2
V R y r y
c
dy
0
y dy
2
4
y 2 y3
y2
8
y
dy
0
4
2 12 0 3
4
Example 3:
2
The region in the first quadrant enclosed by the parabola y x , the y -axis, and
the line y = 1 is revolved about the line x = 3/2 to generate a solid. Find the volume
of the solid.
Solution: We draw the region and sketch a line segment across it perpendicular to the
axis of revolution, in this case the line =
y = 0 and y = 1. The radii of the ring swept out by the
r y 3 / 2 y , R y 3 / 2 .
The limits of integration are
line segment are
The volume is
d
2
2
V R y r y dy
c
2
2
1 3
3
y dy
0 2
2
4
2
1
y
3
3 y y dy 2 y3/ 2
0
2
2
0
PROBLEM SET
IP1. Find the volume of the solid generated by revolving the regions bounded by the
curve = − and line = − about the −axis.
Solution: Given curve = 4 − and line = 2 −
The region between the curve = 4 −
and line = 2 −
is shown in figure and sketch a line segment across it perpendicular to the axis of
revolution about − axis.
When the region is revolved about −axis, it will generate a typical ring cross section
of the generated solid.
Limits on integration:
4−
=2− ⟹
− −2 =0
⟹ ( + 1)( − 2) = 0 ⟹ = −1, = 2
⟹ = −1 and = 2
Outer radius ( ) = 4 − and inner radius ( ) = 2 −
Therefore, the ring’s area of cross section is
( ) = {[ ( )] − [ ( )] } = {[4 − ] − [2 − ] }
= 4 ( − 9 + 4 + 12)
∴ Volume of the solid is
b
V
a
R( x)
2
2
r ( x)
dx
1
x5
1
4
2
4 x 9 x 4 x 12 dx 4 3 x3 2 x2 12 x
0
5
0
1
108
4 3 2 12 0
5
5
P1. Find the volume of the solid generated by revolving the region bounded by the
curve = √ and lines = , = about the −axis.
Solution: Given curve = 2√ and lines = 2, = 0
The region between the curve = 2√ and lines = 2, = 0 is shown in figure and
sketch a line segment across it perpendicular to the axis of revolution about − axis.
When the region is revolved about −axis, it will generate a typical ring cross section
of the generated solid.
Limits on integration:
2√ = 2 ⟹ = 1 and = 0 ⟹ = 0 and = 1
Outer radius ( ) = 2 and inner radius ( ) = 2√
Therefore, the ring’s area of cross section is
( ) = {[ ( )] − [ ( )] }
[2 ] − 2 √
=
= 4 (1 − )
∴ Volume of the solid is
b
V
a
R( x) r (x) dx
2
2
1
x2
4 1 x dx 4 x
0
2
0
1
4 1 0 2
2
1
IP2. The region
enclosed by the curves = and =
is rotated about the
line = . Find the volume of the solid.
Solution: The curves = and =
intersect at the points (0, 0) and (1, 1). The
region between them, the solid of revolution, and a cross section perpendicular to
−axis are shown in figure.
Sketch a line segment perpendicular to the axis of revolution i.e., the line = 2.
Limits of integration:
= ⟹ ( − ) = 0 ⟹ ( − 1) = 0 ⟹ = 0, = 1
⟹ = 0 and = 1
A cross section in the plane has the shape of a washer (an annular ring) with inner
radius ( ) = 2 − and outer radius ( ) = 2 − .
Now, we find the cross sectional area by subtracting the area of the inner circle from
the area of the outer circle
( ) = {[ ( )] − [ ( )] }
= {[2 − ] − [2 − ] } = { − 5 + 4 }
∴ Volume of the solid is
b
V
a
R( x)
2
r ( x)
2
dx
1
x5 5 x3 4 x2
4
2
x 5 x 4 x dx
0
5
3
2
0
1 5
8
2 0
15
5 3
1
The solid of revolution and a typical washer is shown below.
P2. The region enclosed by the curves = and =
is rotated about the
−axis. Find the volume of the resulting solid.
Solution: The curves = and =
intersect at the points (0, 0) and (1, 1). The
region between them, the solid of rotation, and a cross section perpendicular to
−axis are shown in figure.
Sketch a line segment across it perpendicular to the axis of revolution about the
−axis.
Limits of integration:
= ⟹ ( − ) = 0 ⟹ ( − 1) = 0 ⟹ = 0, = 1
⟹ = 0 and = 1
A cross section in the plane has the shape of a washer (an annular ring) with inner
radius ( ) =
and outer radius ( ) = .
Now, we find the cross sectional area by subtracting the area of the inner circle from
the area of the outer circle
( ) = {[ ( )] − [ ( )] }
( ) = {[ ] − [ ] } = { − }
∴ Volume of the solid is
V
b
a
1
2
2
R( x) r ( x)
dx
1
x3 x5
2
4
x x dx
0
3
5
0
1 1 2
0
3 5 15
The solid of revolution and a washer is shown below.
IP3:
Find the volume of the solid generated by revolving the region enclosed by the
triangle with vertices ( , ), ( , ) and ( , ) about − axis.
Solution: Given vertices of a triangle is (0, 1), (1,0) and (1, 1). The equation of
the line
is
( − 1)(1 − 0) = (0 − 1)( − 0) ⟹ = 1 −
The equation of the line
is
( − 0)(1 − 1) = (1 − 0)( − 1) ⟹ = 1
The equation of the line
is
( − 1)(0 − 1) = (1 − 1)( − 1) ⟹ = 1
The triangular region of the solid is plotted between the lines = 1 − , = 1,
= 1 is shown below.
Sketch a line segment across it perpendicular to the axis of revolution – −axis.
When the region is revolved about −axis, it will generate a typical ring cross section
of the generated solid.
Limits of integration:
1 − = 1 ⟹ = 0, and = 1 ⟹ = 0 and = 1
From figure, outer radius is ( ) = 1 and inner radius is
Therefore, the ring’s cross sectional area is
( ) = {[ ( )] − [ ( )] }
= {[1] − [1 − ] } = (2 − )
∴ Volume of the solid is
d
V
c
R ( y )2 r ( y)2
( ) = 1− .
1
dy 2 y y 2 dx
0
1
2 y3
1 2
4 y 4 1 0
3
3 3
0
P3.
Find the volume of the solid generated by revolving the region enclosed by the
triangle with vertices ( , ), ( , ) and ( , ) about − axis.
Solution:
Given vertices of a triangle is (1, 0), (2,1) and (1, 1).
The equation of the line
is
( − 0)(2 − 1) = (1 − 0)( − 1) ⟹ = + 1
The equation of the line
is
( − 1)(1 − 2) = (1 − 1)( − 2) ⟹ = 1
The equation of the line
is
( − 1)(1 − 1) = (0 − 1)( − 1) ⟹ = 1
The triangular region of the solid is plotted between the lines
= + 1, = 1, = 1 is shown below.
Sketch a line segment across it perpendicular to the axis of revolution – −axis.
When the region is revolved about −axis, it will generate a typical ring cross section
of the generated solid.
Limits of integration:
+ 1 = 1 ⟹ = 0, and = 1 ⟹ = 0 and = 1
From figure, outer radius is ( ) = 1 + and inner radius is ( ) = 1.
Therefore, the ring’s cross sectional area is
( ) = {[ ( )] − [ ( )] }
= {[1 + ] − [1] } = ( + 2 )
∴ Volume of the solid is
d
V
c
R ( y )2 r ( y)2
1
dy y 2 2 y dx
0
1
y3
1 4
4 y 2 4 1 0
3 3
3
0
IP4.
Find the volume of the solid generated by revolving the region in the second quadrant
bounded above by the curve = − , below by the −axis and on the right by the line
= − about the line = − .
Solution: The region is plotted in the second quadrant bounded above by the
curve = − , below by the −axis and on the right by the line = −1 is shown in
figure.
Sketch a line segment across it perpendicular to the axis of revolution about
− axis.
When the region is revolved about −axis, it will generate a typical ring cross section
of the generated solid.
Limits on integration:
− ⁄ = −1 ⟹ = 1 and = 0 ⟹ = 0 and = 1
From figure, outer radius is ( ) = − ⁄ − (−2) = 2 − ⁄ and inner radius is
( )=1
Therefore, the ring’s area of cross section is
( ) = {[ ( )] − [ ( )] }
=
2−
⁄
− [1 ]
=
3+
⁄
−4
⁄
∴ Volume of the solid is
d
2
V R ( y ) r ( y)
c
2
dy
3
3 y 4 y
dy 3 y y5 3 3 y 4 3
0
5
0
3 3
3 3 0
5 5
1
23
1 3
3
P4.
Find the volume of the solid generated by revolving the region in the first quadrant
bounded on the left by the circle
+
= , on the right by the line = √ and
above by the line = √ about the −axis.
Solution: The region is plotted in the first quadrant bounded on the left by the
circle +
= 3 , on the right by the line = √3 and above by the line = √3 is
shown in figure.
Sketch a line segment across it perpendicular to the axis of revolution about
− axis.
When the region is revolved about −axis, it will generate a typical ring cross section
of the generated solid.
Limits on integration:
3−
= √3 ⟹ = 0 and = √3 ⟹ = 0 and = √3
From figure, outer radius is ( ) = √3 and inner radius is ( ) = 3 −
Therefore, the ring’s area of cross section is
( ) = {[ ( )] − [ ( )] }
=
√3 −
∴ Volume of the solid is
d
V
c
3−
R( y )2 r ( y )2
3
y3
3 0
3
3
=
dy
0
3
y 2 dy
0 3
3
EXERCISES
1.
Find the volumes of the solids generated by revolving the regions bounded by
the lines and curves in problems 1-4 about the x -axis.
a)
y x , y 1,
x0
y x 2 1,
y sec x,
b)
c)
y x3
y 2, / 4 x 4
2
2. The region in the first quadrant bounded above by the parabola y x , below
by the x -axis, and on the right by the line x = 2. Find the volume of the solid
generated by revolving the region about the y -axis.
3. In the first quadrant, a region is bounded above by the curve y
x 2 , below by
the x -axis, and on the right by the line x 1. Find the volume of the solid
generated by revolving the region about the line x 1.
4. Find the volume of the solid generated by revolving the region bounded by
y x
a.
b.
c.
d.
and the lines y = 2 and
the x -axis;
the y -axis;
the line y = 2;
the line x = 4.
x
= 0 about
5. Find the volume of the solid generated by revolving the region bounded by
y x 2 and the line y 1 about
a.
the line y = 1;
b.
the line y = 2;
c.
the line y 1.
3.4. Shell Method of Finding Volumes
Learning objectives:
To find the volume of the solid generated by shell method.
AND
To practice the related problems.
In finding the volume of a solid of revolution, cylindrical shell method sometimes
works better than rings. We derive the shell formula below.
In the figure below, a solid of revolution is approximated by the cylindrical shells.
To estimate the volume of the solid, we can approximate the region with rectangles
based on a partition P of the interval a, b over which the region stands. The typical
approximating rectangle is xk units wide by f ck units high, where ck is the
midpoint of the rectangle’s base.
A formula from geometry tells us that the volume of the shell swept out by the
rectangle is
Vk 2 average shell radius shell height thickness
2 ck f ck xk
We approximate the volume of the solid by adding the volumes of the shells swept
n
n
out by the n rectangles based on P :
V
Vk
2 ck f ck xk
k 1
k 1
The limit of this sum as P 0 gives the volume of the solid:
n
b
V lim
2 ck f ck xk a 2 x f x dx
P 0 k 1
The volume of the solid generated by revolving the region between the x -axis and
the graph of a continuous function y f x 0,0 a x b , about the y -axis is
b
V 2 shell radius shell height dx
a
b
V 2 x f x dx
a
.............. (1)
Example 1:
The region bounded by the curve y x , the x -axis, and the line x 4 is
revolved about the y -axis to generate a solid. Find the volume of the solid.
Solution:
Sketch the region and draw a line segment across it parallel
to the axis of revolution.
Label the segment’s height, and distance from the axis of revolution.
dx .
The limits of integration are a 0 and b 4 .
The width of the segment is the shell thickness
b
V 2 shell radius shell height dx
a
b
4
a
0
2 x f x dx 2 x x dx
4
4 3/2
128
2 5/2
x
dx 2 x
0
5
5
0
2
Equation (1) is for the vertical axes of revolution. For horizontal axes, we
replace x ’s with y ’s.
d
V 2 shell radius shell height dy
c
d
2 y f y dy
c
Example 2:
The region bounded by the curve y x , the x -axis, and the line x = 4 is
revolved about the x -axis to generate a solid. Find the volume of the solid.
Solution:
Sketch the region and draw a line segment across it parallel to the axis of revolution.
Label the segment’s length (shell height), and distance from the axis of revolution
(shell radius). The width of the segment is the shell thickness dy . The limits of
c = 0 and d = 2.
d
V 2 shell radius shell height dy
c
integration are
4 2
y
2 y f y dy 2 y 4 y 2 dy 2 2 y 2 8
c
0
4
0
d
2
Example 3:
2
The region in the first quadrant bounded by the parabola y x , the y -axis, and
the line y = 1 is revolved about the line x =2 to generate a solid. Find the volume of
the solid.
Solution: Draw a line segment across it parallel to the axis of revolution (the line x =
2).
Label the segment’s height, and distance from the axis of revolution. The width of the
segment is the shell thickness dx .
The limits of integration are a = 0 and b = 1.
b
V 2 shell radius shell height dx
a
1
2 (2 x) 1 x 2 dx
0
13
2 x 2 x 2 x3 dx
0
6
1
2
PROBLEM SET
IP1.
Use the shell method to find the volume of the solid generated by revolving the
region bounded by the curve =
and the line = about the −axis.
Solution: Given curve =
and the line is = .
The region between the curves =
and the line = is shown in figure and draw
a line segment across it parallel to the axis of revolution: − axis.
Limits of integration:
= ⟹ ( − 1) = 0 ⟹ = 0 and = 1
⟹ = 0 and = 1
From the figure,shell radius = , shell height =
shell thickness =
Therefore, the volume of the solid is
−
b
a 2 shell radius shell height dx
1
2
2
x2 x3 dx
2 x x x dx 2
0
0
V
1
x3 x 4
1 1
2 2 0
4
3 4 6
3
0
P1.
Use the shell method to find the volume of the solid generated by revolving the
region bounded by the curve =
− , and the line = about the −axis.
Solution: Given curves are = 2 − and the line is = 0.
The region between the curves = 2 − and the line = 0 is shown in figure
and draw a line segment across it parallel to the axis of revolution: − axis.
Limits of integration:
2 −
= 0 ⟹ (2 − ) = 0 ⟹ = 0 and = 2
⟹ = 0 and = 2
From the figure, shell height = 2 − , shell radius =
shell thickness =
Therefore, the volume of the solid is
b
a 2 shell radius shell height dx
2
2
2
3
2 x3 x 4 dx
2 x 2 x x dx 2
0
0
V
2
x 4 x5
32 16
2 2 8 0
5
5
5
2
0
The solid generated by the revolution is shown below.
IP2.
Use the shell method to find the volume of the solid generated by revolving the region
bounded by the curves = − , =
and the line = about the −axis.
Solution: Given curves are = 2 − , =
and the line is = 0.
The region between the curves = 2 − , =
and the line = 0 is shown in
figure and draw a line segment across it parallel to the axis of revolution: − axis.
Limits of integration:
2−
=
⟹ 2 = 2 ⟹ = 1 and = 0
⟹ = 0 and = 1
From the figure, shell radius =
shell height = (2 − ) −
= 2(1 −
shell thickness =
Therefore, the volume of the solid is
)
b
a 2 shell radius shell height dx
1
1
2
2 x 2 1 x dx 4 x x3 dx
0
0
V
1
x2 x4
1 1
4 4 0
4
2 4
2
0
P2.
Use the shell method to find the volume of the solid generated by revolving the
region bounded by the curve = − and the line = about the line = .
Solution:
Given curve = −
and the line is = 0.
The region between the curves = − and the line = 0 is shown in figure and
draw a line segment across it parallel to the axis of revolution: the line = 2.
Limits of integration:
−
= 0 ⟹ (1 − ) = 0 ⟹
= 0 and
=1
⟹ = 0 and = 1
From the figure, shell radius = 2 −
shell height = −
shell thickness =
Therefore, the volume of the solid is
b
a 2 shell radius shell height dx
1
1
2 2 x x x2 dx 2 x3 3 x2 2 x dx
0
0
V
1
x4
1
3
2
2 x x 2 1 1 0
2
4
4
0
IP3.
Use the shell method to find the volume of the solid generated by revolving the
region bounded by the curve =
and the lines = − , = , ≥ about the
− axis.
Solution:
Given curve is =
and the lines are = − , = 2, ≥ 0.
The region between the curve =
and the lines = − , = 2 is shown in figure
for ≥ 0 and draw a line segment across it parallel to the axis of revolution:
−axis.
Limits of integration:
= − ⟹ ( + 1) = 0 ⟹
= 2 ⟹ = 0 and = 2
= 0,
= 1(not a solution because
From the figure, shell radius =
shell height =
− (− ) =
shell thickness =
+ ,
≥ 0) and
Therefore, the volume of the solid is
d
c 2 shell radius shell height dy
2
2
3
2
2 y y 2 y dy 2
0
0 y y dy
V
2
y 4 y3
8
40
2
2 4 0
3
3
3
4
0
P3.
Use the shell method to find the volume of the solid generated by revolving the
region bounded by the curve =
and the lines = − , = about the
− axis.
Solution: Given curve is =
and the lines are = − , = 2.
The region between the curve =
and the lines = − , = 2 is shown in
figure and draw a line segment across it parallel to the axis of revolution: −axis.
Limits of integration:
= − ⟹ ( − 1) = 0 ⟹ = 0 , = 1 and = 2
⟹ = 0 and = 2
From the figure, shell radius =
shell height =
− (− ) =
+ ,
shell thickness =
Therefore, the volume of the solid is
d
c 2 shell radius shell height dy
2
2
32
2
2 y y y dy 2
0
0 y y dy
V
2
2
y3 2 1 y3
2 5 2 y3
2
2 y
3
3 2 1 3 0
5
0
2 5 2 8 16
2 2 0
3 25
5
3
15
IP4.
Use the shell method or washer method to find the volume of the solid generated
by revolving the region in the first quadrant bounded by the curve = −
and
the lines = , = about
A.
The − axis
B. The −axis
B.
The line =
D. The line =
Solution: Given curve = −
and the lines = 1, = 1.
A.
The region between the curve = −
and the lines = 1, = 1 is as
shown in figure and draw a line segment across it parallel to the axis of revolution:
−axis.
Limits of integration:
Since the region in the first quadrant, so = 0 and = 1
From the figure, shell radius = , shell thickness =
shell height = 1 − ( − ) = 1 − +
Therefore, the volume of the solid is
d
V 2 shell radius shell height dy
c
1
1
2 y 1 y y 3 dy 2 y y 2 y 4 dy
0
0
1
y 2 y3 y5
1 1 1
11
2
2 0
3
5
2 3 5
15
2
0
B.
In this case, we cannot express explicitly in terms of . So, shell method cannot
be used. Here we are using the washer method to solve the problem.
The region is shown in figure. Sketch a line segment across it perpendicular to the axis
of revolution about the −axis.
Limits of integration:
Since the region in the first quadrant, so = 0 and = 1
When the region is revolved about −axis, it will generate a typical washer cross
section of the generated solid.
From the figure, inner radius ( ) = ( − ) and outer radius ( ) = 1
Therefore, the cross section’s area is
( )=
( ) − ( )
= [1 − ( − ) ] = [1 −
−
+2 ]
Therefore, the volume of the solid is
d
1
2
2
V R ( y ) r ( y ) dy 1 y 2 y 6 2 y 4 dy
c
0
3
7
5 1
y
y
2y
1 1 2 97
y
1 0
3
7
5
3 7 5 105
0
C.
In this case, we cannot express explicitly in terms of . So, shell method cannot
be used. Here we are using the washer method to solve the problem.
The region is shown in figure. Sketch a line segment across it perpendicular to the
axis of revolution about the line = 1.
Limits of integration: Since the region in the first quadrant, so = 0 and = 1
When the region is revolved about −axis, it will generate a typical washer cross
section of the generated solid.
From the figure, inner radius ( ) = 0 and outer radius ( ) = [1 − ( − )]
Therefore, the cross section’s area is
( )=
( ) − ( )
=
1−( − ) −0
= [1 +
+
−2 +2 −2
Therefore, the volume of the solid is
d
]
1
2
2
V R( y ) r ( y ) dy 1 y 2 y 6 2 y 2 y 3 2 y 4 dy
0
c
4
5 1
y3 y 7
y
2
y
y
y2
3
7
2
5
0
1 1
1 2 121
1 1 0
2 5 210
3 7
D.
The region is as shown in figure and draw a line segment across it parallel to
the axis of revolution: = 1.
Limits of integration:
= 0 and = 1
From the figure, shell radius = (1 − ), shell thickness =
shell height = 1 − ( − ) = 1 − +
Therefore, the volume of the solid is
d
V 2 shell radius shell height dy
c
1
1
2 1 y 1 y y 3 dy 2 1 y y3 y y 2 y 4 dy
0
0
1
3
4
5
y
y
y
2 1 2 y y 2 y3 y 4 dy 2 y y 2
0
3
4
5
0
1 1 1 23
2 1 1 0
3 4 5 30
1
P4.
Use the shell method to find the volume of the solid generated by revolving the
region bounded by the curve = √ and the lines = , = about
C.
The − axis
B. The −axis
C. The line =
D. The line =
Solution: Given curve = √ and the lines = 0, = 2.
A.
The region between the curve = √ and the lines = 0, = 2 is shown in
figure and draw a line segment across it parallel to the axis of revolution: −axis.
= 0 ⟹ = 0 and = 2
⟹ = 0 and = 2
From the figure, shell radius =
; shell thickness =
Limits of integration:
shell height =
−0 =
Therefore, the volume of the solid is
d
V 2 shell radius shell height dy
c
2
y4
2
3
2 y y dy 2 y dy 2 2 4 0 8
0
0
4 0
2
2
B. The region is as shown in figure. Draw a line segment across it parallel to the axis
of revolution: −axis.
Limits of integration: √ = 2 ⟹ = 4 and = 0
⟹ = 0 and = 2
From the figure, shell radius = , shell height = 2 − √ and shell thickness =
Therefore, the volume of the solid is
b
V 2 shell radius shell height dx
a
4
4
2 x 2 x dx 2 2 x x3 2 dx
0
0
4
2 x3 2 1
64 32
2 x
2 16 0
3 2 1
5
5
0
B.
The region is as shown in figure and draw a line segment across it parallel to
the axis of revolution: = 4
Limits of integration: = 0 and = 4
From the figure, shell radius = 4 − , shell height = 2 − √ ,
shell thickness =
. Therefore, the volume of the solid is
.
b
V 2 shell radius shell height dx
a
4
4
2 (4 x) 2 x dx 2 8 4 x1 2 2 x x3 2 dx
0
0
4
3 2 1
x1 21
2 x
2 8 x 4
x
1
2
1
3
2
1
0
4
52
8x3 2
2 2x
2 8 x
x
3
5
0
64
64 224
2 32 16 0
3
5
15
C.
The region is as shown in figure and draw a line segment across it parallel to
the axis of revolution: = 2.
Limits of integration:
= 0 and = 2
From the figure, shell radius = 2 −
shell height =
−0=
, shell thickness =
Therefore, the volume of the solid is
d
V 2 shell radius shell height dy
c
2
2
2 2 y y 2 dy 2
2 y 2 y3 dy
0
0
3
4 2
2y
y
16 16 8
2
2 0
4
3 4 3
3
0
EXERCISES
1.
Use the shell method to find the volumes of the solids generated by revolving
the regions bounded by the curves and lines about the y -axis.
a)
y x,
b)
y x 2 , y 2 x,
x 0, for x 0
y 1 / x , y 0, x 1 / 2, x 2
c)
y x / 2,
x2
2.
Use the shell method to find the volumes of the solids generated by revolving
the regions bounded by the curves and lines about the x -axis.
b.
x 2 y y2 , x 0
y x , y 1
c.
y
a.
x , y 0,
y x2
3.
Find the volumes of the solids generated by revolving the regions about the
given axes.
A.
The triangle with vertices (1, 1), (1, 2), and (2, 2) about (a) the
y -axis; (c) the line x = 10/3; (d) the line y = 1
B.
x y y 3 , x 1, and
y -axis; (c) the line x = 1; (d) the line y = 1
The region in the first quadrant bounded by
y 1 about (a) the x -axis; (b) the
C.
the
x -axis; (b) the
The region in the first quadrant bounded by
x -axis; (b) the line y = 8
D.
The region bounded by
the line x = 1
y x 3 , and y 4 x about (a)
y 2 x x 2 and y x about (a) the y -axis (b)
3.5. Lengths of plane curves
Learning objectives:
To find the length of a smooth plane curve in Cartesian coordinates.
AND
To practice the related problems.
Suppose we want to find the length of the curve = ( ) from = to = . We
partition [ , ] in the usual way and connect the corresponding points on the curve
with line segments to form a polygonal path that approximates the curve.
The length of a typical line segment PQ is
xk 2 yk 2 .
The length of the curve is therefore approximated by the sum
n
2
2
xk yk .
k 1
We expect the approximation to improve as the partition of
a, b becomes finer.
We will show that the sums above approach a calculable limit as the norm of the
partition goes to zero.
A function with a continuous first derivative is said to be smooth and its graph is
called a smooth curve.
If
f
is smooth, by the Mean Value Theorem there is a point
ck , f ck on the
curve between P and Q where the tangent is parallel to the segment PQ.
At this point,
f ck
yk
xk
yk f ck xk
,
With this substitution, we have
n
2
xk yk
2
n
xk 2 f ck xk
k 1
k 1
n
2
1 f ck xk
2
a Riemann sum
k 1
Because 1 f x
2
is continuous on a, b , the limit of the sums on the right as
the norm of the partition goes to zero is
b
2
1 f x dx .
a
We define the length of the curve to be the value of this integral.
If
f
is smooth on
a, b , the length of the curve y f x from a to b is
2
b
2
dy
L 1 dx 1 f x dx
a
a
dx
b
Example 1:
Find the length of the curve y
4 2 3/2
x 1,0 x 1 .
3
Solution:
y
4 2 3/2
dy 4 2 3 1/2
x 1
x 2 2 x1/2
3
dx
3 2
2
dy
8x
dx
The length of the curve from
x = 0 to x = 1 is
2
b
dy
L
1 dx
1 8 x dx
0
a
dx
1
1
2 1
13
3/2
1 8 x
3 8
0 6
At a point on a curve where
....... (1)
dy
dx
fails to exist,
dx
dy
may exist and we may be
able to find the curve’s length by expressing x as a function of
applying the following analogue of equation (1):
L
d
c
2
d
dx
2
1 dy 1 g y dy .........(2)
c
dy
y say = ( ) and
Example 2:
Find the length of the curve y x / 2
Solution:
The derivative
dy 2 x
dx 3 2
1/3
2/3
from
x = 0 to x 2.
1/3
1 1 2
2 3 x
is not defined at x = 0, so we cannot find the curve’s length with equation (1).
We therefore rewrite the equation to express x in terms of y :
y x / 2
2/3
x
y 3/2 x 2 y 3/2
2
From this we see that the curve whose length we want is also the graph of
x 2 y 3/ 2 from y
The derivative
= 0 to
y = 1.
dx
3
2 y1/2 3 y1/2 is continuous on [0, 1]. We may
dy
2
therefore use equation (2) to find the curve’s length.
2
1
1
dx
1 2
3/2
L 1 dy 1 9 y dy 1 9 y
c
0
9 3
0
dy
2
10 10 1 2.27
27
d
2
2
d
dx
dy
The equations L 1 dx and L 1 dy ..... (3)
a
c
dx
dy
b
are often written with differentials instead of derivatives. This is done formally by
thinking of the derivatives as quotients of differentials and bringing the dx and dy
inside the radicals to cancel the denominators. In the first integral we have
2
2
dy 2
dy
2 dy
1 dx 1
dx dx
dx 2 dx 2 dy 2
2
2
dx
dx
dx
In the second integral we have
2
2
dx
dx 2
2 dx
1 dy 1
dy dy
dy 2 dx 2 dy 2
dy 2
dy 2
dy
Thus the integrals in (3) reduce to the same differential formula:
b
L
a
dx 2 dy 2 ........(4)
The differentials dx and dy must be expressed in terms of a common variable, and
appropriate limits of integration must be found for performing the integration.
We can also view equation (5) as follows. We think of dx and dy as two sides of a
small triangle whose “hypotenuse” is ds dx 2 dy 2 .
The differential ds is then regarded as a differential of arc length that can be
integrated between appropriate limits to give the length of the curve.
The arc length differential and the differential formula for the arc length are given by
ds dx 2 dy 2
L ds
PROBLEM SET
IP1. Find the length of the curve =
Solution: Given curve = ⁄ , 0 ≤
Differentiating w.r.t , we get
⁄
from
≤4
=
to
dy 3 3 21 3 1 2
x
x
dx 2
2
2
2
9
dy
3
Now, 1 1 x1 2 1 x
4
dx
2
Therefore, the length of the given curve is
2
4
9
dy
L 1 dx 1 x dx
a
0
4
dx
9
4
Put 1 x u dx du
4
9
Limits: x 0 u 1 and x 4 u 10
b
=
10
10
1
4 t1 21
4 4 10 1 2
t dt t dt
9 1 2 1
9 9 1
1
10
4 2
8
. t 3 2
10 10 1
9 3
27
1
=
P1. Find the length of the curve
Solution:
Given curve is
Differentiating
= + + +
w.r.t , we get
+
,0 ≤
+
+
,
≤
≤ .
≤2
dy 3 x 2
1
1
2
2x 1
x
2
x
1
2
2
dx
3
4 x 1
4 x 1
dy
1
2
x 1
2
dx
4 x 1
Squaring on both sides, we get
2
2
1
1
1
2
4
dy
x
1
x
1
dx
2
4
2
4 x 1
16 x 1
Now,
2
1
1
4
dy
1
1
x
1
4
2
dx
16 x 1
4
x 1
1
16 x 1
4
1
1
2
x 1
2
2
4
x
1
Therefore, Length of the given curve is
2
b
2
1
dy
2
dx
L 1 dx x 1
2
a
0
dx
4 x 1
2
2
x 13
2
1
1
2
dx
x 1
2
0
3
4
x
1
4 x 1
0
2 13
0 13
1
1 53
3
4 2 1 3
4 0 1 6
2
IP2. Find the length of the curve
=
Solution: Given curve is
Differentiating
w.r.t
=
+
+
=
from
, 2≤
= .
to
≤3
, we get
dx 1 2 1 2 1 2
.3 y y y y 2
dy 6
2
2
2
2
dx
1
1
Now, 1 1 y 2 y 2 1 y 4 y 4 2
4
4
dy
2
1
1 1
y 4 y 4 y 2 y 2
4
2 4
Therefore, the length of the given curve is
L
d
c
2
2
3 1 2
dx
1 dy
y y 2 dy
2 4
dy
1 3 2
y y 2 dy
2 2
3
23
13
1 y3
1 33
1
1
1
y 3 2
3
4
2 3
2 2 3
=
P2. Find the length of the curve
⁄
Solution: Given curve is =
−
Differentiating w.r.t , we get
⁄
⁄
=
from
, 1≤
dx 1 3 1 2 1 1 2 1 1 2
. y y
y y 1 2
dy 3 2
2
2
2
⁄
−
to
≤9
2
dx
1 12
1
1 2
Now, 1 1 y y
1 y y 1 2
4
4
dy
2
1
1 1
y y 1 y1 2 y 1 2
4
2 4
= .
Therefore, the length of the given curve is
L
2
2
9 1 12
dx
1 dy
y y 1 2 dy
1 4
dy
d
c
9
9
9
1
1 2
1
y1 2 y 1 2 dy y 3 2 2 y1 2 y3 2 y1 2
2
2 3
1 3
1
1
1 3 2
1 2 1 3 2
1 2 32
9 9 1 1
3
3
3
IP3. Find the length of the curve y
Solution: Given curve is y
x
0
x
0
≤
sin 2t dt ,
sin 2t dt 0 ≤
≤ .
≤
By the Fundamental Theorem of calculus part 1, we have
dy
sin 2 x
dx
2
dy
1 1
dx
sin 2 x
2
1 sin 2 x
sin 2 x cos 2 x 2sin x.cos x
sin x cos x 2
Therefore, the length of the curve is
2
4
dy
L 1 dx
a
0
dx
b
4
0
sin x cos x 2
dx
4
(sin x cos x) dx cos x sin x 0
cos sin cos0 sin 0
4
4
1
1
1 0 1
2
2
P3. Find the length of the curve y
Solution: Given curve is y
x
2
x
2
3t 4 1 dt , − ≤
3t 4 1 dt , −2 ≤
≤− .
≤ −1
By the Fundamental theorem of calculus part 1, we have
Now,
dy
3x4 1
dx
2
2
dy
4
1 1 3x 1 1 3 x4 1 3x 4
dx
Therefore, the length of the curve is
2
1
dy
L 1 dx
3 x4 dx
a
2
dx
b
1 2
3
x dx
2
1
13 2 3 7 3
x3
3 3
3
3
3
3
2
IP4.
a.
Find a curve through the point ( , ) whose length integral is
L
2
1
1
1
y4
dy
b.
How many such curves are there? Give reasons for your answer.
Solution:
a)
Given the length integral is L
Comparing (1) with L
d
c
and integrating, we get
⟹ = − + , where
2
1
1
1
y
4
dy ..... 1
2
2
dx
dx
1
dx
1
1 dy , we get
dy y 2
y4
dy
dy
arbitrary constant
Since the curve passes through the point (0, 1), we get
Required curve = − + 1 ⟹ =
b)
=
If we take
=1
then there is only one curve because we know the derivative
of the function and the value of the function at one value of
.
P4. Find the arc length of the semi-cubical parabola
=
between the points
( , ) and ( , )
Solution: We have to compute the arc length of the given semi-cubical parabola
=
between the points (1, 1) and (4, 8).
=
⟹ = ⁄ ……………. (1)
Differentiating w.r.t , we get
2
=3
dy 3 x 2
3x 2
dx 2 y 2 x3 2
dy 3
x
dx 2
from (1)
We have
ds
dx
2
dy
2
2
dy
1 dx
dx
2
9
3
ds 1
x dx 1 x dx
4
2
Limits of integration: From the given points (1, 1) and (4, 8), we have
=4
Therefore, the length of the curve is
= 1 and
9x
dx
a
1
4
9x
4
Put 1
u dx du
4
9
13
Limits : x 1 u
and x 4 u 10
4
10
4 4 10
u du u1 2 du
13 4
9 9 13 4
b
L ds
4
1
10
4 u1 21
9 1 2 1
13 4
10
4 2
. u 3 2
13 4
9 3
8
3 2 13
10
27
4
3 2
1
80 10 13 13
27
EXERCISES:
1.
a.
b.
c.
d.
Find the lengths of the curves in problems 1-5.
y 1/ 3 x 2
f.
x 0 to x 3
y 1 to y 3
y 1 to y 2
y 3 / 4 x 4/3 3 / 8 x 2/3 5 from
y
e.
3/2
from
x y 3 / 3 1 / 4 y from
x y 4 / 4 1 / 8 y 2 from
2
0
x
y
0
x
sec 4 t 1 dt
cos 2t dt
1 x 8
/ 4 y / 4
,
,
= 0 to
= ⁄4
2.
a) Find a curve through the point (1, 1) whose length integral is
4
L
1
1
1
dx
4x
b) How many such curves are there? Give reasons for your answer.
3.6. Areas of surfaces of Revolution
Learning objectives:
To find the area of the surface swept out by revolving the graph of a nonnegative function between given limits about axes.
AND
To practice the related problems.
Suppose we want to find the area of the surface swept out by revolving the graph of a
nonnegative function y f x , a x b, about the x -axis. We partition a , b in
the usual way and use the points in the partition to partition the graph into short arcs.
As the arc PQ revolves about the x -axis, the line segment joining P and Q sweeps out
part of a cone whose axis lies along the x -axis.
A piece of a cone like this is called a frustum of the cone. The surface area of the
frustum approximates the surface area of the band swept out by the arc PQ.
The surface area of the frustum of a cone is 2π times the average of the base radii
times the slant height.
Frustum surface area = 2
r1 r2
L r1 r2 L
2
For the frustum swept out by the segment PQ, this works out to be
Frustum surface area f xk 1 f xk
xk 2 yk 2
The area of the original surface is approximated by the frustum area sum
n
f xk 1 f xk
xk 2 yk 2
...... (1)
k 1
We expect the approximation to improve as the partition of a , b becomes finer. We
will show that the sum in (1) approach a calculable limit as the norm of the partition
goes to zero.
If is smooth, by the Mean Value Theorem there is a point ck , f ck on the curve
between P and Q where the tangent is parallel to the segment PQ.
f ck
At this point,
With this substitution for
yk
xk
yk f ck xk
yk , the sums in (1) take the form
n
2
2
f xk 1 f xk xk f ck xk
k 1
n
2
f xk 1 f xk 1 f ck xk ....... (2)
k 1
The sums in (2) are not the Riemann sums of any function because the points
xk 1,
xk , and ck are not the same and there is no way to make them the same. However,
a theorem called Bliss’s theorem, from advanced calculus, assures us that as the norm
a, b goes to zero, the sum in equation (2) converges to
b
2
2
f
x
1
f
x
dx .
a
of the partition of
We therefore define this integral to be the area of the surface swept out by the graph
of f from a to b .
If the function f x 0 is smooth on a, b , the area of the surface
generated by revolving the curve y f x about the
x -axis is
2
dy
S 2 y 1 dx
a
dx
b
b
2
S 2 f x 1 f x dx ..... (3)
a
Example 1:
Find the area of the surface generated by revolving the curve y 2 x , 1 x 2,
about the x -axis.
Solution:
2
2
1
x 1
x 1
dy
1 1
x
x
dx
x
2
2
x 1
S 2 2 x
dx 4 x 1 dx
1
1
x
2
2
8
3/2
4 x 1
3 32 2
3
3
1
y -axis, we interchange x and y in equation (3).
If x g y 0 is smooth on c, d , the area of the surface generated by revolving
the curve x g y about the y -axis is
For revolution about the
S
d
c
2
d
dx
2
2 x 1 dy 2 g y 1 g y dy .... (4)
c
dy
Example 2:
The line segment x 1 y ,
the cone in the figure below.
0 y 1, is revolved about the y -axis to generate
Find its lateral surface area.
Solution:
From geometry, we know
Lateral surface area =
base circumference
slant height 2
2
We can use equation (4) to obtain the same result.
2
dx
2
1 1 1 2
dy
S
d
c
2
1
dx
2 x 1 dy 2 1 y 2dy
0
dy
1
y2
1
2 2 y 2 2 1 2
2
2
0
The equations
2
dy
S 2 y 1 dx and
a
dx
b
S
d
c
2
dx
2 x 1 dy are often written in
dy
terms of the arc length differential ds dx2 dy 2 as
b
S 2 y ds
a
and
d
S 2 x ds
c
y is the distance from the x -axis to an element of arc length
ds . In the second, x is the distance from the y -axis to an element of arc length
ds . Both integrals have the form
S 2 radius band width 2 ds where is the radius
from the axis of revolution to an element of arc length ds .
S 2 ds
In the first of these,
In any particular problem, we will then express the radius function
and the arc
length differential ds in terms of a common variable and supply limits of integration
for that variable.
Example 3:
Find the area of the surface generated by revolving the curve
y x 3 ,0 x 1/ 2, about the x -axis.
Solution:
S 2 ds
2 y ds
For revolution about the x -axis
,
2 y dx 2 dy 2
We then decide whether to express
y x3 ,
2
dy 3x 2 dx,
2
2
2
dx dy dx 3 x dx
With these substitutions,
S
x 1/2
x 0
dy in terms of dx or dx in terms of dy .
2
x
2
1/2
0
3/2
1 9 x 4 dx
becomes the variable of integration and
2 y dx dy
9
1
27 16
2
61
1
1728
2 x3 1 9 x 4 dx
PROBLEM SET
IP1. Find the area of the surface generated by revolving the curve
= √ + , ≤ ≤ about the −axis.
Solution: Given curve = √ + 1 , 1 ≤ ≤ 5
Differentiating w.r.t , we get
2
dy
1
1
4x 4 1 4x 5
dy
and 1 1
dx 2 x 1
4 x 1 4 x 1 4 x 1
dx
Therefore, the area of the surface generated by revolving the curve
= √ + 1 , 1 ≤ ≤ 5 about the −axis is
2
5
4x 5
dy
S 2 y 1 dx 2 y
dx
a
1
4 x 1
dx
b
5
5
4x 5
x 1.
dx
4 x 5 dx
1
1
2 x 1
2
Put 4 + 5 = ⟹
=
Limits: = 1 ⟹ = 9 and
=5⟹
= 25
25
u1 21
du 25 1 2
u u du
4 1 2 1
4 4 9
9
25
9
2 3 2 25
49
. u
125 27
4 3
9
6
3
P1.
The curve = √ −
, − ≤ ≤ is an arc of the circle
+
= . Find the
area of the surface obtained by rotating this arc about the −axis.
Solution: Given curve = √4 −
, −1 ≤ ≤ 1 Differentiating w.r.t , we get
dy 1
x
2 1 2
4 x
and
2 x
2
dx 2
4 x
2
x2
4
dy
1 1
dx
4 x2 4 x2
Therefore, the surface area of the arc about the
−axis is
2
1
4
dy
S 2 y 1 dx 2 y
dx
a
1
dx
4 x2
1
2
2
4 x2 .
dx
1
2
4x
b
1
1
2 2 dx 2 2 x 1 2 2(1) 2(1) 8
1
IP2. The arc of the parabola =
from ( , ) to ( , ) is rotated about the
−axis. Find the area of the resulting surface.
Solution: Given curve =
⟹ =
,1 ≤ ≤ 4
Differentiating w.r.t , we get
2
2
1
dx
dx
1
1
and 1+ 1
1
2 y
dy 2 y
4y
dy
Therefore, the area of the surface generated by revolving the curve
=
, 1 ≤ ≤ 4 about −axis is
2
4
4
dx
1
S 2 x 1 dy 2 y 1
dy 4 y 1 dy
c
1
1
4y
dy
d
du
4
Limits: y 1 u 5 and y 5 u 17
Put 4 y 1 u dy
17
S
17
5
u1 2 1
du 17 1 2
u u du
4 1 2 1
4 4 5
5
3 2 17
u
17 17 5 5
5
6
6
P2. Find the area of the surface generated by revolving the curve
=
− , ≤ ≤ about −axis.
Solution: Given curve = 2 − 1 ,
Differentiating w.r.t , we get
≤
≤1
2
2
dx
dx
1
1
1
1
2y
and 1+ 1
2
1
dy 2 2 y 1
2 y 1 2 y 1
2 y 1
dy
2y 1
Therefore, the area of the surface generated by revolving the curve
= 2 − 1 , ≤ ≤ 1 about −axis is
2
1
dx
2y
S 2 x 1 dy 2 2 y 1
dy
c
5
8
dy
2
y
1
1
1
2
1
1 12
y
2 2
y dy 2 2
58
1
2
1
5 8
d
2 3 2 1
4 2
2 2 . y
5 8
3
3
16 2 5 5
12
5 5
1
8
8
IP3.
Find the surface area of the cone frustum generated by revolving the line segment
= + ,
≤ ≤ about the −axis. Check the answer with its geometry
formula.
Solution: Given = + , 1 ≤ ≤ 3
Since the lateral surface of the cone is generated by revolving about −axis,
= 2 − 1, 1 ≤ ≤ 2
Differentiating w.r.t , we get
dx
2
dy
2
dx
2
and 1 1 2 5
dy
Therefore, the lateral surface area of the cone generated by revolving the line
segment = + , 1 ≤ ≤ 3 about the −axis is
2
2
dx
S 2 x 1 dy 2 2 y 1 5 dy
c
1
dy
d
2
2
2
2
y
1
dy
2
5
y
y
1
1
2 5
2 5 4 2 1 1 4 5
Verification:
From the geometry, we know that
Frustum surface area = π( + ) ×
ℎ ℎ
By the hypothesis, r = 1 , r = 3
) + (ℎ ℎ )
Slant height = (
= (2 − 1) + (3 − 1) = √5
Therefore, the frustum surface area is
= (1 + 3) × √5 = 4 √5
Hence verified
P3.
Find the lateral surface area of the cone generated by revolving the line segment
= ,
≤ ≤ about the −axis. Check the answer with its geometry
formula.
Solution: Given = , 0 ≤ ≤ 4
Since the lateral surface of the cone is generated by revolving about −axis,
=2 , 0≤ ≤2
Differentiating w.r.t , we get
2
dx
dx
2
2
and 1 1 2 5
dy
dy
Therefore, the lateral surface area of the cone generated by revolving the line
segment = , 0 ≤ ≤ 4 about the −axis is
d
dx
S 2 x 1
c
dy
2
2
0
dy 2 2 y 5 dy
2
y2
4 5 y dy 4 5 2 5 4 0 8 5
0
2 0
2
Verification:
From the geometry, we know that
1
× base circumference × slant height
2
Base circumference = 2 × radius of the base circle
= 2 (4) = 8
Slant height = (radius) + (height)
= (4) + (2) = 2√5
Therefore, the lateral surface area is = × 8π × 2√5 = 8π√5
Hence verified
IP4. Find the area of the surface generated by revolving the curve
Lateral surface area =
32
1 2
x 2
, 0 x 2 about the −axis.
3
32
1 2
, 0x 2
Solution: Given curve is y x 2
3
y
Differentiating
w.r.t
, we get
12
dy 1 3 2
. x 2
2x x x2 2
dx 3 2
ds
dx
2
dy
2
2
dy
1 dx
dx
2
1 x x2 2 dx 1 x 2 x2 2 dx
x4 2 x2 1 dx
2
x2 1
dx x2 1 dx
Therefore, the area of the surface generated by revolving the curve
y
32
1 2
x 2
, 0 x 2 about the −axis is
3
b
S 2 x ds
a
0
2
2
2 x x 2 1 dx 2
2
x 4 x2
2 2
2
4
0
0
4
2
2
4
2
2
x3 x dx
0 4
P4. Find the area of the surface generated by revolving the curve
y4
1
x
2 , 1 y 2 about the −axis.
4 8y
y4
1
Solution: Given curve is x
2 , 1 y 2
4 8y
Differentiating w.r.t , we get
dx
1
y3
dy
4 y3
ds
dx
2
dy
2
2
dx
1 dy
dy
2
1
1
1
1 y 3 3 dy 1 y 6
dy
6 2
4
y
16
y
1
1
y6
dy
6 2
16
y
2
3
3
1
1
y 3 dy y 3 dy
4y
4y
Therefore, the area of the surface generated by revolving the curve
y4
1
x
2 , 1 y 2 about the −axis is
4 8y
d
2
1
S 2 y ds 2 y y3 3 dy
c
1
4 y
2 4
1
y
1
4 y2
2
2
y5 1
dy 2
5 4 y 1
25
1 1 1 253
2
20
5 4(2) 5 4
EXERCISES:
1. Find the lateral surface area of the cone generated by revolving the line segment
y x / 2,0 x 4, about the x -axis.
2. Find the surface area of the cone frustum generated by revolving the line segment
y ( x / 2) 1/ 2 , 1 x 3, about the x -axis.
3.
Find the areas of the surfaces generated by revolving the curves in problems 3about the indicated axes.
3
y
x
/ 9,
0 x 2;
x-axis
a)
2
y
2
x
x
,
0.5 x 1.5; x -axis
b)
c)
=√
, 3⁄4 ≤ ≤ 15⁄4 ;
−axis
⁄
⁄
d)
=
−
, 1 ≤ ≤ 3 ; −axis
3
x
y
/ 3,
0 y 1;
y -axis
e)
f)
x 2 4 y , 0 y 15 / 4;
y -axis
4.1. Derivatives of Inverse Functions
Learning objectives:
To study the derivative rule for inverses.
AND
To practice the related problems.
If we calculate the derivatives of ( ) =
we see that
( ) = 2 − 2,
+ 1 and its inverse
d
d 1
d
1 d 1
f x x 1 ;
f x 2 x 2 2
dx
dx 2
dx
2 dx
The derivatives are reciprocals of one another. The graph of
f
is the line
=
+ 1, and the graph of
is the line = 2 − 2. Their slopes are
reciprocals of one another.
This is not a special case. Reflecting any non horizontal or non vertical line across the
line = always inverts the line’s slope. If the original line has slope ≠ 0, the
reflected line has slope 1⁄ .
The reciprocal relation between the slopes of graphs of inverses holds for other
functions as well.
If the slope of = ( ) at the point a , f a is f a 0 , then the slope of
y f 1( x) at the corresponding point f a , a is 1/ f a .
1
Thus, the derivative of f
at f a equals the reciprocal of the derivative of f
at a .
If = ( ) has a horizontal tangent line at ( , ( )) then the inverse function
has a vertical tangent line at ( ( ), ), and this infinite slope implies that
is not
differentiable at ( ). The following theorem gives the conditions under which
is differentiable in its domain which is the same as the range of .
Theorem 1
The Derivative Rule for Inverses
df
is never zero on , then
dx
df 1
1
f is differentiable at every point of the interval ( ). The value of
at any
dx
df
particular point f a is the reciprocal of the value of
at a .
dx
df 1
1
......... (1)
dx
df
x f a
dx x a
1
1
In short notation,
f
.......... (2)
f
If
is differentiable at every point of an interval
and
Example
For
f x x2 , x 0 , and its inverse f 1 x x ,
df
d 2
x 2x
We have
dx dx
df 1 d
dx
dx
x 2 1x ,
x0
The point (4, 2) is the mirror image of the point (2, 4) across the line
df
2 x 2(2) 4
dx
df 1
1
1
1 1
At the point (4, 2):
dx
2 x 2 4 4 df
dx
At the point (2, 4):
= .
df 1
Equation (1) sometimes enables us to find specific values of
without knowing
dx
a formula for
Example
f 1.
df 1
3
Let f x x 2 . Find the value of
at x 6 f 2 without finding a
dx
formula for f 1 x .
Solution
df
3 x 2
12
x2
dx x 2
df 1
1
1
dx
df
12
x f (2)
dx x 2
There is another way to look at theorem 1.
If y f ( x ) is differentiable at x a and we change
the corresponding change in y is approximately
x by a small amount dx ,
dy f a dx
This means that
about 1/
f a times as fast as
changes about
f a times as fast as
and that
changes
.
PROBLEM SET
IP1. If ( ) =
,
a. Find
( )
b. Evaluate
at
≥
;
=
and
=
then
at
= ( ) to show that at these points
=
Solution:
Given ( ) = 2 , ≥ 0
Clearly, is one-to-one and so its inverse exists.
a. Now, solve for in terms of
=2
⟹
=
Interchange and
⟹
:
=
=
The inverse of the function ( ) = 2
is the function
=
( )=
= 4 and
b. Now,
( )=2
⟹ (5) = 2(5) = 50
=
=
( )
∴
P1. If ( ) = +
c. Find
( )
;
at
;
√
=
( )
=
( )
d. Evaluate
= 4(5) = 20
=
(
)
=
=
=−
=
=
then
and
at
= ( ) to show that at these points
=
Solution:
Given ( ) = + 7
Clearly, is one-to-one and so its inverse exists.
c. Now, solve for in terms of
= + 7 ⟹ − 7 = ⟹ = 5 − 35
Interchange and :
= 5 − 35
The inverse of the function ( ) = + 7 is the function
( ) = 5 − 35
=
=
d. Now,
=
and
=
( ) = + 7 ⟹ (−1) = − + 7 =
=5 ;
∴
( )
(
)
=
=5=
(
)
=5
IP2. Let ( ) =
Solution:
Given ( ) =
−
− ,
− 4 − 5,
We have
Now,
> 2 then find
( )
at the point
=
= ( )
>2
=
=2 −4
= 2(5) − 4 = 6
∴
( )
P2. Let ( ) =
−
= − = ( ) is
Solution:
Given ( ) =
−3
We have
Now,
=
− ,
≥
− 1,
≥2
( )
=3
=
then the value of
at the point
=
−6 ,
≥2
= 3(3) − 6(3) = 9
∴
( )
=
=
IP3. Suppose that the differentiable function = ( ) has an inverse and that the
graph of g passes through the origin with slope 2. Then find the slope of the graph
of
at the origin
Solution: Given that = ( ) is a differentiable function and has an inverse.
Since the graph of passes through origin (0,0) and has a slope 2 there,
=2
∴ The slope of the graph of
∴
=
at the origin is
( )
=
=
P3. Suppose that the differentiable function = ( ) has an inverse and that the
graph of passes through the point ( , ) and has a slope of / there. Find the
value of
at = .
Solution:
Given that = ( ) is a differentiable function and has an inverse.
Since the graph of passes through (2, 4) and has a slope 1/3 there,
=
∴
=
( )
=
=
=3
IP4. Show that ( ) = and ( ) = √
are inverses of one another.
a. Find the slopes of the tangents to the graphs of and at ( , ) and
(− , − ) (four tangents in all).
b. What lines are tangents to the curves at the origin?
Solution:
a. Given
( )=
and
⁄
( ) = (4 )
f g ( x) f g ( x)
f 3 4x
3 4x 3
x
4
13
x3 x3
g f ( x) g f ( x) g 4
4 4
f g ( x) g f ( x) x
x
⟹ ( ) and ( ) are inverses of one another.
Note:
The points of intersection of
= √4 ⟹
=
= √4 are given by
and
= 256
⟹
− 256 = 0 ⟹ ( − 256) = 0
⟹ ( − 16 )( + 16 ) = 0
⟹ ( − 2)( + 2)( + 16) = 0 ⟹ = 0, ±2
The curves intersect at (0, 0)(2, 2)(−2, −2)
b. The slopes of the tangents to the graphs of and at (2, 2) and (−2, −2) are:
Given ( ) =
Differentiating
ʹ(
w.r.t , we get
)=
=
( )
= 3;
We have ( ) = √4 and
Now,
=
=
=
(
)
=3
=
( )
(
=
)
=
=
and
=
c. We have
=
⟹
⟹
=0
= ( ) has a horizontal tangent line at (0, 0)
⟹ = 0 is a tangent to the curve = at the origin.
Since =
, has a vertical tangent line at (0, 0)
⟹ = 0 is a tangent to the curve = √4 at the origin.
P4. Show that ( ) =
and ( ) = √ are inverses of one another.
c. Find the slopes of the tangents to the graphs of and at ( , ) and
(− , − ) (four tangents in all).
d. What lines are tangents to the curves at the origin?
Solution:
d. Given ( ) =
and ( ) = √
3
3
3
f g ( x) f g ( x) f x x x
g f ( x) g f ( x) g
13
x3 x3
x
f g ( x) g f ( x) x
⟹ ( ) and ( ) are inverses of one another.
Note:
The points of intersection of =
and
= √ are given by
=√ ⟹
= ⟹
− = 0 ⟹ ( − 1) = 0
⟹ ( − 1)( + 1) = 0
⟹ ( − 1)( + 1)( + 1) = 0 ⟹ = 0, ±1
The curves intersect at (0, 0)(1, 1)(−1, −1)
e. The slopes of the tangents to the graphs of and at (1, 1) and (−1, −1) are:
Given ( ) =
Differentiating w.r.t , we get
ʹ(
)=3
= 3(1) = 3 ;
We have ( ) = √
Now,
=
=
and
= 3(−1) = 3
=
( )
(
=
)
=
=
and
=
f. We have
⟹
=3
=0
⟹ = ( ) has a horizontal tangent line at (0, 0)
⟹ = 0 is a tangent to the curve =
at the origin.
Since =
, has a vertical tangent line at (0, 0)
⟹
= 0 is a tangent to the curve
= √ at the origin.
EXERCISES
1. The formula for ( ) is given below:
1
(a) Find f
x
df
df 1
(b) Evaluate
at x a and
dx
dx
df 1 1
points
.
df
dx
dx
a. f x 2 x 3, a 1
b. f x 5 4 x, a 1/ 2
at
x f a
to show that at these
4.2. Natural Logarithms
Learning objectives:
To define the Natural Logarithm Function.
To practice the related problems.
To derive the derivative of
.
To prove the properties of Natural Logarithms.
To study the graph and range of the
And
.
The function-inverse pair consisting of the natural logarithm function
and the
exponential function
is an important pair.
The Natural Logarithm Function
The natural logarithm of a positive number , written as
, is the value of an
integral.
=∫
>0
If
> 1, then ln is the area under the curve
= from = 1 to = .
For 0 < < 1,
gives the negative of the area under the curve from to 1. The
function is not defined for ≤ 0. We also have
1=∫
=0
(∵upper and lower limits are equal)
We notice that we show the graph of = in the figure but use = in the
integral. Using for everything would have us writing
=∫
with meaning two different things. So we change the variable of integration to .
The Derivative of
:
By the first part of the Fundamental Theorem of Calculus,
= ∫
=
For every positive value of , therefore,
=
If is a differentiable function of whose values are positive, so that
defined, then applying the Chain Rule =
to the function =
=
=
.
,
>0
=
------------ (1)
is
gives
Example1:
(2 ) =
2 =
The function = 2 has the same derivative as the function
for =
for any number .
( )= ( )=
=
-------(2)
Example 2
Equation (1) with =
+ 3 gives
( + 3) =
. ( + 3) =
.2 =
=
. This is true
Properties of Logarithms
The properties of logarithms are listed below.
For any numbers > 0 and > 0,
Product Rule:
Reciprocal Rule:
=
=−
Quotient Rule:
Power Rule:
=
=
+
−
The properties made it possible to replace multiplication of positive numbers by
addition and division of positive numbers by subtraction. They also made it possible
to replace exponentiation by multiplication.
We prove the properties as follows.
(i)
=
+
We noted that
and
have the same derivative. By Corollary 1 of the Mean
Value Theorem, the functions must differ by a constant,
∴
=
+
-------(3)
for some . This equation holds for all positive values of , so it must hold for = 1.
Hence,
( . 1) = 1 + ⟹
=0+
⟹
= ln
Substituting =
in equation (3) we get
=
+
--------(4)
(ii)
=−
In Equation (4) replace
+
so that
=−
by gives
=
.
=
1=0
giving the Reciprocal Rule
(iii)
=
−
Equation (4) with
replaced by gives
=
=
.
−
=
+
Product
Reciprocal
(iv)
=
We assume rational
(
=
)=
)
= . = (
Since
and
have the same derivative,
=
+
for some constant . Taking = 1 we get = 0.
This completes proof.
The rule holds for all , rational and irrational.
Example 3
a) 6 = (2.3) = 2 + 3
Product
b) 4 − 5 =
= 0.8
Quotient
=− 8
= − 2 = −3 2
Example 4
a)
4 + sin = (4 sin )
b)
= ( + 1) − (2 − 3)
c)
c)
sec =
d)
√ +1=
=−
Reciprocal
Power
Product
Quotient
cos
( + 1) =
Reciprocal
( + 1) Power
The Graph and Range of
The derivative
(
)
= is positive for
> 0. Thus
is an increasing function of
, hence it is one-to-one and invertible. The second derivative, − , is negative, so
the graph of
is concave down.
We can estimate ln 2 by numerical integration to be about 0.69. We therefore know
that
2 =
2>
=
and
2
It follows that
lim
→∞
=−
=∞
2<−
=−
and
lim
→
= −∞
The domain of ln is the set of positive real numbers; the range is the entire real
line.
PROBLEM SET
=?
IP1: Simplify:
Solution:
( × )
=
=
=
=
P1:
+
Solution:
7
+5
= 7
= 7[
×
2 −
= 7 [4
)
+
=?
+5
+3
×
(3 × 5)] + 5[
2−
×
(3 × 2 )] + 3[
5 −
5 ] + 5 [2
3−
5−
,
≠
then find ʹ at
=− .
Solution:
Given
=
(
)
√
,
≠1
+ 1) ] −
1−4
1
(1 − 4 )
⟹
= 2 ( + 1) −
2
Differentiating on both sides w.r.t , we get
( + 1) − .
(1 − 4
=2
⟹
=
=
Now,
ʹ|
[(
(2 ) −
(
)
= [(
)
=
+
]
(
. 2(−1) −
( )
)
(−12
[
(
)
)
) ]
(−12 (−1) )
(12) = −2 + =
P2: Find the derivative of ( ) =
Solution:
Given ( ) =
+√ −1
+√
3−
5 + [−7 − 5 + 12 ]
2
=
(5 × 2 )]
3 −
2 ] + 3 [4
3−3
2 + [−7 + 10 − 3]
2+0+0 =
IP2: If
=2
+3
= [28 − 15 − 12 ]
= 1.
(
−
.
5−4
3
2]
Differentiating on both sides w.r.t , we get
ʹ(
)=
+√
√
=
. 1+
√
=
1+
√
−1
(
√
− 1)
(2 )
√
√
=
√
=
√
(
) , then find .
IP3: If =
Solution:
(
)
Given
= ln
Differentiating on both sides w.r.t , we get
=
=
=
(
(
)
)
(
(
)
(
∙
(
)
)
)
(
∙
)
(
)+
(
)], then find
P3: If = [
Solution:
) + cos(
)]
Given
= [sin(
Differentiating on both sides w.r.t , we get
).
(
√ , then find
.
.
) + −sin(
)
(
)
) + cos (
+ [sin(
)]. (1)
) − sin(
) + cos(
=
cos(
) + sin(
)
) − sin(
) + sin(
) + cos(
)
= cos(
= 2cos(
)
= . cos(
IP4: If =
Solution:
√ = ln
Differentiating on both sides w.r.t , we get
Given
=
=
=
∙
∙
=
∙
=
=
P4: If =
Solution:
Given
√
∙
√
, then find
.
√
=
(sin cos ) −
=
= ( (sin ) +
(1 + 2 ln )
(cos )) −
(1 + 2 ln )
Differentiating on both sides w.r.t , we get
=
∙ cos −
= (cot
=
∙ sin
−
(1 + 2
∙
− tan ) −
cot − tan −
(
)
Exercises:
2 and
3.
2. Express the following logarithms in terms of 5 and
(a)
(b) 9.8
(c) 7√7
7
1. Express the following logarithms in terms of
(a)
0.75
(b)
(c)
√9
(d)
1225
(d)
(e)
(e)
3√2
(f)
0.056
√13.5
(f)
3. Use the properties of logarithms to simplify the expressions.
a.
sin −
b.
(3
−9 )+
(4 ) − 2
c.
d.
sec + cos
e. (8 + 4) − 2 2
f. 3 √ − 1 − ( + 1)
)
with respect to , , or , as appropriate.
4. Find the derivative of
a.
b.
c.
=
=
=
3
d.
=
e.
=
f.
g.
h.
i.
j.
k.
l.
=
= ( + 1)
= (2 + 2)
=
)
=(
= ( )
= √
( )
m. =
−
n.
=
−
o.
=
p.
=
q.
=
r.
s.
t.
u.
=
=
=
=
v.
=
w.
x.
=
=
y.
=
z.
= ln
(
)
( (
))
√
(
(
))
√
(
)
(
)
4.3. Logarithmic Differentiation and ∫
Learning Objectives:
To learn logarithmic differentiation
To evaluate ∫ , where is a nonzero differentiable function
AND
To practice the related problems
Logarithmic Differentiation
The derivatives of positive functions given by formulas that involve products,
quotients, and powers can often be found more quickly if we take the natural
logarithm on both sides before differentiating. This enables us to use the properties
of logarithms to simplify the formulas before differentiating. This process is called
logarithmic differentiation.
Example 1
(
)(
)
(
)(
Find if =
, > 1.
Solution: We take the natural logarithm on both sides and simplify using the
properties of logarithms.
=
(
=
)
+ 1)( + 3) −
( − 1)
= ( + 1) + ( + 3) − ( − 1)
( + 3) − ( − 1)
= ( + 1) +
We then take derivatives of both sides with respect to :
=
∙2 + ∙
−
Next we solve for
:
=
+
Finally, we substitute for :
=
(
)(
−
)
+
−
The Integral ∫
(
)=
We have,
, > 0 ------------ (1)
Equation (1) leads to the integral formula
=
+ ------------- (2)
∫
where is a positive differentiable function. If is negative, then − is positive and
= ∫ ( ) (− ) = (− ) + ------ (3)
∫
(equation (2) with
replaced by − )
We can combine equations (2) and (3) into a single formula by noticing that in each
case the expression on the right is | | + . In equation (2),
= | | because
> 0; in equation (3), (− ) = | | because < 0. Whether is positive or
negative, the integral of ∫
is | | + .
If
is a nonzero differentiable function,
= | | + --------------- (4)
∫
We know that
=
+ , ≠ −1
+1
Equation (4) handles the case when equals −1.
Equation (4) says that integrals of a certain form lead to logarithms. That is,
ʹ(
)
= | ( )| +
∫ ( )
whenever ( ) is a differentiable function that maintains a constant sign on the
domain given for it.
Example 2
2
=
−5 ⇒
=2
=
(0) = −5 , (2) = −1
−5
= | |] = |−1| − |−5| = 1 − 5 = − 5
Example 3
∫
=∫
=2
= 3 + 2 sin ⇒
− =1,
| |] = 2 |5| − 2
= 2 cos
=5
|1 | = 2 |5 |
The Integrals of
and
Equation (4) helps us in the evaluation of integrals of the tangent and cotangent
functions.
For the tangent,
=∫
=∫
∫ tan
( = cos ⇒
)
= − sin
=− | |+
|+
=− |
= |
+
(Reciprocal rule)
|
=
For the cotangent,
=∫
∫ cot
|
|+
=∫
( = sin ⇒
= | |+
|+
= |
= cos
)
=−
|
+
|
(Reciprocal rule)
|+
=− |
|+ = |
=− |
|+ = − |
= |
∫
∫
|+
|+
Example 4
∫ tan 2
=2 ⇒
(0 ) = 0 ,
= ∫ tan
= ∫ tan
|
=
|] = (
2−
=2
=
1) =
2
PROBLEM SET
=(
IP1: Find the derivative of
Solution:
)
with respect to
.
)
Given = (
Taking logarithm on both sides, we get
( (
))
=
, we get
Differentiating on both sides w.r.t
⟹
( (
=
⟹
=(
(
)
)
(
=
=
(
(
⟹
P1: If
))
)
)
(
)
(
(
) (
)
(
) (
)
)
)
, then find
.
Solution:
Given,
=
(
) (
)
(
) (
)
Taking logarithm on both sides, we get
=
) (
)
(
) (
)
(1 − 2 ) (1 + 3 )
=
⟹
(
=
2
3
(1 − 2 ) + −
3
4
−
(1 − 6 ) (1 + 7 )
(1 + 3 ) −
5
6
(1 − 6 ) +
Differentiating on both sides w.r.t , we get
(−2) −
(3) −
(−6) +
=
6
7
(7 )
(1 + 7 )
=
=
(
)
−
(
(
) (
)
(
) (
)
)
+
+
+
−
(
)
−
(
)
IP2: If
=( + )
, then find
.
Solution:
Given,
=( + )
Taking logarithm on both sides, we get
(
)= ( + )
⟹
+
=( + ) ( + )
Differentiating on both sides w.r.t , we get
⟹
+ .
=
.
⟹
+ .
=
+
⟹
−
=
⟹
(
.
−
=
)
⟹
( + )
(
)
=
⟹
=
P2: If
= ( + ) and
= , then find the value of
Solution:
Given,
=( + )
Taking logarithm on both sides, we get
( + )
+
=
Differentiating on both sides w.r.t , we get
+
= .
⟹ −
⟹
(
=
)
=
1+
−
(
)
.
Given,
⟹ + −
=
− −
⟹ 2( + ) = ( + )
⟹ =2
IP3: Evaluate ∫
Solution:
Given, ∫
.
=
(
)=2
(−
= − (2
=( − ) 2
+
∴∫
(
)+2
) + (2
)
)
=
∫
=
∫
=
[ |
=
[ | |−
(
)
(
)
|]
+
| |]
=
P3: Evaluate ∫
.
Solution:
= 1 − sin ⇒
= − cos
= − ⇒ = 2, = ⇒ =
= −∫
∫
= −[ | |] = −
IP4: Evaluate ∫
Solution:
Given, ∫ (
Put 1 +
∴ ∫(
(
)
−
|2 | =
.
)
= ⇒
)
=
=∫
=
+
=(
)
+
P4: Evaluate ∫
Solution:
Given, ∫
(sec
+ tan
) = 2 sec
= 4 sec
∴∫
tan + 2 tan sec
tan
= ∫
= ∫
=
(
|sec
)
+ tan
|+
4
Exercises:
I.
Use logarithmic differentiation to find the derivative of
given independent variable.
( + 1)
5. =
6. = ( + 1)( − 1)
7.
=
8.
=
9.
10.
11.
12.
= √ + 3 sin
= tan √2 + 1
= ( + 1)( + 2)
= ( )( )
(
)
13. =
14. =
15. =
√
√
(
)
(
(
16. =
17. =
18. =
II.
)
)
(
)
(
)(
)(
(
)
)
Evaluate the integrals.
a. ∫
;
b. ∫
c. ∫
d. ∫
e. ∫
f. ∫
g. ∫
h. ∫
i. ∫
j. ∫
k. ∫
(
)
√
with respect to the
l. ∫
m. ∫ tan
n. ∫ cot
o. ∫ 2 cot
p. ∫ 6 tan 3
q. ∫
r. ∫
√
(
)
4.4. Exponential Functions
PROBLEM SET
EXERCISE
4.5. General Exponential Functions
Learning objectives:
To define the general exponential functions.
To derive the power rule of differentiation for any real number.
To compute the derivative and integration of the general exponential
functions.
AND
To practice the related problems.
x
1
x
The definition e ln x defines e for every real value of x , irrational as well
as rational. This enables us to raise any other positive number to an arbitrary power
x
and thus to define an exponential function y a for any positive number a . We
tan x
x
will also define functions like x and sin x
that involve raising the values of
one function to powers given by another.
x
The General Exponential Function a
ln a
x
Since a e
for any positive number a , we can think of a as
ln a x
e
e xln a . We therefore make the following definition.
Definition
For any numbers
> 0 and
x , the exponential function with base
…………… (1)
= . =
If = the definition gives
=
=
Example 1
x
x lna
3
3 ln 2
a
e
2 e
a)
2 e ln 2
b)
x
The function a obeys the usual laws of exponents.
For > 0 and any x and y :
x
y
x y
1. a a a
1
ax
2.
a x
3.
ax
x y
a
ay
is
4.
x y
a
a
xy
y x
a
The Power Rule
We can now define
x n for any x 0 and any real number n as x n enln x .
Therefore, the n in the equation ln x
can be any number as long as x 0 .
n
nln x
ln x ln e
n
nln x no longer needs to be rational – it
n ln x ln e n ln x
x n enln x enable us to
n
establish the Power Rule for differentiation. Differentiating x with respect to x
Together, the law
a x / a y a x y
and the definition
gives
d n d nln x
d
n
x e
en ln x n ln x xn nx n1
dx
dx
dx
x
In short, as long as x 0 ,
d n
x nx n1
dx
The Chain Rule extends this equation to the Power Rule.
Power Rule (General form):
If u is a positive differentiable function of
differentiable function of x and
Example 2
x and n is any real number, then u n is a
d n
du
u nun1
dx
dx
d
x 2 2x 2 1 x 0
dx
d
b)
sin x sin x 1 cos x
sin x 0
dx
x
The Derivative of a
x
xln a
We use the definition a e
.
d x d xln a
d
a e
e xln a xln a a x ln a
dx
dx
dx
a)
∴
d
a x a x ln a
dx
With the Chain Rule, we get a more general form.
a 0 and u is a differentiable function of x , then a u
function of x and
If
d u
du
a a u ln a
dx
dx
If
is a differentiable
......... 2
a e , then ln a 1 and equation (2) simplifies to
d x
e e x lne e x
dx
x
This shows why e , is the exponential function preferred in calculus.
Example 3
d x
3 3 x ln 3
dx
d x
d
b)
3 3 x ln 3 x 3 x ln 3
dx
dx
d sin x
d
c)
3
3sin x ln 3 sin x 3sin x ln 3 cos x
dx
dx
x
From equation (2), we see that the derivative of a is positive if lna 0 , or
a)
a 1, and negative if ln a 0 , or 0 a 1. Thus,
a x is an increasing function of x if a 1 and a decreasing function of x if
0 a 1. In each case, a x is one-to-one. The second derivative
d2 x
d x
2 x
a
a
ln
a
ln
a
a
2
dx
dx
x
is positive for all x , so the graph of a is concave up on every interval of the real
line.
Note:
As → ∞, we have
→ 0 if 0 < < 1 and
→ ∞ if > 1
As → −∞, we have
→ ∞ if 0 < < 1 and
→ 0 if > 1
Other Power Functions
The ability to raise positive numbers to arbitrary real powers makes it possible to
x
ln x
define functions like x and x
for x 0 . We find the derivatives of such
functions by rewriting the functions as powers of e .
Example 4
Find
dy
if y x x , x 0 .
dx
Solution
x
Write x as power of
x
e.
y x e xln x
equation (1) with a x
Then differentiate as usual:
dy d xln x
xln x d
dx
dx
e
e
dx
x ln x )
1
x x x ln x x x 1 ln x
x
u
The Integral of a
If a 1, so that lna 0 , we can divide both sides of equation (2) by ln a to
obtain
du
1 d u
au
a
dx
ln a dx
Integrating with respect to
au
du
dx
dx
x then gives
1 d u
1
a dx
ln a dx
ln a
d u
1 u
a dx
a C
dx
ln a
writing the first integral in differential form gives
au
u
a du
C
......... ( 3 )
ln a
Example 5
2x
x
a.
2 dx
C
b.
2
ln 2
sin x cos x dx
2u
2 sin x
u
2 du
C
C
ln 2
ln 2
PROBLEM SET
.
IP1. Solve the equation
Solution:
Given
.
⟹
=
=
.
for
.
=
⟹
=
⟹ (4 − 5 )
= (4 − 3 )
⟹ 4
−4
=5
−3
⟹ [4
−4 ]=5
−3
⟹
=
P1. Solve for :
(
.
)
=
Solution:
.4
Given
=8
⟹ 2 .2 ( ) = 2 (
)
⟹ 2 . 2(
= 2(
⟹2
=2
⟹2
=2
⟹ + 2 = −6 + 3
⟹7 =1⟹
IP2. If
=
)
)
=
then show that
=(
)
Solution:
Given
=
Taking logarithms on both sides, we get
= −
That is, =
Differentiating both sides w.r.t , we get
=
(
).
(
.
)
=(
)
P2. If
+
=
then find
Solution:
Given
+
=
Differentiating both sides w.r.t
+
.
, we get
+
.
=
1+
=
( + )
=[
−
−
(1 −
) =
⟹
=
)(
)(
=(
IP3. If
=[
.
)
.
]
−
[
−1]
(
(
)
)
then find
(
]
−
=
at
)
)
)(
Solution: Given
=(
Taking logarithms on both sides, we get
)
)(
=(
………. (1)
Again taking logarithms on both sides, we get
(
)=
(
)+ (
)
Differentiating both sides w.r.t , we get
.
=
=
.
=
.
+
+1+(
)(
= (
Now, at
+
)
⟹
(
= 1 and
=
=
P3. If
then find
Solution: Given = ( )
Taking logarithms on both sides, we get
=
. ( )
Differentiating both sides w.r.t , we get
(
=
=
(
)+
)
+
.
=(
(
)
)
(
)
2
IP4. Evaluate
x2 x 1 ln x dx
1
2
Solution: To evaluate
1
1=0
= 1.1.2[0 + 1] = 2
∴
.
)
.
[
.
.
x 2 x 1 ln x dx
+ 1) +
]
Put
=
Taking logarithms on both sides, we get
=2
⟹ . =2
+2 .
(
⟹ =
+ 1)
Limits: = 1 ⟹ = 1 ;
= 2 ⟹ = 2 = 16
2
16
1
1
15
x2 x 1 ln x dx
du u 16
2
2 1
2
1
1
4
P4. Evaluate the
1
3
tan t
sec2 t dt
0
Solution:
4
To evaluate
1
3
tan t
sec2 t dt
0
Put =
⟹
=
Limits: = 0 ⟹ = 0 ; = ⟹ = 1
Now,
4
1
tan
t
u
1
1
2
sec t dt du
3
3
0
0
1
1 3 u
1 1 1 0
1
2
ln 1 3 3 3 3 ln 3
ln 1 3
0
4
tant
2
1
sec 2 t dt
3 ln 3
3
0
EXERCISES
1. Find the derivative of
x
a. y 2
b.
y 5
s
y with respect to the given independent variable.
c.
y x
d.
y cos
2
y 7 sec ln 7
y 2sin 3t
e.
f.
2. Use logarithmic differentiation to find the derivative of
independent variable.
a) = ( + 1)
b) = ( )
c) = √
d) = √
)
e) = (
f)
=
g) =
w.r.t the given
3. Evaluate the integrals given below:
1
a.
5
2
b.
x
dx
;0
x2
x2
d
1
4
c.
dx ;
x
1 .3
2 d
x
2
5 d
x
dx
1
0
d.
2
2
e.
1
f.
g.
x dx
x.2
2
2
;
7
0
2
cos t
sin t dt
;
1
2ln x
dx
x
4.6. General Logarithmic Functions
Learning objectives:
To define the general logarthmic function.
To derive the Derivative of log .
To evaluate integrals involving log .
To study the uses of common logarithm(log
And
To practice the related problems.
).
Logarithms with Base :
As we saw earlier, if is any positive number other than 1, the function
is oneto-one and has a nonzero derivative at every point. It therefore has a differentiable
inverse. We call the inverse the logarithm of with base and denote it by log .
Definition
For any positive number ≠ 1,
log
= inverse of
The graph of = log can be obtained by reflecting the graph of =
across
the line = .
Since log and
are inverses of one another, composing them in either order
gives the identity function.
=
( > 0)
------(1)
log
=
Example 1:
a)
log 2 = 5
b)
log (10 ) = −7
c)
2
=3
d)
10
=4
(
)
-------(2)
The Evaluation of
The evaluation of log is simplified by the observation that log
multiple of
.
log
=
∙
=
------(3)
We can derive equation (3) from equation (1):
=
⟹ ln
=
⟹ log ∙
=
⟹ log
=
is a numerical
.
Example 2: log 10 =
≈ .
≈ 0.3010
The arithmetic properties of log are the same as the ones for
For any numbers > 0 and > 0,
1. Product Rule :
log
= log + log
2. Reciprocal Rule: log = − log
log
3. Quotient Rule:
4. Power Rule:
log
= log
=
.
− log
log
These rules can be proved by dividing the corresponding rules for the natural
logarithm function by
.
The Derivative of
To find the derivative of a base logarithm, we first convert it to a natural
logarithm. If is a positive differentiable function of , then
(log ) =
=
(
)=
∙
(log
Example 3:
log (3 + 1) =
)=
∙
(3 + 1) = (
∙
Integrals involving
To evaluate integrals involving base
logarithms.
Example 4: ∫
=
∫
=
=
---------(4)
)(
logarithms, we convert them to natural
, ℎ
∫
+
)
=
=
(
)
+
=
(
)
+
Base
Logarithms
Base 10 logarithms, often called common logarithms, appear in many scientific
formulas. For example, earthquake intensity if often reported on the logarithmic
Richter scale. Here the formula for magnitude
is
= log
+
where is the amplitude of the ground motion in microns at the receiving station,
is the period of the seismic wave in seconds, and is an empirical factor that
allows for the weakening of the seismic wave with increasing distance from the
epicenter of the earthquake.
Example 5:
For an earthquake 10,000
from the receiving station, = 6.8. If the recorded
vertical ground motion is = 10 microns and the period is = 1 , the
earthquake’s magnitude is
= log
+ 6.8 = 1 + 1.68 = 7.8
An earthquake of this magnitude does great damage near its epicenter.
The pH scale for measuring the acidity of a solution is a base 10 logarithmic scale.
The pH value (hydrogen potential) of the solution is the common logarithm of the
reciprocal of the solution’s hydronium ion concentration, [H O ].
pH = log
= − log [H O ]
[
]
The hydronium ion concentration is measured in mole per liter. The total scale
ranges from about 0.1 for normal hydrochloric acid to 14 for a normal (1 )
solution of sodium hydroxide.
Another example of the use of common logarithm is the decibel or db scale for
measuring loudness. If is the intensity of sound in watts per square meter, the
decibel level of the sound is
= 10 log ( × 10 )
-------(5)
Example 6:
Doubling in equation (5) adds about 3 .
Writing log for log (a common practice), we have
sound level with doubled = 10 log(2 × 10 )
= 10 log(2 ∙ × 10 )
= 10 log 2 + 10 log( × 10 )
= original sound level + 10 log 2
≈ original sound level + 3 ( log 2 ≈ 0.30)
PROBLEM SET
IP1: If
=
(
(
)
)
,then
=?
Solution:
Given
= log
(
(
)
)
∙
⟹ =
=
= ( + 1) −
Differentiating both sides w.r.t , we get
=
−
= ( )( ) =
( − 1)
(
P1: If =
Solution:
), then find
(
.
)
Given = 3 log (log ) =
=
Differentiating both sides w.r.t , we get
=
=
∙
(
∙
)(
)(
)
)
IP2: If =
Solution:
Given
=(
, then find
√
= log
?
√
(
=
)
√
√
=
(
=
)
Differentiating both sides w.r.t , we get
=
− ∙
=
=
P2: If
(
)
(
)
=
=
(
)
?
then find
Solution:
= log
=
=
=
[ 7 −
(3 + 2)]
Differentiating both sides w.r.t , we get
(7 ) −
=
=
(7) −
=
−
=
(
)
(3 + 2 )
.
(3)
=
(
)
∴
=
(
)
=?
IP3: ∫
Solution:
log
Let
1
=
ln 2
=
⟹
=
= 1 ⟹ = 0; = 2 ⟹
1
=
log
∴
=
4
1
2
=
ln 2
=
=
(
)
=
(
)
=
4
4
=?
P3: ∫
Solution:
log
1
=
10
10 1
=
let
=
∴
IP4: ∫
Solution:
⟹
10 1
(
)
2 log ( − 1)
−1
=
=
10
=
=
=?
10
10
=
∴
2 log ( − 1)
−1
=
P4:
=?
| |+
|
|+
( − 1)
−1
2
2
=
1
2
( ( − 1))
2
2
2
=
1
[(
2
2) − (
1) ] =
2
Solution:
2
10 log
2
=
10
10
1
=2
=2
=(
) −(
(
)
2
1) = 1
Exercises:
1. Find the derivative of with respect to the given independent variable.
4. = log 5
5. = log (1 +
3)
6. = log
− log √
7. = log + log
8. = log ∙ log
9. = log ∙ log
10. = sin(log )
11. = log
12. = log
13. = 3
14. = log 8
15. = log
(
2. Evaluate the integrals.
a. ∫
b. ∫
(
c. ∫
d. ∫
e. ∫
f. ∫
g. ∫
h. ∫
(
)
(
)
(
)
)
)(
)
4.7. L’Hôpital’s Rule
Learning Objectives:
To discuss the indeterminate form
and its evaluation by L’Hôpital’s Rule
To state and prove Cauchy’s Mean Value Theorem and apply it to prove the
stronger form of L’Hôpital’s Rule
To discuss the other indeterminate forms and their evaluation by L’Hôpital’s
Rule
AND
To practice the related problems
If the continuous functions ( ) and ( ) are both zero at = , then
( )
lim → ( ) cannot be found by substituting = . The substitution produces , an
expression known as an indeterminate form. L’Hôpital’s rule enables us to evaluate
the limits that otherwise lead to indeterminate forms.
Theorem: L’Hôpital’s Rule (First form)
Suppose that ( ) = ( ) = 0, that ( ) and ( ) exist, and that ( ) ≠ 0.
Then
( )
( )
lim
=
… … … … (1)
→
( )
( )
Proof
f x f a
lim
f a x a
f x f a
xa
lim
x a g x g a
g a
g x g a
lim
x a
xa
f x 0
f x
lim
lim
x a g x 0
x a g x
Example 1
3x sin x 3 cos x
2
x 0
x
1
x 0
a)
lim
b)
1
1 x 1 2 1 x
lim
x0
x
1
1
2
x 0
Some times after differentiation, the new numerator and denominator both equal
to zero at = . In this case, we apply a stronger form of L’Hôpital’s Rule.
The proof of the stronger form of L’Hôpital’s Rule is based on Cauchy’s Mean Value
theorem, a Mean Value Theorem that involves two functions instead of one.
Theorem: Cauchy’s Mean Value Theorem
If the functions and are continuous on [ , ], differentiable throughout
( , ) and ( ) ≠ 0 throughout ( , ), then there exists a number in ( , ) such
that
( )
( )− ( )
=
( )
( )− ( )
Proof: We apply Lagrange’s Mean Value Theorem twice
(i) First we use it to show that ( ) ≠ ( ). If ( ) = ( ), then by Lagrange’s
Mean Value Theorem
( )− ( )
( )=
=0
−
for some ∈ ( , ), which is not possible, since ( ) ≠ 0 in ( , ).
(ii) We next apply Mean Value Theorem to the function
( )− ( )
( ) = ( )− ( )−
[ ( ) − ( )]
( )− ( )
Notice that is continuous on [ , ], differentiable on ( , ) and
( ) = 0 = ( ). Therefore, by Rolle’s theorem, there exists a ∈
( , ) such that ( ) = 0,
( ) ( )
i.e., ( ) = ( ) − ( ) ( ) ( ) = 0
⇒
( )
( )
=
( )
( )
( )
( )
Hence the theorem.
Note: Lagrange’s Mean Value Theorem is a special case of Cauchy’s Mean Value
Theorem.
If ( ) = , then Cauchy’s Mean Value Theorem reduces to
( )− ( )
( )=
−
Theorem: L’Hôpital’s Rule (Stronger form)
If and are differentiable on an open interval I containing
0 and ( ) ≠ 0 on I when ≠ , then
lim
→
( )
( )
= lim
→
, ( )= ( )=
( )
( )
assuming that the limit on the right hand side exists.
Proof: We first establish the limit equation for the case → .
Suppose that lies to the right of . Then ( ) ≠ 0 and we can apply Cauchy
Mean Value Theorem to the closed interval [ , ]. Thus there exists ∈ ( , ) such
that
( )
( )
=
( )
( )
( )
( )
=
( )
( )
(Since ( ) = ( ) = 0)
As approaches , approaches , since it lies between and . Therefore
( )
( )
( )
lim
= lim
= lim
→
→
→
( )
( )
( )
Thus, L’Hôpital’s Rule is proved for the case when approaches from above.
The case when approaches from below can be proved by applying Cauchy’s
Mean Value Theorem to the closed interval [ , ], < . The combination of these
two cases now establishes the result.
Hence the theorem.
Example:
lim
x 0
x sin x 1 cos x
3
x
3x 2
Still
?
x0
0
0
This example can be solved by using a stronger form of L’Hôpital’s Rule which says
that whenever the rule gives we can apply it again, repeating the process until we
get a different result. With this stronger rule we get
lim
x 0
x sin x
x3
lim
0
; apply the rule again
0
0
Still ; apply the rule again
0
1 cos x
3 x2
sin x
lim
x 0 6 x
cos x 1
lim
x 0 6
6
Still
x 0
0
Not ; limit is found
0
Example 2
lim
x 0
1 x 1 x / 2
x
0
0
2
lim
1/ 21 x 1/2 (1/ 2)
2x
x 0
lim
x 0
0
0
1/ 41 x
2
3/2
Not
0
0
1
8
When you apply L’Hôpital’s Rule, look for a change from to something else. This is
where the limit is revealed.
Example 3
1 cos x
0
lim
x 0 x x 2
0
sin x
0
lim
Not
x 0 1 2 x
0
0
0
1
If we continue to differentiate in an attempt to apply L’Hôpital’s Rule once more, we
get
1 cos x
sin x
cos x 1
lim
lim
x 0 x x2
x0 1 2 x
x0
2
2
lim
which is wrong.
L’Hospital’s rule applies to one-sided limits also.
Example 4
lim
sin x
0
0
x2
cos x
lim
2x
x0
x0
Not
0
0
L’Hôpital’s Rule also applies to quotients that lead to the indeterminate form
( ) and ( ) both approach infinity as → , then
lim
xa
f x
f x
lim
g x x a g x
provided the limit on the right exists. The
infinite.
Example 5
here may itself be either finite or
sec x
x /2 1 tan x
sec x tan x
lim
lim sin x 1
x /2
x /2
sec2 x
lim
a)
b)
. If
ln x
1/ x
1
lim
lim
0
x 2
x x 1 / x x x
lim
Indeterminate Products and Differences
We can sometimes handle the indeterminate forms 0 ∙ ∞ and ∞ − ∞ by using
algebra to get or instead. It must be remembered that all these forms are not
some numbers but only notations for functional behaviors when considering limits.
Example 6
lim x cot x
x 0
0
1
x 0
tan x
x
0
lim
Now
x 0 tan x
0
1
1
lim
1
2
x 0 sec x
1
lim x
Example 7
1
1
lim
Find x 0
sin
x
x
Solution
If
x 0 , then sin x 0 and
If
x 0 , then sinx 0
1
1
sin x x
1
1
()
and
sin x x
Neither form reveals what happens in the limit. To find out, we first combine the
fractions.
1
1 x sin x
sin x x
x sin x
and then apply L’Hôpital’s Rule to the result:
1
x sin x
1
lim
lim
x 0 sin x
x x 0 x sin x
1 cos x
lim
x 0 sin x x cos x
sin x
0
lim
0
x 0 2 cos x x sin x
2
0
0
0
0
Indeterminate Powers
Limits that lead to the indeterminate forms 1 , 0 and ∞ can sometimes be
handled by taking logarithms first. We use L’Hôpital’s Rule to find the limit of the
logarithm and then exponentiate to find the original function behavior.
ln f
If lim
x a
x L , then
lim f x lim e ln
x a
f x
x a
eL
Here may be either finite or infinite.
1/ x
Example 8 : Show that lim 1 x
x 0
e
Solution: The limit leads to the indeterminate form 1∞
1/ x
. We let f x 1 x
ln f x .
and find xlim
0
Since ln f x ln 1 x
1/ x
1
ln 1 x
x
L’Hôpital’s Rule gives
lim lnf x lim
x 0
x 0
ln1 x
x
1
1
lim 1 x 1
x0
1
1
0
0
Therefore,
1/ x
lim 1 x
x 0
lim f x lim e ln f x e1 e
x 0
x0
Example 9
1/ x
lim
x
Find x
.
Solution
The limit leads to the indeterminate form ∞ . We let
f x x1/ x
lim ln f x . Since
x 0
ln f x ln x1 / x
ln x
x
,
L’Hôpital’s Rule gives
ln x
x
x
1
0
lim x 0
x 1
1
lim ln f x lim
x
Therefore,
lim x1/ x lim f x lim e ln f x e 0 1
x
x
x
PROBLEM SET
IP1: Find lim
t 0
10 sin t t
t3
by using L’Hôpital’s Rule.
Solution:
lim
10 sin t t
t 0
t3
10 co s t 1
0
form
0
0
form
t0
0
3t
10 sin t
0
lim
form
t 0
6t
0
10 c os t
10
5
lim
t 0
6
6
3
lim
2
5x2 3x
P1: Find xlim
by using L’Hôpital’s Rule.
7x2 1
5x2 3x
lim
form
Solution: x
2
7x 1
10 x 3
10
5
lim
form lim
x 1 4 x
x 14
7
and find
IP2: Find lim 2 x tan x by using L’Hôpital’s Rule.
x
2
Solution:
lim x tan x
2
x
0 . form
2
2 x sin x
lim
cos x
x
0
form
0
2
x cos x sin x 1 1
2
lim
1
sin x
1
x
2
csc x cot x by using L’Hôpital’s Rule.
P2: Find lim
x 0
Solution:
lim csc x cot x
fo rm
x 0
c os x
1
lim
x 0 sin x
sin x
1 cos x
lim
x0
sin x
sin x
0
lim
0
x 0 cos x
1
0
fo rm
0
IP3:
3
1
Find xlim
x
x
by using L’Hôpital’s Rule.
Solution:
3
lim 1
x
x
x
1 fo rm
Th e limit le ad s to the indete rminate form 1 .
x
3
Let f x 1 and find lim ln f x .
x
x
3
ln 1
x
3
3
x
ln f x ln 1 x ln 1
1
x
x
x
L'H opita l's rule gives
3
ln 1
x
0
lim ln f x lim
form
1
x
x
0
x
1 3
2
1 3 x
3x
x
lim
lim
x
x x 3
1
2
x
3
lim 3
x 1
form
x
3
ln f x
lim 1 lim f x lim e e 3
x
x
x
x
P3: Find lim x
1
ln x
by using L’Hôpital’s Rule.
x 0
Solution:
lim x
1
ln x
0 0 fo rm
x 0
T h e lim it le a d s to th e in d e te rm in a te fo rm 0 0 .
Let f
x
ln f
x
x
1
ln x
x .
a n d fin d lim ln f
x0
ln x
1
ln x
ln x
1
ln x
L 'H o p i t a l 's r u l e g iv e s
x
lim ln f
x 0
l i m x
x 0
1
ln x
lim
x 0
1
li m f
x 0
1 2 x
IP4: Find lim
x
x
1
2 ln x
1
l i m e ln
f
x
e 1
x 0
1
e
by using L’Hôpital’s Rule.
Solution:
lim 1 2 x
x
1
2 ln x
0 fo r m
T h e lim it le a d s t o th e in d e te r m i n a te f o r m 0 .
1
Le t f
ln f
x 1 2 x 2 ln x a n d fi n d lim ln f
x
x
ln 1 2 x
1
2 ln x
ln 1 2 x
2 ln x
x .
L 'H o p ita l's ru le g iv e s
lim ln f
x
x
ln 1 2 x
lim
fo r m
2 ln x
2
x
li m 1 2 x li m
x
x 1 2 x
2
x
1
1
li m
x 2
2
x
lim 1 2 x
1
2 ln x
x
lim f
x
x
lim e
fo rm
ln f
x
x
e
1
2
e
P4:
Find all values of , that satisfy the conclusion of Cauchy’s Mean Value Theorem for
the functions ( ) = , ( ) = , and interval ( , ) .
Solution:
We have ( ) = , ( ) =
and interval ( , ). Now, ( ) = 1 and
( )=2 .
By Cauchy’s Mean Value Theorem, we seek in the interval ( , ) so that
( )
( )
⇒
⇒
=
( )
( )
( )
( )
=
=(
)(
)
=
=
Exercises:
1. Use L’Hôpital’s Rule to find the limits.
x2
x2 4
t 3 4t 15
b. tlim
3 t 2 t 12
a. lim
x2
c. lim
x
5 x2 3x
7 x2 1
sin t 2
d. lim
t0
t
8x2
e. lim
x 0 cos x 1
2
f. lim
/ 2 cos 2
=
g.
h.
i.
j.
k.
1 sin
/ 2 1 cos 2
x2
lim
x 0 ln sec x
t 1 cos t
lim
t 0 t sin t
lim x sec x
x / 2
2
3sin 1
lim
lim
x2x
l. lim
x 0 2 x 1
0
m. lim
x
ln x 1
log 2 x
ln x 2 2 x
n. lim
x0
ln x
5 y 25 5
o. lim
y 0
y
ln 2 x ln x 1
p. lim
x
1
1
x
sin
x
1
1
lim
r. x1
x 1 ln x
q. xlim
0
1
lim
s. x x dt
t
cos 1
t. lim
0 e 1
2x
et t 2
u. lim
t et t
2. Find the limits.
a.
lim x 1 / 1 x
x 1
1/ x
lim ln x
b. x
c. lim
x 0
x
x
4.8. Relative Rates of Growth
Learning Objectives:
1) To discuss the comparison of functions as → ∞.
2) To define order, little - o and big – O of functions.
And
3) To practice the related problems.
If we look at the graphs, we notice that exponential functions like 2 and
grow
more rapidly as gets large than the polynomials and rational functions. The graphs
of , 2 , and
are shown below. In fact, as → ∞, the functions 2 and
grow
faster than any power of .
In contrast, logarithmic functions like = log
as → ∞ than any positive power of .
and
=
grow more slowly
These important comparisons of exponential, polynomial, and logarithmic functions
can be made precise by defining what it means for a function ( ) to grow faster
than a function ( ) as → ∞.
Definition
Rates of Growth as → ∞
Let ( ) and ( ) be positive for sufficiently large.
1. grows faster than as → ∞ if
lim ( )
=∞
→∞ ( )
or, equivalently, if
lim ( )
= 0.
→∞ ( )
We also say that
grows slower than
as
→ ∞.
grow at the same rate as → ∞ if
lim ( )
= , where
→∞ ( )
According to these definitions, = 2 does not grow faster than
functions grow at the same rate because
lim →
= lim 2 = 2
2.
and
= . The two
→
which is a finite, nonzero limit. This is sensible if we want “ grows faster than ” to
mean that for large -values is negligible when compared with .
Example 1
grows faster than
as → ∞ because
lim
= lim
→
= lim
→
→
=∞
using L’Hôpital’s Rule twice
Example 2
a)
3 grows faster than 2 as → ∞ because
lim
= lim
→
=∞
→
(∵ > 1)
b) As part (a) suggests, exponential functions with different bases never grow at the
same rate as → ∞. If > > 0, then
grows faster than . Since ( > 1),
lim
→
= lim
=∞
→
Example 3
grows faster than
lim
→
= lim
→
/
as
= lim 2
→
→ ∞ because
= ∞ (L’Hôpital’s Rule)
Example 4
grows slower than as → ∞ because
/
lim
= lim
= lim = 0
(L’Hôpital’s Rule)
→
→
→
Example 5
In contrast to exponential functions, logarithmic functions with different bases
and always grow at the same rate as → ∞.
/
lim
= lim
=
→
/
→
The limiting ratio is always finite and never zero.
If
grows at the same rate as as → ∞, and grows at the same rate as ℎ as
→ ∞, then grows at the same rate as ℎ as → ∞. The reason is that
lim =
lim =
→
→
Together imply
If
and
lim
→
= lim . =
→
are finite and nonzero, then so is
.
Example 6
Show that √ + 5 and 2√ − 1 grow at the same rate as → ∞.
Solution
We show that the functions grow at the same rate by showing that they both grow
at the same rate as the function ( ) = .
lim
√
→
→
√
lim
= lim 1 +
= lim
→
√
=1
= lim 2 −
→
→
=4
√
Order and O-Notation
We introduce the “little-o” and “big-O” notation invented by number theorists a
hundred years ago and now commonplace in mathematical analysis and computer
science.
Definition
( )
A function is of smaller order than as → ∞ if lim
= 0. We indicate this by
( )
→
writing
= ( ) (“ is little-o of ”).
We note that saying = ( ) as → ∞ is another way to say that
than as → ∞.
Example 7
= ( )
= (
→ ∞ because lim
→
+ 1) as
lim
= lim
→
→
= lim
→
=0
→ ∞ because
= lim
→
=0
Definition
Let ( ) and ( ) be positive for sufficiently large. Then
(
of as → ∞ if there is a positive integer for which
)
( )
sufficiently large. We indicate this by writing
= ( ) (“ is big-O of ”).
Example 8
+
= ( ) as → ∞ because
≤ 2 for
Example 9
+
= (
= (
) as
) as
grows slower
→ ∞ because
→ ∞ because
→ 0 as
→ 1 as
is of at most the order
≤
for
sufficiently large.
→∞
→∞
From the definitions, we see that = ( ) implies = ( ) for functions that are
positive for sufficiently large. Also, if and grow at the same rate, then
= ( ) and = ( ) .
PROBLEM SET
IP1.
Which of the following statement is true
I. If ( ) = log and ( ) = log ,
as → ∞.
II. If ( ) = √
Solution:
≠
then
+ 5 and ( ) = 2√ − 1
and
grow at the same rate
grow at the same rate as
I.
log a x
f(x)
lim
x g( x ) x logb x
ln x / ln a lnb
lim
x ln x / ln b ln a
Since limiting ratio is finite and never zero, and
as → ∞.
lim
grow at the same rate
II.
2
f(x)
x 5
lim
lim
2
x g( x ) x 2 x 1
x
lim
⟹
1
(finite and nonzero)
4
and grow at the same rate as
Thus, I and II are true.
x 1
5
x2
1
x 2
x
2
→ ∞.
P1.
Which of the following statements is false?
A.
grows faster than
as → ∞
B. 2 grows faster than 7 as → ∞
C. 5 grows faster than 3 as → ∞
D.
grows faster than ln as → ∞
Solution:
A.
ex
F
orm
x x2
by L ' H opital ' s rule
ex
lim
2
x
and still
form
x
lim
ex
lim
by L ' H opital ' s rule
2
x 2
→ ∞.
⇒
grows faster than
→∞
as
B.
x
2x
2
lim
lim 0 but < 1
x 7 x
x 7
⟹ 2 grows slower than 7 as
→∞
c.
x
5x
5
lim
lim and > 1
x 3 x
x 3
⟹ 5 grows faster than 3 as → ∞
D.
x2
lim
x ln x
Form
2x
x 1 / x
by L' Hopital' s rule
lim
lim 2 x 2
x
⟹
grows faster than
as → ∞
Thus only the statement B is false.
IP2. Which of the following functions grow slower than
a)
b)
10 c) ln (ln ) d) ln (2 + 5)
Solution:
a)
lim
2
= lim
=
1
lim
2
=
2 →
grows at the same rate as
→
→
Thus
as
b)
lim
10
→
=
= lim
→
1
lim
10 →
Thus
10
as ⟶ ∞.
10
=
10
10
10
1
10
lim
1
10 →
=
grows at the same rate as
1
10
as
2
2
⟶∞
⟶∞
c).
1
(
lim
)
→
= lim
1
→
Thus
(
= lim
1
=0
→
) grows slower than
⟶ ∞.
as
c)
(2 + 5)
lim
→
2
= lim 2 + 5
1
→
lim
lim
=
=1
→∞
→∞
Thus (2 + 5) grows at the same rate as
as ⟶ ∞.
Therefore, (
) grows slower than
as ⟶ ∞.
P2.
Which of the following functions grow faster than
as ⟶ ∞
b)
b)
c)
d)
Solution:
d)
e
1
lim
= lim
=0
→
→ e
e
=
⇒
grows slower than
as
⟶ ∞.
e)
e
= lim = ∞
→
→
e
⇒ e grows faster than e as ⟶ ∞.
f)
For all real values of , we have
−1 ≤
≤1⇒
≤
≤
lim
⇒
Further,
lim
→∞
≤
≤
= 0 and
lim
→∞
=0
lim
→∞
⟶ ∞.
By Sandwich theorem we conclude that
Thus
grows slower than
as
g)
lim
→∞
Thus,
lim 1 1
=
→∞
grows at the same rate as
=
Therefore, only
grows faster than
as
.
⟶ ∞.
= 0.
IP3.
Order the following from functions from slowest growing to fastes growing as
⟶∞
a) 2
b)
c) ( 2) d)
Solution:
I.
( 2)
( ( 2)) ( 2)
( 2) ( 2)
lim
= lim
= lim
→
→
→
2
2
( (
))
lim (
=
2) = 0 (∵ 2 < 1)
→∞
Thus (
2) grows slower than
lim
II.
= lim
→
→
(
as
= lim
)
grows slower than 2 as
Thus
lim
III.
= lim
→
⟶ ∞.
→
(
)
= 0;
⟶ ∞.
= 0;
→
Thus 2 grows slower than
⟶ ∞.
as
Therefore, the order of the functions from slowest growing to fastest growing as
⟶ ∞ is: (
2) ,
, 2 and
.
P3.
Order the following functions from fastest growing to slowest growing as
a)
c)(
b)
)
d)
Solution:
lim
→∞
I.
⟹
II.
For
⟹(
as
>
>
, we have
)
=
lim
→ ∞(
>
)
for
=
⟶ ∞.
=∞
as
⟶ ∞.
> 0 . Now,
lim
→∞
grows faster than (
>1
and
lim
→∞
) grows faster than
We have
⟹
=∞
grows faster than
lim (
→∞
Now,
III.
lim
→∞
=
= ∞ (∵
) ;
>1
> 0)
⟶ ∞.
Therefore, the order of the functions from fastest growing to slowest growing as
⟶ ∞ is
,(
) ,
and
.
IP4.
Which of the following statements are true.
I.
+
= ( )
II.
+
= ( )
Solution:
I.
x sin x
sin x
lim
lim 1
x
x
x
x
sin x
lim 1 lim
2
x
x x
⟹ +
= ( )
II.
ex x2
lim
lim 1
x e x x
1 lim
x2
e x
x2
x e x
1 lim
2x
x e x
1 lim
2
x e x
by L' Hopital' s rule
by L' Hopital' s rule
1
e x x2 O( e x )
So, both I and II are true
P4.
Which of the following statement is false
I.
= ( + 5)
II.
= ( 2 )
Solution: We recall,
lim ( )
= ( ) if
=0
⟶∞ ( )
I.
We have,
x
lim
Form
x x 5
1
lim 1
by L' Hopital' s rule
x 1
Therefore, = ( + 5) is false.
II.
We have,
ln x
lim
Form
x ln 2 x
1
lim x
by L' Hopital' s rule
x 1 .2
2x
lim 1 1
x
Therefore,
= ( 2 ) is false.
Hence, both I and II are false.
Exercises
1. Which of the following functions grow faster than
the same rate as ? Which grow slower?
a. + 3
b.
+
c. √
d. 4
e. (3/2)
/
f.
g.
/2
h. log
as
→ ∞? Which grow at
2. Which of the following functions grow faster than
the same rate as ? Which grow slower?
as
→ ∞? Which grow at
a.
b.
c.
d.
e.
f.
g.
h.
+4
−
√ +
( + 3)
ln
2
8
3. Which of the following functions grow faster than ln
the same rate as ln ? Which grow slower?
a. log
b. ln 2
c. ln √
as
→ ∞? Which grow at
d. √
e.
f. 5 ln
g. 1/
h.
4. True, or false? As → ∞
a. = ( )
b. = ( + 5)
c. = ( + 5)
d. = (2 )
e.
= ( )
f.
+ ln = ( )
g. ln = (ln 2 )
h. √ + 5 = ( )
i.
= ( )
j.
+
= ( )
k. − = ( )
l. 2 +
= (2)
m.
+ = ( )
n.
= ( )
( )= ( )
o.
( ) = ( ( + 1))
p.
4.9. Derivatives of Inverse Trigonometric Functions
Learning objectives:
To find the derivatives of inverse trigonometric functions.
And
To solve the related problems.
Inverse trigonometric functions provide anti derivatives for a variety of functions
that arise in engineering. The derivatives of the inverse trigonometric functions are
given below.
1.
2.
3.
4.
5.
6.
(
)
(
)
(
)
(
)
(
)
(
)
=√
| |<1
,
= −√
,
| |<1
,
| |>1
,
| |>1
=
=−
=|
=
|√
| |√
We derive formulas 1 and 5 below.
The Derivative of =
:
The function = sin is differentiable in the interval
< < and that its
derivative, the cosine, is positive there. Therefore, the inverse function = sin
is differentiable throughout the interval −1 < < 1. We cannot expect it to be
differentiable at = 1 or = −1 because the tangents to the graph are vertical at
these points.
We find the derivative of = sin
as follows.
(sin ) = 1 ⟹ cos
sin = ⟹
=1
⟹
=
=
=
√
Therefore, the derivative of = sin
(
)=
with respect to
is
with | | < 1, we apply the Chain Rule
If
is a differentiable function of
=
to = sin
to obtain
=
| |<1
,
The Derivative of =
:
We find the derivative of = sec
, | | > 1, in a similar way.
(sec ) = 1 ⟹ sec tan
sec =
⟹
=1
⟹
=
=±
√
From the figure below, we see that for | | > 1 the slope of the graph of
= sec
is always positive.
Therefore,
(sec
)=
if
√
−
√
if
>1
< −1
With absolute values, we can write this equation as a single formula.
(
)=
| |>1
| |
with | | > 1, we apply the Chain Rule to
If is a differentiable function of
obtain
(
)=
| |>1
,
| |
We find the derivative of = tan
as follows.
(tan ) = 1 ⟹ sec
tan = ⟹
=1
⟹
=
=
=
The derivation of formula 3 is similar to the above derivations.. Formulas 2, 4, and 6
can be derived from formulas 1, 3, and 5 by differentiating appropriate identities.
(cos
)=
=
2
(sin
−
= 0−
Example 1:
(a)
sin (
(b)
tan
)=
(
)
√ +1 =
∙
=
sec (−3 ) =
=
−
√ −
√
√ +1
( + 1)
)
|
| (
)
|
|| | (
)
=
)=
√
(
√
)=
∙
√
=
(c)
(
∙
)
√ −
(cos
− sin
∙ (−3)
=|
| |√
(−3 )
∙
|√
Example 2:
1+
=[ ]
=
⁄
=
, where
⁄
= tan
and
−1
PROBLEM SET
IP1: If =
Solution:
Given = tan
⟹ ( )=
√
−
√
− 1 + csc
tan
√
+
> 1 , then find
,
−1
,
+
>1
(csc
)
.
=
1+
⟹
=
√
=
∙
=
∙
=
∙
=
=
P1: If =
Solution:
Given =
=
(
− 1) −
(2 − 0) −
√
(2 ) −
−
=
|√
(∵
√
> 1)
√
√
+√ −
, then find
.
+ √1 −
(sin
∙
)
=0
√
sin
(
√
(2 ) −
√
√
−1 +|
√
√
∙
∙
√
∙√
) + sin
+ sin
∙ (1) +
=√
+ sin
+
=
+ sin
+
=√
+ sin
−√
=√
+ sin
−√
√
( ) +
∙
√1 −
(1 −
√
)
(−2 )
√
(−2 )
√
= sin
IP2: If ( ) = (
ʹ(
)=( +
Solution:
Given ( ) = (
ʹ(
)=
=
⁄
)
−
, then show that
) .
)
−
⁄
cos
∙
√
∙
(
√
(
)
)
(
)
(
)(
∙
) (
)(
(
=
)
[
√
]
(
)
(
)(
)
)
=
(
)
√
∙(
√
)
= ( + cos )
=?
P2:
Solution:
=
√
√
[
]=
[
=
.
=
√1 −
And
√
√
[
Now,
]
=
+
√
.1
(1 −
√
)
. (−2 ) =
√
√
∴
( )
+
√
=
]+
√
.
=
√
√
√
∙
∙
.
=
√
∙
.
=
√
=
=
=
=
=
=
=
IP3: If ( ) =
Solution:
Given ( ) = cot
Let
=
=
√
(
)√
=
, then find ʹ( ).
.
√
(
)
⟹
=
Differentiate both sides w.r.t , we get
=
∙ +
(
⟹
)=
(1 +
)
=
Let
⟹
=−
Differentiate both sides w.r.t , we get
=− ∙ −
∴
ʹ(
(
⟹
)=−
(1 +
)
)=−
(
=−
ʹ(
1) = −
+
(1 + 1)(1 +
)(1 +
)
1) = −1
P3: Find the derivative of
w.r.t
Solution:
Derivative of a function w.r.t another function
.
Suppose ( ) and ( ) are two functions and we have to find the derivative of
( ) w.r.t ( ). Now let
∴
Let
= tan
Put
= tan
= ( ),
=
=
and
⟹
and
⟹
= tan (tan 2 ) and
⟹
= 2 and
=2
⟹
= 2 tan
and
=
⟹
=
and
=
ʹ(
)
.
)
= sin
.
= tan
= tan
⟹
ʹ(
= ( ). Then
= sin
= sin (sin 2 )
= 2 tan
=
= 1.
=
ʹ(
),
=
ʹ(
).
=?
IP4: ∫√
Solution:
Let = sec
⟹
= √2 ⟹
=
(sec
∴
√
√
=
;
)
=2⟹
=
−1
∵
√
∈ √2, 2
=
sec
= [tan ]
⁄
⁄
= tan − tan
= √3 − 1
P4:
( +
=?
)
Solution:
1
(1 +
tan
Let
= tan
1
1+
tan
∴
tan
=
)
⟹
=
1
(1 +
1
1+
tan
=
1
)
=
| |+
=
=
|tan
|+
|tan
|+
Exercises:
1. Find the derivative of
a. = cos ( )
b. = cos
c.
d.
e.
f.
g.
h.
= sin √2
= sin (1 − )
= sec (2 + 1)
= sec 5
= csc ( + 1) ,
= csc
i.
= sec
j.
= sin
k.
l.
m.
= cot √
= cot √ − 1
)
= ln(tan
,0 <
w.r.t appropriate variable.
>0
<1
n.
o.
p.
q.
r.
s.
t.
= tan (ln )
= csc ( )
= cos ( )
= √1 − cos
= √ − 1 − sec
= tan √ − 1 + csc
= cot
− tan
u.
=
(
,
>1
+ 4) − tan
2. Find the derivative of the following w.r.t .
a. sin (3 − 4 )
b. sin
c. cos (4
d. sin
−3 )
√
e. tan
f. tan
g. tan
h. tan
3. Find the derivative of
a. sec
w.r.t √1 −
√
b. tan
w.r.t tan
w.r.t sin
c.
d. cos
w.r.t tan
4. Evaluate the integrals.
a. ∫
b. ∫
c. ∫
d. ∫
√
√
(
)
√
√
e. ∫
(
f. ∫
g. ∫ √
.
)
(
)
(
√
)
.
4.10. Integration Formulas involving Inverse
Trigonometric Functions
Learning objectives:
To study the Integration formulas of inverse trigonometric functions.
AND
To practice the related problems.
The derivative formulas for the inverse trigonometric functions, given in the
previous module, yield three useful integration formulas given below for any
constant ≠ 0.
du
u
sin1 C
valid for u2 a2
1.
a
a2 u2
du
1
u
tan1 C
a
a u
a
2.
2
valid for all u
2
du
u
3.
2
u a
2
1 1 u
sec
C
a
a
valid for u
2
a2
They are readily verified by differentiating the functions on the right-hand sides.
Note:
du
1 u
1 u
sin
C
cos
valid for u 2 a2
C
1.
2
2
a
a
a u
(since cos
2.
+ sin
= )
du
1
1
u
u
tan1 C cot1 C valid for all u
a
a
a u
a
a
(Since tan
+ cot
= )
2
2
u
3.
du
u 2 a2
(since sec
Example 1
a)
1
1 1 u
1
u
sec
C csc 1 C
a
a
a
a
+ csc
3/2
dx
2/ 2
1 x2
valid for u
= )
sin1x
3
2
1
1
sin
sin
2/ 2
2
2 3 4 12
3/ 2
1
tan 1 x tan 1 1 tan 1 0 0
b)
2
0
4
4
0 1 x
c)
dx
2
2/ 3
dx
x x2 1
2
2
sec1 x
2 /
3
4
6
12
a2
Example 2
a)
b)
dx
dx
x
sin1 C
3
32 x2
9 x2
dx
2
3 4x
1
du
2 a2 u2
a 3 , u 2x
1
1
2x
u
sin1 C sin1 C
2
2
a
3
Example 3
Evaluate
dx
4x x2
Solution
4 x x 2 by completing the square.
2
4x x 2 x2 4 x x 2 4x 4 4 4 x 2
The we substitute a 2, u x 2 , and du dx to get
dx
dx
4x x2 4 ( x 2)2
We rewrite
du
2
a 2, u x 2
2
a u
u
x2
sin1 C sin 1
C
a
2
Example 4
a)
dx
1
1 x
tan
2
10 x 10 10 C
b)
dx
dx
7 3x2
7 3x
2
1
du
2
2
3 a u
a 7, u 3x
3x
1 1 1 u
1 1
tan C
tan 1
C
a
3 a
3
7
7
3x
1
tan 1
C
21
7
Example 5
dx
Evaluate
4 x2 4 x 2
Solution
We complete the square on the binomial 4 x
2
4x .
1
4
4 x2 4 x 2 4 x2 x 2 4 x2 x 2
4
4
2
1
2
4 x 1 2 x 1 1
2
Then we substitute a 1, u 2 x 1 du 2 dx to get
dx
dx
4 x2 4x 2 2 x 12 1
a 1, u 2x 1
1
du
2 u 2 a2
1 1
u
tan1
2 a
a
1
tan 1 2x 1 C
2
Example 6
Evaluate
x
dx
4x2 5
Solution
x
dx
2
4x 5
du / 2
u / 2
2
2
u 2x, a 5
u a
du
u u2 a2
2 x
1
u
1
sec1 C sec1
C
a
a
5
5
Example 7
Evaluate
dx
e2x 6
Solution
dx
e2 x 6
du / u
u ex , du ex dx, a 6
u2 a2
du
u u 2 a2
1
u
sec1 C
a
a
x
1
1 e
sec
C
6
6
PROBLEM SET
IP1.
0
Evaluate
6 dt
2
1 3 2t t
Solution:
0
6 dt
2
1 3 2t t
0
6 dt
2
1 4 t 2t 1
0
6
dt
2
2
1 2 t 1
0
1 t 1
6 sin
2 1
1
6 sin 1 sin1 0
2
π
6 0 π
6
P1.
1
6 dt
Evaluate
1 2 3 4t 4t
2
Solution:
1
1
6 dt
2
1 2 3 4t 4t
6 dt
2
1 2 4 4t 4t 1
1
6
dt
2
2
1 2 2 2t 1
1
6
2t 1
sin 1
2
2 1 2
1
3 sin 1 sin 1 0
2
π
π
3 0
6
2
IP2.
2
Evaluate
2
x 2 6 x 10 dx
1
Solution:
4
2
x 2 6 x 10
4
dx 2
2
1
x 2 6 x 9 1 dx
2
4
2
1
1 x 3 2
dx
2
4
2 tan 1 x 3
2
2 tan 1 4 3 tan 1 2 3
2 tan 1 1 tan 1 1
π π
2 π
4 4
P2.
2
Evaluate
8
x 2 2 x 2 dx
1
Solution:
2
2
8
1
x 2 2 x 2 dx 8 x2 2x 1 1 dx
1
1
2
8
1
1 x 12
dx
1
2
8 tan 1 x 1
1
8 tan 1 2 1 tan 1 1 1
8 tan 1 1 tan1 0
π
8 0 2π
4
IP3.
Evaluate
dx
x 1 x 2 2 x
Solution:
dx
x 1 x 2 2 x
Put
+1=
dx
x 1 x2 2 x
x 1
dx
x 1 x 12 1
⟹
=
du
sec 1 u C sec1 x 1 C
u u2 1
where is an arbitrary constant
P3.
2 2
Evaluate
1
Solution:
dx
x2 2 x 1 1
dy
y 4 y2 1
2 2
dy
To evaluate
y 4 y2 1
1
=2 ⟹
=2
Put
= −1 ⟹
Limits:
= −2 and
2 2
1
√
⟹
= −√2
2
dy
2
y 4y 1
=−
du
u u2 1
2
2
sec1 u
sec1 2 sec1 2
2
12
4 3
IP4.
e 4
Evaluate
4 dt
t 1 ln 2 t
1
Solution:
e 4
To evaluate
t 1 ln 2 t
1
⟹
=
Put
4 dt
=
Limits: = 1 ⟹
= 0 and =
e 4
1
⁄
⟹
=
4
4 dt
2
t 1 ln t
4
4du
4 tan 1 u
0
1 u2
0
4 tan 1 tan 1 0 4 tan 1
4
4
P4.
ln 3
Evaluate
e x dx
1 e2 x
0
Solution:
ln 3
To evaluate
e x dx
1 e2 x
0
Put
= ⟹
=
Limits: = 0 ⟹ = 1 and = √3 ⟹
ln 3
3
x
3
e dx
du
tan 1 u
1
1 e2 x
1 u2
0
1
tan 1 3 tan 11
= √3
3
4
12
EXERCISES
1. Evaluate the integrals:
a.
c.
e.
1 4 r 1
dx
17 x 2
dx
x
0
2
i.
0
b.
2
d.
f.
2
25 x 2
dx
1 4x2
dx
9 3x 2
dx
x
5 x2 4
3 2 4
1
g.
3 dr
4 ds
4 s2
dt
8 2t 2
h.
0
2
j.
2
ds
9 4 s2
dt
4 3t 2
m.
o.
3 dr
k.
1 4 r 1
dx
2 x 1
q.
2
2
2
2 x 1 4
csc 2 x dx
1 cot x
sec 2 y
2
2
2 cos d
2
r.
2
2
1 sin
p.
6
s.
4 r 1
dx
1 3x 1
n.
2
6 dr
dx
2x 1
4
l.
y dy
1 y4
dy
1 tan y
2. Evaluate the integrals:
a.
c.
e.
g.
i.
e
y
cos 1 x
dx
1 x2
dy
2
2y 5
x 2
e cos
1
x
d.
tan1 x dx
1 x2
2x x
dy
y 6 y 10
2
f.
x2 4 x 3
2
dx
2
dx
dx
1 x
b.
e
sin 1 x
1 x2
sin x
1
h.
j.
dx
2
dx
1 x2
dy
tan y 1 y
1
2
2
sin
k.
m.
dy
1
y 1 y2
l.
sec 2 sec 1 x dx
x x2 1
cos sec 1 x dx
x x2 1
4.11. Hyperbolic Functions
Learning objectives:
1. To find derivatives and integrals of hyperbolic functions.
And
2. To practice the related problems.
Every function that is defined on an interval centered at the origin can be written
in a unique way as the sum of one even function and one odd function. The
decomposition is
f x f x f x f x
f x
2
2
Even function
If we write
Odd function
this way, we get
e x e x e x e x
e
2
2
x
The even and odd parts of , called the hyperbolic cosine and hyperbolic sine of ,
respectively, are useful in their own right. They describe the motion of waves in
elastic solids, the shapes of hanging electric power lines, and the temperature
distributions in metal cooling fins.
Definitions
The hyperbolic cosine and hyperbolic sine functions are defined by the first two
equations in the table below. The table also lists the definitions of the hyperbolic
tangent, cotangent, secant, and cosecant. We will see the hyperbolic functions bear
a number of similarities to the trigonometric functions after which they are named.
1.
2.
e x e x
cosh x
2
ex ex
sinh x
2
3.
4.
5.
6.
sinh x e x e x
tanh x
x x
cosh x e e
cosh x e x e x
coth x
x x
sinh x e e
1
2
sec hx
x x
cosh x e e
1
2
csc hx
x x
sinh x e e
The graphs of the six hyperbolic functions are shown below.
Identities
Hyperbolic functions satisfy the identities in the table below. Except for differences
in the sign, these are identities already known for trigonometric functions.
1. sinh 2 x 2sinh x cosh x
2.
3.
4.
5.
cosh 2 x cosh 2 x sinh 2 x
cosh 2 x 1
cosh 2 x
2
cosh 2 x 1
sinh 2 x
2
cosh 2 x sinh 2 x 1
6.
7.
tanh 2 x 1 sec h 2 x
coth 2 x 1 csc h 2 x
Derivatives and Integrals
The six hyperbolic functions, being rational combinations of the differentiable
functions and
, have derivatives at every point at which they are defined.
Again, there are similarities with trigonometric functions.
1.
2.
3.
4.
5.
6.
d
du
sinh
u
cosh
u
dx
dx
d
du
cosh u sinh u
dx
dx
d
du
tanh u sec h 2u
dx
dx
d
du
coth u csc h 2u
dx
dx
d
du
sec
h
u
sec
h
u
tanh
u
dx
dx
d
du
csc hu csc hu coth u
dx
dx
The derivative formulas in the table above lead to the integral formulas in the table
below.
1.
2.
3.
4.
5.
6.
sinh u du cosh u C
cosh u du sinh u C
sech u du tanh u C
csch u du coth u C
sech u tanh u du sech u C
csch u coth u du csch u C
2
2
Example 1
d
d
tanh 1 t 2 sec h 2 1 t 2
1 t2
dt
dt
t
sec h 2 1 t 2
2
1 t
Example 2
cosh 5 x
dx
u sinh 5 x
sinh 5 x
1 du
5 u
1
1
ln u C ln sinh 5 x C
5
5
coth 5 xdx
Example 3
1
0
cosh 2 x 1
dx
0
2
sinh 2 x dx
1
1
1 1
1 sinh 2 x
cosh 2 x 1 dx
x
2 0
2 2
0
sinh 2 1
4
2
Example 4
ln 2
0
x
4e sinh x dx
ln 2
0
ln 2
e x e x
4e
dx 2e2 x 2 dx
0
2
x
ln 2
e2 x 2 x 0 e2 ln 2 2 ln 2 1 0
4 2 ln 2 1 3 2 ln 2
PROBLEM SET
IP1. If =
Solution:
Given, =
ℎ ( ) then
=?
ℎ ( ) = (sech( ))
=2
ℎ( )
(sech( ))
=2
ℎ( )
(sech( ))
=2
ℎ( ) (−
ℎ( )
ℎ( ))
=2
ℎ( ) (−
ℎ( )
ℎ( )) ( )
= −2
= −2
= −2 (
ℎ ( )
1−
ℎ( )
ℎ ( )
ℎ( ) −
ℎ( )
ℎ ( ))
( )
P1. If =
Solution:
Given, =
( −
), then
ℎ (1 −
ℎ )
(1 −
ℎ ) + (1 −
=
ℎ
=
ℎ
=
ℎ
=
ℎ (
=
ℎ
ℎ −
ℎ
=
ℎ
ℎ
ℎ
1
×−
ℎ
0−
×
ℎ
ℎ
ℎ
(
ℎ )
ℎ
ℎ )(−
ℎ )(−
ℎ +
ℎ )
+ (1 −
+ (1 −
ℎ ) + (1 −
IP2. If =
Solution:
Given =
=?
ℎ
−
ℎ
ℎ
ℎ )
ℎ )
ℎ
ℎ
ℎ )(−
ℎ
=?
ℎ −
ℎ
Differentiating on both sides, w.r.t , we get
(
=
(
=
(
=
ℎ +
=
ℎ (1 +
=
ℎ (
=
ℎ
P2. If = (
− )
Solution:
Given
= (4 − 1)
= (4 − 1 ).
= (4
ℎ ) − (2
=
ℎ )−
ℎ )
(
ℎ
ℎ ) )
ℎ )
)
)
)
)
(
)
ℎ(
2 )
=?
∵ csch
− 1).
− 1).
ℎ
ℎ
∵
∵
= (4
((
ℎ (−
ℎ (
ℎ
1
2
ℎ )−
= ;ℎ
=
1
;ℎ
=
=2
=2
ℎ )
= (4
(
− 1 ). (
)
)
=4
∴ =4
Differentiating on both sides w.r.t
IP3.
sech (√ )tanh (√ )
=?
√
Solution:
sech (√ )tanh (√ )
=
√
Put, √ = ⟹
=
√
sech (√ )tanh (√ )
=2
√
= 4.
, we get
⟹
√
=2
sech tanh
= −2sech +
= −2sech (√ ) +
P3.
ℎ(
Solution:
ℎ(
)
ℎ(
)
)
ℎ(
)
Put,
=
=
ℎ .
=−
= −
⟹
=?
=
=
ℎ
ℎ +
)+
ℎ(
IP4.
4
ℎ
∫
Solution:
4
ℎ
2
ℎ
We have,
∫
4
=?
ℎ
=
=
=2∫
⟹4
(
ℎ
ℎ − 1)
ℎ − 1)
= 2(
ℎ − )
= 2(
ℎ(
= 10 −
= 2(
= 2(
10) −
−2
ℎ − )
10) =
10 = 9.9 − 2
−
10
−2
10
P4.
2
ℎ
=?
ℎ
=
Solution:
2
+
2
2.
= ∫
+1
= ∫
+1
=
−
=
−
2 −
2−
=
−
+
=
+
2
=
+
−
+
4
4
Exercises
I. Rewrite the expression in terms of exponentials and simplify the results.
a) 2cosh (
)
b) sin (2
)
c)
d)
cosh 5 x sinh 5 x
cosh 3x sinh 3x
(sinh x cosh x)
4
e)
( ℎ +
f)
ℎ )+ ( ℎ −
ℎ )
II. Find the derivative of with respect to appropriate variable.
a) = 6 ℎ
b)
=
ℎ(2 + 1)
c) = 2√
ℎ√
d) =
ℎ
e) = ( ℎ )
f)
= ( ℎ )
g) =
ℎ (1 −
ℎ )
h) =
ℎ (1 −
ℎ )
i)
=
ℎ −
ℎ
j)
= ( + 1) ℎ( )
III. Evaluate the following indefinite integrals
a) ∫
ℎ2
b) ∫
ℎ
c) ∫ 6
d) ∫ 4
ℎ
−
ℎ(3 −
3
2)
e) ∫
ℎ
f) ∫
ℎ
√
g) ∫
ℎ
−
h) ∫
ℎ (5 − )
IV. Evaluate the following definite integrals
a) ∫
ℎ
ℎ2
b) ∫
c) ∫
4
ℎ
ℎ(
d) ∫
e) ∫ 2
ℎ(
(
f) ∫
g) ∫
h) ∫
)
)
)
√
√
ℎ
4.12. Inverse Hyperbolic Functions
Learning objectives:
To define six basic Inverse hyperbolic functions
To practice the related problems.
To find the derivatives and integrals of inverse hyperbolic functions
AND
We use the inverses of the six basic hyperbolic functions in integration. Since
d sinh x
cosh x 0 , the hyperbolic sine is an increasing function of x .We denote
dx
1
its inverse by y sinh x
1
For every value of x in the interval x , the value of y sinh x is
the number whose hyperbolic sine is x . The graphs of
y sinh1 x are shown in figure (a) below.
y sinh x and
Fig (a)
Fig (b)
Fig (c)
y cosh x is not one-to-one as we can see from its graph in the
previous module. But the restricted function y cosh x , x 0 , is one-to-one and
The function
therefore has an inverse, denoted by
1
y cosh x
1
For every value of x 1, y cosh x is the number in the interval 0 y
whose hyperbolic cosine is x . The graphs of y cosh x , x 0 , and
y cosh1 x are shown in figure (b) above.
Like y cosh x , the function y sech x 1 / cosh x fails to be one-to-one,
but its restriction to nonnegative values of x does have an inverse, denoted by
y sech -1 x
x in the interval (0, 1), y sech -1 x is the nonnegative
number whose hyperbolic secant is x . The graphs of y sech x , x 0 , and
For every value of
y sech -1 x
are shown in figure (c) above.
The hyperbolic tangent, cotangent, and cosecant are one-to-one on their domains
and therefore have inverses, denoted by
1
1
1
y tanh
x
,
y coth
The functions are graphed below.
x
,
y csch x
The following identities for inverse hyperbolic functions are useful.
1
sech 1x cosh 1
x
1
csch 1x sinh 1
x
1
coth1 x tanh 1
x
Derivatives and Integrals
The derivatives of inverse hyperbolic functions are given below.
d sinh 1 u
1
du
1.
dx
1 u 2 dx
d cosh1 u
2.
dx
d tanh 1 u
3.
4.
5.
dx
d coth1 u
dx
d sech 1u
dx
d csch 1u
1
du
u 2 1 dx
,
u 1
du
1 u 2 dx
u 1
du
1 u 2 dx
u 1
1
1
du
u 1 u 2 dx
1
0 u 1
du
u0
dx
dx
2
u 1 u
The chief use of inverse hyperbolic functions lies in integrations that reverse the
derivative formulas above.
1
The restrictions u < 1 and u > 1 on the derivative formulas for tanh u and
coth 1 u come from the natural restrictions on the values of these functions. The
6.
distinction between
1
u
< 1 and
u
> 1 becomes important when we convert the
derivative formulas into integral formulas. If
tanh 1 u + C. If u
> 1, the integral is
u
< 1, the integral of
coth 1 u
+ C.
1
1 u
2
is
Example 1
Show that if
then
u
is a differentiable function of
x whose values are greater than 1,
d
1
du
cosh1 u
dx
u 2 1 dx
Solution
y cosh1 x for x > 1:
x x cosh y
First we find the derivative of
1
y cosh
Differentiating both sides w.r.t , we get
dy
dy
1
1
1
1 sinh y
dx
dx sinh y
cosh2 y 1
x2 1
d
1
In short,
cosh1 x
. The Chain Rule gives the final result:
dx
2
x 1
d
1
du
cosh1 u
dx
u 2 1 dx
With appropriate substitutions, the derivative formulas, given earlier, lead to the
integration formulas given below:
1.
2.
3.
4.
du
u
sinh1 C ,
a
a2 u2
a0
du
u
cosh1 C
a
u2 a2
ua0
1
1 u C if
tanh
a
du
a
a 2 u 2 1 coth1 u C if
a
a
du
1
u
sech 1 C
a
a
u a2 u2
du
1
u
csch 1 C
a
a
u a2 u2
Example 2
1
2 dx
Evaluate
0 3 4x2
5.
u2 a2
u2 a2
0 u a
u0
Solution
The indefinite integral is
2 dx
du
3 4 x2
a2 u2
,
u 2 x, a 3
2x
u
sinh 1 C sinh 1
C
a
3
Therefore,
1
2dx
1
2 x
1
1 2 sinh 1 0
sinh
sinh
3 0
3
0 3 4x 2
2
sinh 1
3
Domain and Range of Inverse Hyperbolic Functions:
Domain ( )
Inverse Hyperbolic
function = ( )
=
ℎ
=
ℎ
=
ℎ
=
ℎ
=
ℎ
=
ℎ
Range ( )
ℝ
[1, ∞)
(−1, 1)
ℝ − [−1, 1]
(0, 1]
ℝ − {0}
ℝ
[0, ∞)
ℝ
ℝ − {0 }
[0, ∞)
ℝ − {0}
Formulas for Inverse Hyperbolic Functions in terms of Natural logarithms.
1.
ℎ
=
+ √ + 1 , −∞ < < ∞
2.
ℎ
=
+√ −1 , ≥ 1
3.
ℎ
=
4.
ℎ
=
5.
ℎ
=
6.
ℎ
=
Proof of 1:
Let ∈ ℝ and =
=
⟹
ℎ =
=
(
)
⟹2
, | |<1
√
+
,0 <
√
| |
,
≤1
≠0
,| | > 1
ℎ
. Then
=
=
=
(
)
=(
) −1
⟹ ( ) − (2 )
This is a quadratic equation in
. Therefore,
−1 =0
=
Since
Thus,
Hence,
±√
=
±√
> 0 for all ∈ ℝ and
= +√ +1
=
+√ +1
<√
=
Proof of 2:
Let ∈ [1, ∞) and
=
⟹
ℎ =
=
(
)
=
ℎ
=
=
=(
−1=
=
=
=
=
+√
+√
(
)
) − (2 )
+1 =0
. Therefore,
±√
−1
< 1 , since
√
≥ 1 for all
Further
Thus,
±√
, −∞ <
) +1
This is a quadratic equation in
−√
+
. Then
⟹(
But
+ 1, we get
+
⟹2
=
+1
>1
≥0
−1
−1
Hence,
=
Proof of 3:
Let ∈ (−1, 1) and
=
⟹
=
ℎ =
⟹ [
⟹
+
=
ℎ
=
+
=
,
. Then
=
+ 1] =
+
(
)
−1
−1
⟹ 1+
=
−
≥
<∞
⟹
⟹
−
= 1+
⟹
(1 − ) = 1 +
=
⟹2 =
(Note that
⟹
> 0 for
∈ (−1, 1))
=
Hence,
+
−
=
, | |<1
Similarly, we can prove the other formulas.
Example
13
6 dx
Evaluate the integral
in terms of Natural logarithms.
2
0
1 9x
Solution:
13
6 dx
To evaluate
0
1 9x 2
Put = 3 ⟹
=3
Limits:
=0⟹ =0 ;
= ⟹ =1
13
1
6 dx
du
2
0
0 1 u2
1 9 x2
1
2 sinh 1 u 2 sinh 11 2 sinh 1 0
0
2 sinh 11 2 ln 1 1 12 2 ln 1 2
Example
2
dx
Evaluate the integral
in terms of Natural logarithms.
5 4 1 x2
Solution:
2
2
5
c oth 1 x
coth 1 2 coth 1
2
5 4
4
5 4 1 x
dx
1 2 1 1 5 4 1 1
1 1
ln
ln
ln 3 ln 9 ln
2 2 1 2 5 4 1 2
2 3
PROBLEM SET
IP1. If =
√ + then find
Solution:
Given =
ℎ 2√ + 1
Differentiating both sides w.r.t , we get
=
2√ + 1
√
=
=
(
.
√
=
2.
)
(
( + 1)
√
(1 + 0)
√
)(
)
=√
∴
=√
P1.
Find the derivative of
a. =
−
b. = ( − )
Solution:
a. Given =
w.r.t the appropriate variable
−
ℎ
Differentiating w.r.t
=
(
)−
(
= −√
−
= −√
− −√
=−
+
√
=−
∴
on both sides, we get
−
ℎ
+
, 0<
=−
+ (1)
√
−
√
ℎ
)
ℎ
ℎ
ℎ
,0<
ℎ
,0 <
, 0<
<1
<1
<1
, 0<
<1
b. Given = (1 − )
ℎ
Differentiating w.r.t on both sides, we get
]
= [(1 − )
ℎ
= (1 −
)
= (1 −
)(
= 1−2
[
ℎ
)
ℎ
]+
+ (0 − 2 )
, | |>1
(1 −
ℎ
ℎ
)
, | |>1
<1
∴
= 1−2
, | |>1
ℎ
(
) then find
IP2. If =
Solution:
Given
=
ℎ ( ℎ )
Differentiating w.r.t on both sides, we get
= [ ℎ ( ℎ )]
= −|
[
|√
=−
ℎ ]
√
=−
(∵
√
ℎ
−
ℎ
= 1)
=−
=−
=−
ℎ
=−
ℎ
∴
P2. If = (
Solution:
Given = (
)
+
( + ) then find
+2 )
ℎ ( + 1)
Differentiating both sides w.r.t
=
[(
=(
+2 )
=(
+2 )
=(
+2 )
=
(
(
+2 )
)
+
)
ℎ ( + 1)]
[
ℎ
(
)
( + 1)] +
2( + 1)
= 2( + 1)
ℎ ( + 1)
ℎ ( + 1)
ℎ ( + 1) − 1
3 13
15
( + 1)
( + 1 ) + (2 + 2 )
IP3.
Evaluate
ℎ
(1 ) + 2 ( + 1 )
= −1 + 2( + 1)
∴
, we get
1
x 1 16 x 2
dx
(
+2 )
ℎ ( + 1)
ℎ ( + 1)
Solution:
3 13
To evaluate
1
x 1 16x
2
dx
15
=4 ⟹
Put
Limits: = ⟹
Now,
=4
=
⟹
=
;
3 13
=
⟹
=
12 13
12 13
du
1
dx
sec h u
u
4
4 5
2
2
x 1 16 x
1 u
15
45 4
1
1
12
4
sec h1 sec h 1
13
5
3 13
15
1
dx sech 1
x 1 16 x2
12
4
sec h 1
13
5
P3.
Evaluate
1
dx
2
9 x 25
Solution:
1
dx
2
9 x 25
Put
=
⟹
1
9 x 2 25
1
5
5
1
9 x2
25
dx
1
1
2
3x 1
5
dx
=
dx
1
5
5 du
3 u2 1
1
1
3x
cosh 1 u C cosh 1 C
3
3
5
1
3x
dx cosh 1 C
3
5
9 x 2 25
1
where
is an arbitrary constant
IP4.
Evaluate
dx
2x 2 5x 6
Solution:
dx
2x 2 5x 6
1
2
1
2
1
2
dx
5
x2 x 3
2
dx
2
x 5 3 25
4
16
dx
2
x 5 23
4 4
2
x5 4
1
1
4x 5
sinh 1
sinh1
C
C
2
23
4
2
23
where C is an arbitrary constant
P4.
x 1
Evaluate
dx
2
x x 1
Solution:
To evaluate
Put
Now,
=
−
∴
x 1
dx
x2 x 1
+1⟹
= (2 − 1)
+ 1 = (2 − 1) +
1
3
2
x
1
x 1
2 dx
dx 2
x2 x 1
x2 x 1
1
2x 1
3
1
dx
dx
2
2
2
2
x x 1
x x 1
1
2
d 2
x x 1
3
dx
dx
2
x2 x 1
1
3
. 2 x2 x 1
2
2
1
x 1 2 2
3
4
1
x 1 2 2
3 2
2
dx
dx
x 1 2
3
x2 x 1 sinh1
C
2
32
3
2x 1
x2 x 1 sinh1
C
2
3
x 1
3
2x 1
dx x2 x 1 sinh1
C
2
2
3
x x 1
where
is an arbitrary constant
EXERCISES
1. Find the derivatives of with respect to the appropriate variable:
a. =
ℎ √
b. = (1 − )
ℎ
c. = (1 − ) ℎ √
d. =
+ √1 −
ℎ
e.
=
ℎ
f.
g.
h.
=
=
=
ℎ (2)
ℎ (
ℎ (
)
)
2. Evaluate the following integrals:
13
2 3
a.
0
dx
4 x2
dx
1 x
dx
1 x
e.
2
54
2
2
0
x
f.
1 9 x2
0
12
2
d.
6 dx
b.
1
dx
g.
4 x2
cos x dx
x 1 sin
2
0
x
e
h.
dx
x 1 ln x
i.
2
1
j.
l.
n.
2x 5
x2 2x 10
dx
2x2 5x 6
dt
t 2 t 1
dx
k.
m.
dx
x 2 2 x 10
dx
2x2 x 1
dx
x2 6 x 13
5.1. Fundamental Principle of Counting
Learning objectives:
To study the Product rule and its Extended Version.
To study the Sum rule and its Extended Version.
AND
To practice the related problems.
The mathematical theory of counting is formally known as combinatorial analysis. The
principle of counting is basic to the combinatorial analysis.
We present two basic counting principles, the Product Rule and the Sum Rule.
Product Rule:
Suppose that a procedure can be broken down into a sequence of two tasks. If there
are ways to do the first task and for each of these ways of doing the first task,
there are ways to do the second task, then there are
ways to do the
procedure.
Proof:
The product rule may be proved by enumerating all the possible ways of the two tasks
as follows:
1,1 ,1,2 ,
2,1 , 2 ,2 ,
,1,n
, 2,n
m,1 , m,2 ,
, m,n
We say that the outcome is i, j if task 1 results in its
th
possible way and task 2
then results in the th of its possible ways. Hence the set of possible outcomes consists
of m rows, each row containing n elements, which proves the result.
Example 1:
A small community consists of 10 women, each of whom has 3 children. If one woman
and one of her children are to be chosen as mother and child of the year, how many
different choices are possible?
Solution:
By regarding the choice of the woman as the outcome of the first task and the
subsequent choice of one of her children as the outcome of the second task, we see
from the product rule that there are 10 × 3 = 30 possible choices.
When there are more than two tasks to be performed, the product rule can be
extended as follows:
Extended Version of the Product Rule:
Suppose that a procedure is carried out by performing the tasks , , … ,
in
sequence. If each task , = 1, 2, … , can be performed in ways, regardless of how
the previous task were done, then there are . . … .
ways to carried out the
procedure.
This can be proved by mathematical induction and the product rule for two tasks.
Example 2: An engineering college planning committee consists of 3 first year
students, 4 second year students, 5 pre-final year students, and 2 final year students.
A subcommittee of 4, consisting of 1 person from each class, is to be chosen. How
many different subcommittees are possible?
Solution:
We may regard the choice of a subcommittee as the combined outcome of the four
separate tasks of choosing a single representative from each of the classes. Hence, it
follows from the extended version of the product rule that there are
3 4 5 2 120 possible subcommittees.
Example 3:
How many different 7-place license plates are possible if the first three places are to
be occupied by letters and the later 4 by numbers? How many license plates would be
possible if repetition among letters or numbers were prohibited?
Solution:
By the extended version of the product rule, the number of license plates is
26 26 26 10 10 10 10 17 ,57,60,000
In the second case there would be
26 25 24 10 9 8 7 7 ,86,24,000
possible license plates.
Example 4:
How many functions defined on n points, are possible if each function value is either
0 or 1?
Solution:
Let the points be 1, 2, . … , n . Since f i must be either 0 or 1 for each
i 1,2, ,n , it follows that there are
2 ×2 ×2 ×…×2 = 2
possible functions.
We now introduce the Sum Rule
The Sum Rule:
If a task can be done either in one of ways or in one of ways, where none of the
set of ways is the same as any of the set of
ways, then there are + ways
to do the task.
Example5: Suppose there are 6 male professors and 3 female Professors teaching
calculus. In how many ways a student can choose a calculus Professor?
Solution:
By sum rule, a student can choose a calculus professor in 6 + 3 = 9 ways.
We can extend the Sum Rule to more than two tasks.
Extended Version of Sum Rule:
Suppose that a task can be done in one of ways, in one of ways, … , or in one of
ways, where none of the set ways of doing the tasks is the same as any of the
set ways, for all pairs and with 1 ≤ < ≤ . Then the number of ways to do
the task is + + ⋯ +
.
The extended version of the sum rule can be proved using mathematical induction
from the sum rule of two sets.
Example6:
A student can choose a computer project from one of three lists. The three lists
contain 13, 5 and 9 possible projects and no project is on more than one list. In how
many ways a student can choose a computer project?
Solution:
A student can choose a project by selecting a project from the first list, the second list
or the third list. Since no project is on more than one list, by extended sum rule a
student can choose 13 + 5 + 9 = 27 ways.
PROBLEM SET
IP1. Find the number of 4 letter words, with or without meaning, which can be
formed out of the letters of the word ROSE, where the repetition of the letters is not
allowed.
Solution:
There are as many words as there are ways of filling in 4 vacant places by the 4 letters
without repetition as follows.
The first place can be filled in 4 different ways by anyone of the 4 letters.
The second place can be filled by anyone of the remaining 3 letters in 3 different ways
The third place can be filled by anyone of the remaining 2 letters in 2 different ways
The fourth place can be filled in 1 way. Hence by the extended product rule, the
required number of words is = 4 × 3 × 2 × 1 = 24
P1. A code word consists of two distinct English alphabets followed by two distinct
numbers between 1 and 9. For example C A 2 3.
I.
How many such code words are there?
II.
How many such code words end with an even integer?
Solution:
I.
By the hypothesis, a code word consists of two distinct English alphabets followed
by two distinct numbers from 1 to 9.
We have 26 English alphabets and 9 digits (1 to 9).
To form such a code word, we have to choose the first alphabet in 26 ways and
the second alphabet in 25 ways.
Again out of 9 digits, first digit can be chosen in 9 ways and the second digit can
be chosen in 8 ways.
Hence by extended product rule, the number of such distinct codes
= 26 × 25 × 9 × 8 = 46800
II. Two distinct alphabets can be selected in 26 × 25 ways and the Unit’s place
can be filled in 4 ways (i.e., by 2, 4, 6, 8). Tenth place can be filled in 8 ways
(since one of the digits is already used).
Thus, the number of desired codes is
= 26 × 25 × 4 × 8 = 20800.
IP2. Find the number of different signals that can be generated by arranging at least
2 flags in order (one below the other) on a vertical staff, if five different flags are
available.
Solution:
A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us count the
possible number of signals consisting of 2 flags, 3flags, 4 flags and 5 flags separately.
There will be as many 2 flag signals as there are ways of filling in 2 vacant places in
succession by the 5 different flags available.
5
4
By product rule, we have 5 × 4 = 20
Similarly, there will be as many 3 flag signals as there are ways of filling in 3 vacant
places in succession by the 5 different flags.
5
4
3
By product rule, we have 5 × 4 × 3 = 60
Continuing the same way, we find that
The number of 4 flag signals= 5 × 4 × 3 × 2 = 120 and the number of 5 flag
signals= 5 × 4 × 3 × 2 × 1 = 120
Now, by sum rule, the number of desired signals
= 20 + 60 + 120 + 120 = 320
P2. How many 3-digit even numbers can be formed by the digits , , , , without
the repetition of the digits?
Solution:
The unit’s place can be filled either by 2 or by 4 to get 3-digit even number with digits
1, 2, 3, 4, 5.
Suppose the unit’s place is filled by 2. Since the digits cannot be repeated the ten’s
place can be filled in 4 ways and having filled ten’s place, the hundred’s place can be
filled by the remaining 3 digits. Thus, by the product rule, the number of 3-digit even
numbers ending with 2 formed from the digits 1, 2, 3, 4, 5 is 4 × 3 = 12.
Similarly, the number of 3-digit even numbers ending with 4 formed from the digits
1, 2, 3, 4, 5 is 12.
By sum rule, the number of 3-digit even numbers formed by the digits 1, 2, 3, 4, 5
without the repetition of the digits is
12 + 12 = 24
IP3. In forming 5 letter (with distinct letters) words using the letters of the word
EQUATIONS.
a. How many begin with a consonant
b. How many in which the vowels and consonants alternate.
c. How many in which Q is immediately followed by U.
Solution:
The given word EQUATIONS contains totally 9 letters in which we have 5 vowels and 4
consonants.
a.
C
4
8
7
6
5
After filling the first position with any one of the 4 consonants, there are 8
letters remaining.
By extended product rule, the number of 5 letter words that begin with
consonants is
= 4 × 8 × 7 × 6 × 5 = 6720
b. By
extended product rule, the number of 5 letter words that contains the
vowels and consonants alternately is
V
C
V
C
V
OR
C
V
C
5
4
4
3
3
4
5
3
= (5 × 4 × 4 × 3 × 3) + (4 × 5 × 3 × 4 × 2) = 1200
V
C
4
2
c. First we place Q so that U may follow it (Q may occupy any one of the first four
positions but not the last). Next we place U (in only 1 way), and then we fill the
three other positions from among 7 letters remaining.
Q
U
By the extended product rule, the number of 5 letter words in which Q is
immediately followed by U is
= (4 × 1 × 7 × 6 × 5) = 840
P3. In forming 5 letter words (with distinct letters) using the letters of the word
EQUATIONS?
d. How many consists only of vowels.
e. How many contain all of the consonants
f. How many begin with E and end in S
Solution:
The given word EQUATIONS contains totally 9 letters in which we have 5 vowels and 4
consonants.
d. There are five places to be filled and 5 vowels are at our disposal.
5
4
3
2
1
By extended product rule, the number of 5 letter words that consists only
vowels is
= 5 × 4 × 3 × 2 × 1 = 120
e. Each word is to contain 4 consonants and one of the 5 vowels. A vowel (V)
has 5 choices. Now, the vowel can be placed in any one of the 5 places and
we can fill the remaining 4 positions with consonants (C).
For example:
V
C
C
C
C
By extended product rule, the number of 5 letter words that consists of all
consonants is
= 5(5 × 4 × 3 × 2 × 1) = 600
f.
E
S
7
6
5
Now, there are just 3 positions to be filled and 7 letters are
at our disposal.
By extended product rule, the number of 5 letter words that begin with E and
end with S is
= 7 × 6 × 5 = 210
IP4. If repetitions are not allowed.
a. How many 3-digit numbers can be formed with the digits
, , , , , , , , , ?
b. How many of these are odd numbers?
c. How many of these are even numbers?
d. How many are divisible by 5?
e. How many are greater than 600?
Solution:
a.
≠0
The first position can be filled in 9 ways, since 0 cannot be used. The middle
position can be filled in 9 ways since 0 can be used and the last position can
filled in 8 ways.
Thus, by extended product rule, the required number of 3-digit numbers is
9 × 9 × 8 = 648
b. We have
≠0
Odd
First, we fill the last position with the 5 digits 1,3,5,7,9 in 5 ways.
Now, we fill the first position with the remaining 8 digits in 8 ways.(since one
odd digit and 0 are excluded)
The middle position can be filled with the remaining 8 digits in 8 ways (since
two digits are now excluded)
Thus, the number of 3-digit odd numbers is
= 8 × 8 × 5 = 320
c. The number of 3-digit even numbers
= 648 − 320 = 328 (How?)
(The students are encouraged to do it by the direct method)
d. A number is divisible by 5 if and only if it ends with 0 or 5
First we form all numbers ending in 0 as follows
9
8 1
0
Next, we form all the numbers ending in 5 as follows
8
8
1
5
≠0
Hence the 3-digit numbers that are divisible by 5 is
(9 × 8 × 1) + (8 × 8 × 1) = 72 + 64 = 136
e. To get the numbers greater than 600,
4
9
8
6
First, we fill the first position with any one of the digits 6, 7, 8 or 9 in 4 ways.
The middle position can be filled with the remaining 9 digits in 9 ways and the
last position can be filled with remaining 8 digits in 8 ways.
Thus, the number of 3-digit numbers greater than 600 is = 4 × 9 × 8 = 288.
P4. How many numbers are there between 100 and 1000 which have exactly one of
their digits as 7?
Solution:
The numbers between 100 and 1000 having 7 as exactly one of their digits can be
classified into three types.
(i) When the unit’s place has 7:
Here the ten’s place can have any one of the digits except 7. It can be filled in 9
different ways. The hundred’s place can have any one of the digits except 0 and
7. So hundred’s place can be filled in 8 different ways. Therefore, there are
9 × 8 = 72 such numbers.
(ii) When the ten’s place has 7:
The unit’s place can be filled in 9 different ways. It can have any one of the
digits except 7. The hundred’s place can have any one of the digits except 0
and 7. So hundred’s place can be filled in 8 different ways. So, there are
9 × 8 = 72 such numbers.
(iii) When the hundred’s place has 7:
Here the unit’s place can be filled by 9 different ways (except 7) and ten’s place
can be filled by 9 different ways (except by 7). So there are 9 × 9 = 81 such
numbers.
Hence the number of desired numbers is
= 72 + 72 + 81 = 225
EXERCISES
1. Given 4 flags of different colors, how many different signals can be generated, if
the signal requires the use of 2 flags one below the other?
2. How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the
digits can be repeated?
3. Using the letters of the word MARKING and calling any arrangement a word,
a. How many different 7-letter words can be formed,
b. How many different 3-letter words can be formed?
4. If repetitions are allowed:
a. How many three-digit numbers can be formed with the digits 0, 1, 2, 3, 4, 5,
6, 7, 8, 9?
b. How many of these are odd numbers?
c. How many are even numbers?
d. How many are divided by 5?
5. Find the number of 4 letter words that can be formed using the letters of the word
EQUATION. How many of these words begin with E? How many end with N? How
many begin with E and end with N?
6. Find the number of 4-digit numbers that can be formed using the digits 2, 3, 5, 6, 8
(without repetition).
a. How many of them are divisible by 2?
b. How many of them are divisible by 5?
c. How many of them are divisible by 25?
5.2. Permutations
Learning objectives:
To define a permutation and a circular permutation on a finite set.
To determine the number of permutations and circular permutations on a set
with elements.
AND
To practice the related problems.
We wish to know the possible ordered arrangements of the letters
direct enumeration we see that there are 6: namely
a, b and c . By
abc,acb,bac,bca,cab,cba
Each arrangement is known as a permutation. Thus there are 6 possible permutations
of a set of 3 objects. The result could also have been obtained from the product rule,
since the first object in the permutation can be any of the 3, the second object in the
permutation can then be chosen from any of the remaining 2, and the third object in
the permutation is then chosen from the remaining 1. Thus there are 3 2 1 6
possible permutations.
Suppose that we have n objects. Reasoning similar to that we have just used for the
three letters shows that there are
n n 1 n 2 3 2 1 n!
different permutations of the n objects.
n! is read “ n factorial” and is defined by
n! n n 1 n 2 3 2 1
It is the product of all the consecutive integers from n down to 1.
The notation
For example,
6! 6 5 4 3 2 1 720
Note:
The expression 0! is defined to be 1.
Example 1: How many different batting orders are possible for a baseball team
consisting of 9 players?
Solution:
There are 9! = 3,62,880 possible batting orders.
Example 2:
A class in probability theory consists of 6 boys and 4 girls. An examination is given, and
the students are ranked according to their performance. Assume that no two students
obtain the same score.
(a)
How many different rankings are possible?
(b) If the boys are ranked just among themselves and the girls among themselves,
how many different rankings are possible?
Solution:
(a) As each ranking corresponds to a particular ordered arrangement of the 10
people, we see that the answer to this part is 10! = 3,628,800.
(b) As there are 6! possible rankings of the boys among themselves and 4! possible
rankings of the girls among themselves, it follows from the product rule that there are
(6!)(4!) = (720)(24) = 17,280 possible rankings in this case.
Example 3:
Mr. Jones has 10 books that he is going to put on his bookshelf. Of these, 4 are
mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language
book. Jones wants to arrange his books so that all the books dealing with the same
subject are together on the shelf. How many different arrangements are possible?
Solution:
There are 4! 3! 2! 1! arrangements such that the mathematics books are first in line,
then the chemistry books, then the history books, and then the language book.
Similarly, for each possible ordering of the subjects, there are 4! 3! 2! 1! possible
arrangements. Hence, as there are 4! possible orderings of the subjects, the desired
answer is 4! 4! 3! 2! 1! = 6912.
Circular permutation of objects:
An arrangement of distinct objects in definite order in a circle is called a circular
permutation.
Example 4:
In how many ways can 10 boys be arranged (a) in a straight line, (b) in a circle?
Solution:
(a) The boys may be arranged in a straight line in 10! Ways.
(b) We first place a boy at any point on the circle. The other 9 boys may then be
arranged in 9! Ways.
This is an example of a circular permutation.
In general n objects may be arranged in a circle in n 1 ! ways.
Example: 5
If the letters of the word MASTER are permuted in all possible ways and the words
thus formed are arranged in dictionary order, find the ranks of the words i) REMAST
ii) MASTER.
Solution:
i) The letters of the given word in dictionary order
A, E, M, R, S, T
In the dictionary order, first we write all words that begin with A. If we fill the first
place with A, then the remaining 5 places can be filled with remaining 5 letters in 5!
ways. That is, there are 5! words that begin with A. Proceeding like this, after writing
all words that begin with A, E, M we have to write the words that begin with R. Among
them first come the words with first two letters R, A. As above there are 4! such
words. On proceeding like this we get
A
⟶
5! words
E
⟶
5! words
M
⟶
5! words
R
A
⟶
4! words
R
E
A
⟶
3! words
R
E
M
A
S
T
⟶
1 word
The rank/serial number of the word REMAST is
= 3(120) + 24 + 6 + 1 = 391
ii) The letters of the given word in dictionary order
A, E, M, R, S, T
In the dictionary order, first we write all words that begin with A. If we fill the first
place with A, then the remaining 5 places can be filled with remaining 5 letters in 5!
ways. That is, there are 5! words that begin with A. Proceeding like this, after writing
all words that begin with A, E we have to write the words that begin with M. Among
them first come the words with first three letters M, A, E. As above there are 3! such
words. On proceeding like this we get
A
E
M
M
M
M
M
A
A
A
A
A
E
R
S
S
S
E
R
T
E
⟶
⟶
⟶
⟶
⟶
⟶
⟶
R
The rank/serial number of the word MASTER is
= 2(120) + 2(6) + 2(2) + 1 = 257
PROBLEM SET
IP1. Find the value of
(
if
)!
)!
!(
:
!
)!
!(
=
:
Solution:
(
Given that
We have
(
(
⟹
(
)!
!(
(
!(
!
:
)!
=
!
)!
)(
!(
) (
(
)
)
)(
)!
)
)!
⟹
II.
=
= 121
×
(
)!
)(
)!
=
=6
!
then find
II.
If ( + )! =
Solution:
I.
+ != !
!
= 11!
!(
×
=
⟹
=
⟹ 2 − 1 = 11 ⟹
P1.
I.
If ! + ! =
⟹
= 44: 3
)!
!(
)!
!(
⟹
)!
!
!
+
× !
( !) then find
!
+
=
!
×
!
!
+
!
!
!
= 110 + 11 = 121
( + 2)! = 2550( !)
⟹ ( + 2)( + 1) ! = 2550( !)
⟹ ( + 2)( + 1) = 2550
⟹
+ 3 − 2548 = 0
⟹ ( − 49 )( + 52 ) = 0
⟹ = 49, since = −52 ∉
5! words
5! words
3! words
3! words
2! words
2! words
1 word
IP2. Find the number of ways of arranging History, Economics and Civics books
in a shelf (in a row) such that the books on the same subject are together.
Solution:
Consider 5 History books as one unit, 4 Economics books as a second unit and 4 Civics
books as a third unit.
The three units can be arranged in 3! ways. Moreover,
5 History books are arranged themselves in 5! ways
4 Economics books are arranged themselves in 4! ways
4 Civics books are arranged themselves in 4! ways
The number of the required arrangements
= 3! × 5! × 4! × 4! = 6 × 120 × 24 × 24 = 4,14,720
P2. Find the number of ways of arranging the letters of the word KRISHNA in which
all the vowels come together
Solution:
The word KRISHNA has 7 letters in which there are two vowels namely I and A. Treat
the vowels as one unit.
Thus, we have 5 consonants + 1 unit vowels = 6 things, which can be arranged in 6!
Ways. Now the vowels can be arranged among themselves in 2! Ways. Therefore, by
product rule, the number of arrangements in which the 2 vowels come together is
6! × 2! = 720 × 2 = 1440
IP3. Ten guests are to be seated in a row. Three of them are to be seated together.
Of the remaining two guests do not wish to sit side by side. Find the number of
possible arrangements.
Solution:
Treat the 3 guests who are to be seated together as one unit. Then we have 7 guests
+ 1 unit of 3 guests. They can be arranged in 8! ways and the one unit of 3 guests can
be permuted in 3! ways. Therefore,
Number of permutations of 10 guests in which 3 guests are always sit together is
= 8! × 3!
Now, if the two guests who donot wish to sit side by side, are considered to be sitting
side by side, then the number of permutations is = 7! × 3! × 2!
Therefore, the number of permutations of 10 guests so that 3 particular guests are
seated together and two particular guests do not sit side by side is
= 8! × 3! − 7! × 3! × 2!
= 7! × 3! (8 − 2) = 1,81,440
P3. Find the number of ways of arranging
of these arrangements
a. All the girls are together
b. Boys and girls come alternately
boys and
girls in a row. In how many
Solution:
The number of ways of arranging 6 boys and 6 girls in a row is 12! ways.
a. Treat 6 girls as one unit. Then we have6 boys + 1 unit girls
They can be arranged in 7! ways and the one unit of girls can be permuted in 6!
ways. Hence the number of arrangements in which all 6 girls are together is 7! × 6!
b. Let us take 12 places. The row may begin with either a boy or a girl which can be
arranged in 2 ways.
If it begins with a boy, then all the odd places (1, 3, 5, 7, 9, 11) will be occupied by
boys and the even places (2, 4, 6, 8, 10, 12) occupied by girls.
The 6 boys can be arranged in the 6 odd places in 6! Ways and the 6 girls can be
arranged in the 6 even places in 6! Ways. Thus the number of arrangements in
which boys and girls come alternately is 2 × 6! × 6!
IP4. The countries Japan, China and Russia sent 15, 14 and 13 representatives for a
round table conference to discuss the international cooperation among them. Find
the number of ways of these representatives sits at a round table so that
a. All Japanese are together
b. Representatives of same nationality together
Solution:
a. Treat 15 Japanese as one unit. Then we have
14 Chinese + 13 Russians + 1 unit of Japanese = 28 entities.
They can be arranged at a round table conference in
(28 − 1)! = 27! ways
Now, the 15 Japanese among themselves can be arranged in 15! ways.
Hence the required arrangement is 27! × 15!
b. Treating 3 nationalities as three units, can arrange at a round table in (3 − 1)! =
2! ways.
Now, 15 Japanese among themselves can be permuted in 15! ways.
Similarly, 14 Chinese in 14! ways
13 Russians in 13! ways
Hence the required number of ways arrangements is
2! × 15! × 14! × 13!
P4. Find the number of ways of arranging 8 persons around a circular table if two
particular persons were to sit together.
Solution:
Treat the two particular persons as one unit. Then we have 6 + 1 = 7 entites. They
can be arranged around a circular table in (7 − 1)! = 6! ways.
Now, the two particular persons can be permuted among themselves in 2!
Therefore, the number of arrangements is 6! × 2! = 1,440
EXERCISES
1. Evaluate ( ) 5!
( ) 7!
( ) 7! – 5!
7!
12!
(b)
5!
10 ! 2 !
n!
when n = 5, r
r ! n r !
2. Compute (a)
3. Evaluate
= 2.
4. Simplify
n!
n 1 !
n 2 !
b.
n!
r 1 !
c.
r 1 !
1 1
x
8! 9 ! 10!
a.
5. If
then the value of
6. Find the number of ways that 4 people can sit in a row of 4 seats.
7. A family has 3 boys and 2 girls.
a. Find the number of ways they can sit in a row.
b. Find the number of ways they can sit in a circle.
c. Find the number of ways the boys and girls sit in a row alternately.
8. A debating team consists of 3 boys and 3 girls. Find the number n of ways they
can sit in row where:
a. there are no restrictions,
b. the boys and girls sit alternately,
c. just the girls are to sit together.
d.
9. Find the number n of ways 5 large books, 4 medium-size books, and 3 small books
can be placed on a shelf so that all books of the same size are together.
10.Find the number of different 8-letter arrangements that can be made from the
letters of the word DAUGHTER so that
a. All vowels occur together
b. All vowels do not occur together.
5.3. Permutations with Repetitions
Learning objectives:
To find the number of permutations of a set with
elements when certain
elements are indistinguishable from each other.
AND
To practice the related problems.
We shall now determine the number of permutations of a set of
certain of the objects are indistinguishable from each other.
n
objects when
Example 1: How many different letter arrangements can be formed using the letters
P E P P E R?
Solution: We first note that there are 6! permutations of the letters P1E1P2 P3 E2 R
when the 3P’s and the 2E’s are distinguished from each other. However, consider any
one of these permutations – for instance, P1P2 E1P3 E2 R . If we now permute the P’s
among themselves and the E’s among themselves, then the resultant arrangements
would still be of the form P P E P E R. That is, all 3! 2! permutations
P1 P2 E1 P3 E2 R
P1P2 E2 P3 E1 R
P1 P3 E1P2 E2 R
P1P3 E2 P2 E1 R
P2 P1 E1 P3 E2 R
P2 P1E2 P3 E1 R
P2 P3 E1 P1 E2 R
P2 P3 E2 P1 E1 R
P3 P1E1P2 E2 R
P3 P1E2 P2 E1 R
P3 P2 E1 P1 E2 R
P3 P2 E2 P1 E1 R
are of the form P P E P E R. Hence there are 6!/(3! 2!) = 60 possible letter
arrangements of the letters P E P P E R.
In general, the same reasoning as that used in example 1 shows that there are
n!
n1 ! n2 ! nr !
different permutations of n objects, of which n1 are alike, n2 are alike, …, nr
are
alike.
Example 2: A chess tournament has 10 competitors of which 4 are from Russia, 3 are
from the United States, 2 from Great Britain, and 1 from Brazil. If the tournament
result lists just the nationalities of the players in the order in which they placed, how
many outcomes are possible?
Solution: There are
10 !
12 ,600 possible outcomes.
4 ! 3! 2 !1!
Example 3: How many different signals, each consisting of 9 flags hung in a line, can
be made from a set of 4 white flags, 3 red flags, and 2 blue flags if all flags of the same
color are identical?
Solution: There are
9!
1,260 different signals.
4! 3! 2 !
Example 4: Find the number m of seven-letter words that can be formed using the
letters of the word “BENZENE”.
Solution: We seek the number of permutations of seven objects of which three are
alike, the three E’s, and two are alike, the two N’s. Therefore,
m
7!
420
3! 2!
Example 5: A set of snooker balls consists of a white, a yellow, a green, a brown, a
blue, a pink, a black and 15 reds. How many distinguishable permutations of the balls
are there?
Solution: In total there are 22 balls, the 15 reds being indistinguishable.
Thus, the number of distinguishable permutations is
22 !
22 !
85 ,95 ,41,760
1! 1! 1! 1! 1! 1!1! 15! 15 !
PROBLEM SET
IP1. Find the number of 5 digit even numbers that can be formed using the
digits , , , , .
Solution: To find 5 digit even numbers, fill the units place by 2 and the remaining
!
4 places can be arranged using the remaining digits 1, 1, 2, 3 in
= 12 ways.
!
Thus, the number of 5 digit even numbers that can be formed using the digits
1, 1, 2, 2, 3 is 12.
P1. In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S
(ii) All the vowels are occur together
Solution: Notice that the given word PERMUTATIONS has 12 letters in which there are
2 T’s and all the other letters occur only once.
(i) If P and S are fixed at the extreme ends (P at the left end and S at the right end),
then 10 letters are left in which we have 2 T’s
P
S
10 letters
!
Hence, the required number of arrangements is = !
(ii) There are 5 vowels in the given word namely E, U, A, I, O
each appearing only once.
Treat 5 vowels as one unit. Thus, we have 8 objects (i.e., 7 letters + 1 unit).
i.e.,
E U A IO
7 letters
!
In these 8 objects there are 2 T’s, which can be arranged in ! ways.
Corresponding to each of these arrangements, the 5 different vowels can be
arranged in 5! ways.
!
Therefore, the required number of arrangements is × 5!
!
IP2. Find the number of ways of arranging the letters of the word SPECIFIC. In how
many ways can be arranged that the two C’s come together.
Solution:
a. The given word has 8 letters in which there are 2 I’s and 2 C’s. Hence, they can
be arranged in
!
=
= 10,080 ways
! !
b. Teat the 2 C’s as one unit, then we have, 6 letters+1 unit = 7 objects in which
two letters (I’s) are alike.
!
Hence, they can be arranged in = ! = 2520 ways
!
!
Now, the 2 C’s among themselves can be arranged in
Thus, the number of required arrangements is 2520.
= 1way.
P2. Find the number of ways of arranging the letters of the word SINGING so that
a. They begin and end with I
b. The two G’s come together
Solution:
a. First we fill first and last places with I’s in
!
!
= 1 way as shown below
Now, we fill the remaining 5 places with the remaining
!
5 letters S, N, G, N, G in
= 30 ways.
! !
I
I
Hence, the number of required permutations is 30.
b. Treat the two G’s as one unit. Then we have, 5 letters + 1 unit = 6 objects, in
which we have 2 I’s and 2N’s.
Hence, they can be arranged in
!
! !
= 180 ways
Now, the G’s among themselves can be arranged in
number of required permutations is 180.
!
!
= 1 way. Hence the
IP3. How many numbers greater than , ,
can be formed by using the
digits , , , , , , ?
Solution:
Notice that 10,00,000 is a 7-digit number and the number of given digits is 7.
Therefore, the numbers to be counted will be 7-digit only.
Also, the numbers have to be greater than 10,00,000, so they have to begin with
1, 2 or 4:
The numbers begins with 1:
1
0
2
2
2
4
4
1 is fixed at the extreme left position and the remaining digits to be rearranged will
be 0,2,2,2,4,4
!
Therefore, the number of numbers begins with 1 is =
= 60
! !
The numbers begins with 2:
2
1
0
2
2
4
4
2 is fixed at the extreme left position, the remaining digits to be rearranged will be
1,0,2,2,4,4
!
Therefore, the number of numbers begins with 2 is = ! ! = 180
The numbers begin with 4:
4
1
0
2
2
2
4
4 is fixed at the extreme left position, the remaining digits to be rearranged will be
1,0,2,2,2,4
!
Therefore, the total number of numbers begins with 4 is = ! = 120
Thus, the required number of numbers is
= 60 + 180 + 120 = 360
P3. In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a
row if the discs of the same color are indistinguishable.
Solution:
Total number of discs is 4 + 3 + 2 = 9
Out of 9 discs, 4 are red, 3 are yellow and 2 are green
!
∴ The number of arrangements is
= 1260
! ! !
IP4. How many arrangements can be made with the letters of the word
MATHEMATICS? In how many of them vowels are together?
Solution:
There are 11 letters in the given word MATHEMATICS of which we have 2 M’s, 2 A’s, 2
T’s and all other are distinct.
!
Therefore, the required number of arrangements is = !× !× !
There are 4 vowels i.e., A, E, A, I.
Treat 4 vowels as one unit and in the remaining 7 letters (i.e., M,T,H,M,T,C,S) we have
(1 unit + 7 letters) 2 M’s, 2 T’s and the rest are different.
!
These 8 objects can be arranged in !× ! ways.
!
But the 4 vowels (A, E, A, I) can be put together in ways.
!
Hence, the total number of arrangements in which vowels are always together is
!
!
= !× ! × ! = 1,20,960
P4. How many different words can be formed by using all the letters of the word
ALLAHABAD?
a. In how many of them vowels occupy the even positions?
b. In how many of them both L do not come together?
Solution:
There are 9 letters in the given word ALLAHABAD, in which 4 A’s, 2 L’s and the rest are
all distinct.
!
So, the required number of words is =
= 7560
! !
a. There are 4 vowels i.e., 4 A’s and all are alike. Also, there are 4 even places i.e.,
2, 4, 6, 8.
!
These 4 even places can be occupied by 4 vowels in
= ! = 1way
Now, we are left with 5 places in which 5 letters, of which 2 L’s are alike and
!
other are distinct, can be arranged in ways.
!
Hence, the total number of words in which vowels occupy the even places is
!
!
!
= ! × ! = ! = 60
b. Treat 2 L’s as one unit and we have 8 objects (7 letters + 1 unit), out of which
!
repeats 4 times and others are distinct. These 8 letters can be arranged in
!
ways.
!
So, the number of words in which both L’s come together is ! = 1680
Hence, the number of words in which both L’s do not come together
= Total no. of words – no. of words in which both L’s
come together
= 7560 − 1680 = 5880
EXERCISES
11.Find the number n of distinct permutations that can be formed from all the letters
of each word
a. THOSE
b. UNUSUAL
c. SOCIOLOGICAL
d. QUEUE
e. COMMITTEE
f. PROPOSITION
g. BASEBALL
12.Find the number n of different signals, each consisting of 6 flags hung in a vertical
line, which can be formed from 4 identical red flags and 2 identical blue flags.
13.Find the number n of different signals, each consisting of 8 flags hung in a vertical
line, which can be formed from 4 identical red flags, 2 identical blue flags, and 2
identical green flags.
14.Find the number of arrangements of the letters of the word INDEPENDENCE. In
how many of these arrangements,
a. do the words start with P
b. do all the vowels always occur together
c. do the vowels never occur together
d. do the words begin with I and end in P?
15.How many permutations can be made of the letters, taken all together, of the
“word” MASSESS?
a. In how many ways will the four S’s be together?
b. How many will end in SS?
16.Find the number of 5-digit numbers that can be formed using the digits 1, 1, 2, 2, 3.
( Ans: 230)
17.Garlands are formed using 4 red roses and 4 yellow roses of different sizes. In how
many of them
a. All 4 red roses come together.
b. Red roses and yellow roses come alternately.
(Ans: 288, 72)
18.How many ways can the letters of the word ENGINEERING be arranged so that the 3
N’s come together but the 3 E’s do not come together?(Ans: 13,860)
19.How many permutations can be made of the letters of the word ARRANGEMENT? In
how many of these the vowels occur together? (Ans: 2491800, 10800)
20.Find the number of 5-digit numbers that can be formed using the digits 2, 2, 3, 3, 4.
How many of them are greater than 30,000? (Ans: 30, 18)
5.4. Ordered Samples
Learning objectives:
To define a − permutation of a set with
for the number of −permutations of
elements and to derive a formula
To introduce the concept of sampling and to derive the formulae for the
number of ordered samples of size with and without replacement from a set
with elements.
AND
To practice the related problems.
a, b and c , taken all at a time, were discussed in
a previous module and they are abc,acb,bac,bca,cab,cba .
The permutations of a set of letters
Any arrangement of a set of objects in a definite order is called a permutation of the
set taken all at a time.
If a set contains n objects, any ordered arrangement of any r n of the objects is
called permutation of the n objects taken r at a time. For example,
ab,ac,ad ,bc,bd ,cd ,
ba,ca,da,cb,db,dc
are permutations of n = 4 letters a,b,c,d
The number of permutations of
n
taken
objects taken
r
r
= 2 at a time
at a time is denoted by
nP ,
r
n
where r n . We now derive the general formula for Pr .
There are n different ways to choose the first element; following this, the second
element can be chosen in n − 1 ways; and, following this, the third element can be
th
chosen n − 2 ways. Continuing in this manner, we have that the r element can be
chosen in n r 1 n r 1 ways. Thus, by the fundamental principle of
counting, we have
n
Pr n n 1 n 2 n r 1
n n 1 n 2 n r 1 n r !
n r !
n!
n
Pr
n r !
r n , then n Pn n n 1 n 2 3 2 1 n!
There are n! permutations of n objects taken all at a time and the same result we
If
obtained in the previous module.
If
n
r 0,
P0
n!
1
n 0 !
n
The symbol P0 is the number of arrangements which have no objects at all in the
arrangement. It means all the objects are left behind, and there is only one way of
n
P0 = 1.
n
n
n
Example 1: Solve for n , given (a) P2 110 , (b) P4 30 P2
doing so. Therefore,
Solution:
n
(a)
P2 n n 1 110
n 2 n 110 0
, Since
n
is positive,
n
= 11.
n 11 n 10 0
(b) We have
n n 1 n 2 n 3 30 n n 1
n n 1 n 2 n 3 30 n n 1 0
n n 1 n 2 n 3 30 0
n n 1 n 2 5n 24 0
n n 1 n 8 n 3 0
Since n 4 , the required solution is n = 8.
Example 2:
How many different numbers can be formed by using six out of nine
digits 1, 2, 3, … , 9?
Solution:
Here we have 9 different things and we have to find the number of permutations of
them taken 6 at a time. Therefore,
The desired answer 9 P6 9 8 7 6 5 4 60480
Many problems in combinatorial analysis are concerned with choosing an element
from a set S consisting of n elements. A card from a deck or a person from a
population is an example. When we choose one element after another from the set
S , say r times, we call the choice as an ordered sample of size r . The following
two cases are of common occurrence.
(i)
Sampling with replacement
(ii) Sampling without replacement
In sampling with replacement, the element is replaced in the set S before the next
element is chosen. Since there are n different ways to choose each element
(repetitions are allowed), the product rule principle tells us that there are
r
times
n n n n n r
different ordered samples with replacement of size r .
In sampling without replacement, the element is not replaced in the set S before the
next element is chosen. Thus, there are no repetitions in the ordered sample.
Accordingly, an ordered sample of size r without replacement is simply a r permutation of the elements in the set S with n elements. Thus, there are
n!
n r !
different ordered samples without replacement of size r from a population (set) with
n elements. In other words, by the product rule, the first element can be chosen in n
ways, the second in n − 1 ways, and so on.
n
Pr n n 1 n 2 n r 1
Example 3:
Three cards are chosen in succession from a deck of 52 cards. Find the number of
ways this can be done: (a) with replacement
(b) without replacement.
Solution:
(a)
Since each card is replaced before the next card is chosen, each card can be
chosen in 52 ways. Thus,
52 52 52 523 140 ,608
is the number of different ordered samples of size r 3 with replacement.
(b) Since there is no replacement, the first card can be chosen in 52 ways, the
second card in 51 ways, and the last card in 50 ways. Thus,
52
P3 52 51 50 132 ,600
is the number of different ordered samples of size
r 3
without replacement.
Example 4:
Suppose repetitions are not allowed. (a) Find the number n of three-digit numbers
that can be formed from the six digits: 2, 4, 5, 8, 7, 9. (b) How many of them are even?
(c) How many of them exceed 400?
Solution:
There are 6 digits, and we have to form three-digit numbers.
(a) There are 6 ways to fill the first position, 5 ways for the second position, and 3
ways for the third position. Thus,
n 6 5 4 120
Alternatively, n is the number of permutations of 6 things taken 3 at a time,
and so
n 6 P3 6 5 4 120
(b)
Since the numbers must be even, the last digit must be either 2 or 4. Thus, the
third position is filled first and it can be done in 2 ways. Then there are now 5
ways to fill the middle position and 4 ways to fill the first position.
Thus, 4 5 2 40 of the numbers are even.
(c)
Since the numbers must exceed 400, they must begin with 5, 8, 7, 9. Thus,
we first fill the first position and it can be done in 4 ways. Then there are 5 ways
to fill the second position and 4 ways to fill the third position. Thus,
4 5 4 80 of the numbers exceed 400.
Example 5:
A class contains 8 students. Find the number of ordered samples of size 3: (a) with
replacement, (b) without replacement.
Solution:
(a)
Each student in the ordered sample can be chosen in 8 ways; hence there
3
are 8 8 8 8 512 samples of size 3 with replacement.
The first student in the sample can be chosen in 8 ways, the second in 7
ways, and the last in 6 ways. Thus, there are 8 7 6 336 samples of size
3 without replacement.
(b)
PROBLEM SET
IP1. If
Solution:
Given
⟹
⟹
⟹
⟹
⟹
⟹
⟹
P1. If
5
Pr 2 6 Pr 1
5
Pr 2 6 Pr 1
!
(
)!
!
(
)!
!
(
then find the value of .
!
= 2 .(
=
=
)!
(
× × !
(
)!
(
))!
× !
)(
)(
(By definition)
)!
(7 − )(6 − ) = 12
− 13 + 30 = 0
( − 10)( − 3) = 0
= 3 , (∵ ≤ = 5)
( 2 n 1 )
Pn 1 :
( 2 n1 )
Pn 3 : 5
then find the value of
Solution:
( 2 n1 )
Given
Pn 1 :
( 2 n 1 )
Pn 3 : 5
( 2 n 1 )
Pn1 3
( 2 n 1 )
Pn
5
⟹
⟹
(
(
(
(
)!
×
)!
)(
)(
(
)!
= (By definition)
(
)!
)(
)!
(
)!
×
=
) (
)!
(
)!
⟹ 5 (4 + 2) = 3[( + 2)( + 1)]
⟹ 3 − 11 − 4 = 0
⟹ (3 + 1)( − 4) = 0
(∵ ∈ )
⟹ =4 ,
IP2. 18 guests have to be seated half on each side of a long table. 4 particular guests
desire to sit on one particular side and 3 other on the other side. Determine the
number of ways in which the sitting arrangements can be made?
Solution:
Assume that two sides of a long table be and .
Now, 4 particular guests to be sit on the side of the table in 9 chairs in 9 P ways.
4
After this arrangement, again 3 particular guests to be sit on the side
9 chairs in 9 P ways.
of the table in
3
Now, the remaining 11 guests can be permuted on 11 chairs on both sides of the
table, in 11! ways.
Hence, by the product rule, the total number of ways in which 18 guests can be
seated 9 P 9 P 11!
4
3
P2. There are 8 students appearing in an examination of which 3 have to appear in
mathematics paper and the remaining 5 in different subjects. In how many ways can
they be made to sit in a row if the candidates in mathematics cannot sit next to each
other?
Solution:
The total number of candidates is = 8.Now, the 5 different subjects candidates can
5
be seated in P5 5 ! ways.
In between 5 candidates there are 6 places for 3 mathematics candidates. Therefore,
the mathematics candidates can be seated in 6 P ways.
3
Hence, by product rule, the required number of ways they can sit in a row such that
no two candidates in mathematics sit side by side is
5! 6 P 14, 400
3
IP3: Find the number of 5- letter words that can be formed using the letters of the
word EXPLAIN that begin and end with a vowel when repetitions are allowed.
Solution:
We can fill the first and last places with vowels each in 3 ways (A or E or I).
Now, each of the remaining 3 places can be filled in 7 ways (using any letter of the
given 7 letters).
Hence the number of 5 letter words which begin and end with vowels when the
repetitions are allowed is
3 × 7 = 9 × 343 = 3087
P3. Find the number of 4 letter words that can be formed using the letters of the
word PISTON in which at least one letter is repeated.
Solution:
The given word has 6 letters. The number of 4 letter words that can be formed using
these 6 letters
i.
when repetition is allowed is 6 = 1296
ii.
when repetition is not allowed is 6 P 6 5 4 3 360
4
Hence, the number of 4 letter words in which at least one letter repeated is
64 6 P 1296 360 936
4
IP4: Find the number of 4- digit numbers that can be formed using the digits 1, 2, 3,
4, 5, 6 that are divisible by
(i) 2 (ii) 3
when repetition is allowed
Solution:
(i) Numbers divisible by 2
For a number to be divisible by 2, the unit’s place should be filled with an even
digit. This can be done in 3 ways (2 or 4 or 6)
X
Now, each of the remaining 3 places can be filled in 6 ways.
Hence the number of 4- digit numbers that are divisible by 2 is
3 × 6 = 3 × 216 = 648
(ii) Numbers divisible by 3
Fill the first 3 places with the given 6 digits in 6 ways.
X X X
Now, after filling up the first 3 places with three digits, if we fill up the units
place in 6 ways, we get 6 consecutive positive integers. Out of any six
consecutive integers exactly two are divisible by 3. Therefore, the units place
can be filled in 2 ways.
Hence the number of 4 − digit numbers divisible by 3 is = 2 × 216 = 432.
P4. Find the sum of all − digit numbers that can be formed using the
digits , , , , . (if the repetition of digits not allowed).
Solution: We know that the number of 4 − digits numbers that can be formed using
5
the given 5 digits is P4 120 . Now we find their sum.
We first find the sum of the digits in the unit place of all these 120 numbers. If we fill
the units place with 1 as shown below, then the
1
4
remaining 3 places can be filled with the remaining 4 digits in P3 24 ways.
4
This means, the number of 4 digit numbers having 1 in units place is P3 .
Similarly, each of the digits 3, 5, 7, 9 appears 24 times in units place. By adding all
these digits we get the sum of the digits in unit’s place of all 120 numbers is
4
P3 1 4 P3 3 4 P3 5 4 P3 7 4 P3 9 4 P3 25
Similarly, we get the sum of the digits in Ten’s place is
4
P3 25 . Since it is in 10’s
4
place, its value is P3 25 10 .
Similarly, the values of the sum of the digits in 100’s place and 1000’s place are
4
P3 25 100
and
4
P3 25 1000
respectively.
Hence the sum of all the 4-digit numbers formed by using the digits 1, 3, 5, 7, 9
4
P3 25 1 4 P3 25 10 4 P3 25 100 4 P3 25 1000
4 P3 25 1111 ...(1) 24 25 1111 6,66, 600 .
Note:
1. From (1) in the above example, we can derive that the sum of all - digit numbers
that can be formed using the given ‘ ’ non – zero digits (1 ≤ ≤ ≤ 9) is
n 1
Pr 1
Sum of the given digits × 111…1 ( times)
2. In the above, if ‘0’ is one digit among the given digits, then we get that the sum
of the − digit numbers that can be formed using the given digits (including ‘0’)
={
−{
n 1
Pr 1
n 2
Pr 2
sum of the given digits × 111…1 ( times)}
sum of the given digits × 111…1 (( -1) times)}
EXERCISES
1. Find
n
if :
(a)
n
P2 72
(b)
2 n P2 50 2 n P2
n such that
n
n
(a) P5 42 P3 , n 4
n
P
5
n4
(b) n 1 4
P4 3
2. Find the value of
3. Find
r , if 5 4 Pr 6 5 Pr 1 , n 4
4. Find the number n of ways a judge can award first, second, and third places in a
contest with 18 contestants.
5. A box contains 12 light bulbs. Find the number
of ordered samples of size 3:
(a) with replacement
(b) without replacement.
6. A class contains 10 students. Find the number n of ordered samples of size 4:
(a) with replacement (b) without replacement.
7. How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of
digits is not allowed?
8. How many numbers lying between 100 and 1000 can be formed with the
digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?
9. In how many ways can 5 prizes be given away to 4 boys, when each boy is eligible
for all the prizes?
10.How many different words can be formed of the letters of the word MALENKOV so
that
a. no two vowels are together
b. The relative position of the vowels and consonants remains unaltered?
5.5. Combinations
Learning objectives:
To introduce the concept of combinations.
To derive a formula for the number of possible combinations of
taken
AND
elements
at a time
To practice the related problems.
We are often interested in determining the number of different groups of r objects
that could be formed from a total of n objects. For instance, how many different
groups of 3 could be selected from the 5 items A,B,C,D,E ? We reason as follows.
Since there are 5 ways to select the initial item, 4 ways to then select the next item,
and 3 ways to select the final item, there are thus 5 4 3 ways of selecting the group
of 3 when the order in which the items are selected is relevant. However, since every
group of 3, say, the group consisting of items A,B,C , will be counted 6 times (that
is, all of the permutations ABC , ACB , BAC , BCA , CAB , CBA will be
counted when the order of selection is relevant), it follows that the total number of
groups that can be formed is
543
10
3 2 1
In general, as n n 1 n r 1 represents the number of different ways that
a group of r items could be selected from n items when the order of selection is
relevant, and as each group of r items will be counted r ! times in this count, it
follows that the number of different groups of r items that could be formed from a
set of n items is
n n 1 n r 1 n! n r !
r!
r!
Notation and terminology
n
, for r n , by
r
n
n!
r n r !r!
We define
n
represents the number of possible combinations of n objects
r
n
taken r at a time. The notation Cr is also used for the number of possible
n
n
combinations of n objects taken r at a time. The two notations and Cr are
r
and say that
equivalent.
By convention, 0! is defined to be 1. Thus
n n
0 n 1
We also take
n
i 0
when either i 0 or i n .
n
Thus represents the number of different groups of size r that could be selected
r
from a set of n objects when the order of selection is not considered relevant.
The number of combinations of n things taken
n things taken n r at a time.
r
at a time is equal to the number of
This is seen as follows. To each group of r things we select, there is left a
corresponding group of n r things; that is, the number of combinations of n things
r at a time is the same as the number of combinations of things n r at a time.
Therefore,
n n
r n r
n n
r
n
If we put
, then 1
0 n
Example 1: A committee of 3 is to be formed from a group of 20 people. How many
different committees are possible?
Solution: There are
20
C3
20 19 18
1140 possible committees.
3 2 1
Example 2: From a group of 5 women and 7 men, how many different committees
consisting of 2 women and 3 men can be formed? What if 2 of the men are feuding
and refuse to serve on the committee together?
3 men, it follows from the basic principle that there are =
5
7
Solution: As there are 2 possible groups of 2 women, and 3 possible groups of
5
2
5 4 7 6 5
2 1 3 2 1
7
3
= 350 possible committees consisting of 2 women and 3 men.
On the other hand, if 2 of the men refuse to serve on the committee together, then
possible groups of 3 men not containing either of the 2 feuding
men and groups of 3 men containing exactly 1 of the feuding men, it follows
that there are + = 30 groups of 3 men not containing both of the
feuding men. Since there are ways to choose the 2 women, it follows that in this
case there are 30 = 300 possible committees.
there are 0
2
5
3
2
1
5
2
2
0
5
3
2
1
5
2
5
2
5
2
Example 3: Consider a set of n antennas of which m are defective and n m are
functional and assume that all of the defectives and all of the functionals are
considered indistinguishable. How many linear orderings are there in which no two
defectives are consecutive?
Solution:
Imagine that the n m functional antennas are lined up among themselves. Now, if
no two defectives are to be consecutive, then the spaces between the functional
antennas must each contain at most one defective antenna. That is, in the n m + 1
possible positions, represented by , we must select m of these in which to put the
defective antennas.
1 1 1 1
Hence there are
n m1
m
1 functional
place for atmost one defective
possible orderings in which there is at least one functional
antenna between any two defective ones.
Example 4:
A hand of 13 playing cards is dealt from a well-shuffled pack of 52 (a) what is the
number of distinct hands? (b) What is the number of hands containing two aces?
Solution:
The order of the cards in the hand is immaterial.
(a) The total number of distinct hands is equal to the number of combination of 13
objects drawn from 52:
The desired answer is
52
13 .
4
2
(b) The number of hands containing two aces is equal to the number of ways,
, in which the two aces can be drawn from the four available, multiplied by the
48
, in which the remaining 11 cards in the hand can be
11
number of ways,
drawn from the 48 cards that are not aces:
The desired answer is
4 48
2 11 .
Example 5:
A class contains 10 students with 6 men and 4 women. Find the number n of ways:
a) A 4-member committee can be selected from the students.
b) A 4-member committee with 2 men and 2 women.
c) The class can elect a president, vice president, treasurer and secretary.
Solution:
(a)
This concerns combinations, not permutations, since order does not count in
a committee.
10 10 9 8 7
n
210
4 4 3 2 1
(b)
6
ways and the 2 women
2
The 2 men can be chosen from the 6 men in
4
can be chosen from the 4 women in ways. Thus, by the product rule,
2
6 4 6 5 4 3
n
90
2 2 2 1 2 1
(c)
This concerns permutations, not combinations, since order does count. Thus,
n 6 5 4 3 360
Example 6:
A box contains 7 blue socks and 5 red socks. Find the number n of ways two socks
can be drawn from the box if: (a) They can be any color; (b) They must be the same
color.
Solution:
(a)
There are “12 choose 2” ways to select 2 of the 12 socks.
12 12 11
n
66
2
2
1
(b)
7
5
There are 21 ways to choose 2 of the 7 blue socks and 10
2
2
ways to choose 2 of the 5 red socks. By the sum rule,
n 21 10 31
PROBLEM SET
n 1C 8: 5 then find the value of
IP1. If nC3 :
4
?
n 1C 8: 5 nC 8 n 1C
Solution: We have nC3 :
4
3
4
⟹5(
⟹(
)(
!
)! !
= 8(
(
)!
)(
)!
(
=(
)!
)! !
(
)!
)!
⟹ 5 = 2[( − 3)( − 4)]
⟹ (2 − 3)( − 8) = 0
⟹
= 8, (∵
∈
⟹2
− 19 + 24 = 0
⟹ 2 − 3 = 0,
−8=0⟹
= 8,
)
P1. If n Pr 840 and nCr 35 then find the value of
?
n
Solution: We have
Cr
35
n
Pr 840
!
1
( − )! !
⟹
=
⟹ ! = 24 ⟹ ! = 4! ⟹
!
24
( − )!
⟹ n P4 840 7 6 5 4
⟹
n
P4 7 P4 ⟹
=4
=7
IP2. Find the number of ways of selecting 11 member cricket team from 7 bats men,
6 bowlers and 2 wicket keepers so that the team contains atleast 4 bowlers and one
wicket keeper?
Solution:
Therefore, the number of ways of selecting the required cricket team is
= 210 + 252 + 70 + 315 + 210 + 35 = 1092
P2. A committee of 5 members is to be formed out of 6 gentlemen and 4 ladies. In
how many ways that can be done when at least two ladies are included? In how
many ways that can be done when at most two ladies are included?
Solution:
a. Given that there are 6 gentlemen and 4 ladies. To select a committee of 5
members there are 3 possibilities included.
a) 3 gentlemen and 2 ladies
b) 2 gentlemen and 3 ladies
c) 1 gentleman and 4 ladies
Hence, the required number of ways (by product and sum rules) is
6C3 4C2 6C2 4C3 6C1 4C4
6!
4! 6!
4! 6!
4!
3! 3! 2! 2! 4! 2! 1! 3! 5!1! 0! 4!
120 60 6 186
b. A committee of 5 persons, consisting of at most two ladies, can be constituted
in the following ways:
6
I. Selecting 5 gentlemen out of 6 (no ladies). This can be done in C5 ways.
II. Selecting 4 gentlemen out of 6 and one lady out of 4. This can be done in
6
C3 4C1 ways.
III. Selecting 3 gentlemen out of 6 and two ladies out of 4. This can be done is
6
C3 4C2
ways.
Therefore, the total number of ways of forming the committee is
6
C5 6C4 4C1 6C3 4C2 6 60 120 186
IP3: How many diagonals are there in a polygon with sides?
Solution: A polygon of sides has vertices. By joining any two vertices of a polygon,
we obtain either a side or a diagonal of the polygon. Number of line segments
obtained by joining the vertices of a − sided polygon taken two at a time is
n C2
= Number of ways of selecting 2 out of
n( n 1 )
2
Out of these lines, n lines are the sides of the polygon.
∴ Number of diagonals of the polygon =
(
)
−
=
(
)
Application: If a polygon has 44 diagonals then find the number sides of the
polygon.
Solution:
(
)
We know that the number of diagonals of sided polygon is
By the hypothesis, we have
(
)
= 44
⇒
− 3 − 88 = 0
⇒ ( − 11 )( + 8) = 0
⇒ = 11 (∵ ∈ )
Hence, there are 11 sides for the polygon with 44 diagonals.
P3: A box contains 5 red balls and 6 white balls. In how many ways can 6 balls be
selected so that there are at least two balls of each color?
Solution:
The selection of 6 balls, consisting of at least two balls of each colour from 5 red and 6
white balls, can be made in the following ways:
I.
By selecting 2 red balls out of 5 and 4 white balls out of 6. This can be done in
5
C2 6C4
II.
ways.
By selecting 3 red balls out of 5 and 3 white balls out of 6. This can be done in
5
C3 6C3 ways.
III.
By selecting 4 red balls out of 5 and 2 white balls out of 6. This can be done in
5
C4 6C2
ways.
By the sum rule, the total number of ways to select the 6 balls with at least two balls
of each color is
5C2 6C4 5C3 6C3 5C4 6C2
10 15 10 20 5 15 425
IP4. What is the number of ways of choosing 4 cards from a pack of 52 playing
cards?
In how many of these
a) four cards are of the same suit,
b) four cards belong to four different suits,
c) are face cards,
d) two are red cards and two are black cards,
e) Cards are of the same color?
Solution:
The number of ways of choosing 4 cards from a pack of 52 playing cards is
52
C4
52!
52 !
52 51 50 49
2,70 ,725
4 ! 52 4 ! 4 ! 48!
4 3 2 1
Observe the following diagram:
a) There are four types of suits namely diamond, club, spade, heart and there are
13
13 cards in each suit. Therefore, there are C4 ways of choosing 4 diamonds.
13
13
Similarly, there are C4 ways of choosing 4 clubs, C4 ways of choosing 4
13
spades and C4 ways of choosing 4 hearts. Therefore,
the required number of ways
13C4 13C4 13C4 13C4
4
by sum rule
13!
2 ,860
4! 9 !
b) There are13 cards in each suit.
13
Therefore, there are C1 ways of choosing 1 card from 13 cards of diamond,
13
C1 ways of choosing 1 card from 13 cards of hearts, 13C1 ways of choosing
13
1 card from 13 cards of clubs, C1 ways of choosing 1 card from 13 cards of
spades. Hence, by product rule, the required number of ways
13C1 13C1 13C1 13C1 134
( By product rule)
c) There are 12 face cards and 4 are to be selected out of these 12 cards. This can
12
be done in C4 ways. Therefore, the required number of ways
12C4
12!
495
4 ! 8!
d) There are 26 red cards and 26 black cards. Therefore, the required number of
ways
2
26!
2
26
26
C2 C2
325
1,05,625
2! 24 !
e) 4 red cards can be selected out of 26 red cards in
26
be selected out of 26 black cards in C4 ways.
Therefore, the required number of ways
26
C4 26C4 2
26
C4 ways. 4 black cards can
26 !
29,900
4 ! 22!
P4. A person invites 13 guests to a dinner party and arranging 8 of them at one
circular table and the remaining 5 at the other. In how many ways can he arrange
the guests?
Solution: 13 guests in the party, can be divided into two groups of 8 and 5 in
13
C8 or
13!
.
C5 , i.e.,
5! 8 !
13
The first group of 8 guests can be arranged around at one table in (8 − 1)! = 7! ways.
The second group of 5 guests can be arranged around at the other table in (5 − 1)! =
4! ways.
Therefore, the number of arrangements is
13!
13 !
13 !
4! 7!
4! 7 !
15 ,56 ,75 ,520
5! 8!
5.4 ! 8.7 !
40
EXERCISES
1.
a. If
n
C9 nC8 then
5
b. Find the value of
n
C17
5
Cr
(Ans: 31)
r 1
n
n
n
c. If Cr : Cr 1 1 : 2 and
Cr 1 : nCr 2 2 3 then find
the values of and (Ans: = 14, = 4)
2. A committee of 3 persons is to be constituted from a group of 2 men and 3
women. In how many ways can this be done? How many of these committees
would consist of 1 man and 2 women?
3. A student is to answer 8 out of 10 questions on an exam.
a. Find the number n of ways the student can choose the eight questions.
b. Find n if the student must answer the first three questions.
4. Find the number n of committees of 5 with a given chairperson that can be
selected from 12 persons.
5. From 12 books in how many ways can a selection of 5 be made,
a. when one specified book is always included,
b. when one specified book is always excluded?
6. From 7 Englishmen and 4 Americans a committee of 6 is to be formed; in how
many ways can this be done,
a. when the committee contains exactly 2 Americans,
b. at least 2 Americans?
7. Out of 7 consonants and 4 vowels, how many words can be made each containing
3 consonants and 2 vowels?
8. A man has 7 friends. In how many ways can he invite one or more friends to a
party?
(Ans: 127)
9. A man has 7 relatives, 4 of them are ladies and 3 gentlemen, his wife has 7
relatives and 3 of them are ladies and 4 gentlemen. In how many ways can they
invite them to a dinner party of 3 ladies and 3 gentlemen so that there are 3 of
man’s relatives and 3 of wife’s relatives? (Ans: 485)
10.A question paper contains 12 questions, divided into three parts A, B and C. Part A
contains 6 questions while B and C contain 3 questions each. A candidate is
required to attempt 6 questions selecting atleast two from part A and atleast one
from each of part B and part C. In how many ways can the candidate select 6
questions?
(Ans: 720)
6.1 Binomial Theorem for Positive Integral Indices
Learning objectives:
To study the Binomial Theorem for Positive Integral Indices.
And
To practice the related problems.
By actual multiplication
( + ) = +
( + ) =
+2 +
( + ) =
+3
+3
( + ) =
+4
+6
( + ) =
+5
+ 10
+
+4
+
+ 10
+5
These expansions suggest that, when
( + ) =
+
+
(
+
(
)(
+
is a positive integer,
)
∙
)
+⋯+
∙ ∙
+
--(1)
The coefficients in the expansion are denoted by
= ,
1
2
(
=
)
∙
,
3
=
(
)(
)
∙ ∙
,...
The coefficient of any term may be expressed as
=
(
)(
)…(
∙ ∙ ∙∙∙(
)
)
The numbers denoted by the symbol
=(
!
)! !
, where and
are positive integers with
n
≤ [read: “ Cr ” or “ choose r”], are called the binomial coefficients, since they
appear as the coefficients in the expansion of ( + ) .
The expansion in (1) is known as binomial theorem expansion. Thus, the binomial
theorem for positive integral indices is given by
( + ) =
=
=
+
+
1
−2
+
2
+
+⋯+
−1
+
+
…+
) .
Example 1: Expand ( +
(3 + 2 ) = (3 ) + (3 ) (2
∙
∙
)+
∙
(3 ) (2
)
+ ∙ (3 ) (2 ) + (3 )(2 ) + (2 )
= 243 + 810
+ 1080
+ 720
+240
Example 2:
How many subsets are there of a set consisting of elements?
Solution:
There are
subsets of size , 0 ≤ ≤ . Therefore
1 1
= (1 + 1)
=2
The desired answer is 2 .
We consider some special cases of the binomial expansion.
( − ) =
=
−
(− )
+
1
−
2
+ ⋯ + (−1)
3
(1 + ) =
=1+
1
2
+
3
+⋯+
(− )
(1 − ) =
=1−
+
1
+
2
−
3
+ ⋯ + (−1)
Example 3:
Evaluate (1.02) correct to four decimal places.
Solution: (1.02) = (1 + 0.02)
∙
∙ ∙
= 1 + 12 (0.02) + ∙ (0.02) + ∙ ∙ (0.02)
∙
∙
∙
+ ∙ ∙ ∙ (0.02) + ⋯
= 1 + 0.24 + 0.0264 + 0.00176 + 0.0008 + ⋯
= 1.26824
Thus, (1.02) = 1.2682 correct to four decimal places.
+ 32
PROBLEM SET
IP1: Find the sum, √ +
Solution:
1 2
6
+ √ −
2
3
2 C 2
C 2 C 2 C 2
1 2 1 C 2 C 2 C 2
C 2 C 2 C 2
Now, 1 2 2 1 2 1 C (2) C (4) C (8)
1 6C1
2
=?
6
C2
6
3
4
6
5
6
4
6
6
5
2
6
1
6
6
3
6
2
3
4
6
5
6
4
6
6
6
5
6
6
6
6
2
6
6
4
6
2 1 30 60 8 198
P1. Find √ +
− √ −
=?
Solution:
5
2
1 3 5C0 5C1 3 5C 2 3 5C3
3
2
3
5
3
5C4
3
1 3 5C 5C 3 5C 3 5C 3 5C 3
5
5
Now, 1 3 1 3 2 1 5C (3) 5C (9)
5
5
3 1 3 1 2 1 30 45 152
0
1
2
3
2
IP2: For ≠ ,
Solution:
−
4
2 3
4
2
x x C0 x
4
4
4
=?
2 3
4
C1 x
3 4
2
x C2 x
3
2
3
x
4
2
3
3
C3 x 4C4
x
x
3
9
27 81
x8 4 x6 . 6 x 4 2 4 x2 3 4
x
x
x x
108 81
x8 12 x5 54 x 2
4
x
x
4
2
4
4
5C5
5C
5
3
3
5
5
in the expansion ( +
P2: The coefficient of
Solution:
4
) ( + ) is
1 x 1 x
1 C x C x C x C x .
1 C x C x C x C x C x C x
1 4 x 6 x 4 x x
1 7 x 21x 35 x 35x 21x 7 x x
2
7
4
2
4
4
1
7
4
6
2
7
2
1
4
8
4
7
3
2
2
4
3
7
4
3
6
2
7
5
4
7
5
6
6
7C7 x 7
8
3
4
5
6
7
Now, the coefficient of in the expansion is
1 + 4(21) + 6(35) + 4(7) = 323
IP3: Which is larger ( . )
or ,
.
Solution:
Splitting 1.01 and using binomial theorem to write the first few terms we have
1000000
1.01
1000000
1 0.01
1 1000000C1 (0.01) positive terms
1 1000000 0.01 positive terms
1 10000 positive terms
1000000
10001
1.01
1000000
1.01
10000
10000
P3. Find the approximation of ( .
Solution:
We have
(0.99) = (1 − 0.01)
) using the first three terms of its expansion.
= 1−
2
3
4
1 5
1 5
1 5
1 5
1
C0 C1
C2
C3
C4
C5
100
100
100
100
100
5
5
=1−
+(
)
−(
)
+(
)
−(
th
)
= 1 − 0.05 + 0.0001(Neglecting 4 and other terms)
(0.99) = 0.951
5
IP4: Using Binomial theorem, prove that
divided by 25
Solution:
−
always leaves remainder 1 when
n
n
We have 1 x
C x
n
r
r
r 0
Put x 5, we get
n
1 5
n
n
6
n
Cr 5r
r 0
6n nC0 nC1 5 nC2 52 nC3 53 ....... nCn 5n
6n 1 5n 52 nC2 nC3 5 ....... nCn 5n2
6n 5n 1 25 k , where k nC2 nC3 5 ....... nCn 5n2
6n 5n 25 k 1
Thisshows that when divided by 25, 6n 5n leavesremainder 1.
P4: Using Binomial theorem, prove that
Natural numbers .
Solution:
5 + 52 − 1
= 25 + 52 − 1
= (26 − 1) + 52 − 1
2n
+
−
is divisible by
C0 (26) 2n 2nC1 (26)2n1 2 nC2 (26)2n2 ...
2nC2n2 (26)2 2nC2 n1 (26) 2nC2n 52n 1
(26)2 (26)2 n2 2nC1 (26) 2n3 ... 2 nC2n2 (2n)26 1 52n 1
676 (26)2 n2 2nC1 (26)2n3 ... 2nC2n2
which is divisible by 676
Exercises:
1. Expand the following using binomial theorem.
a. (2 + 3 )
b. (4 + 5 )
c.
+
d.
−
e. (3 +
−
)
, for all
+
f.
2. Find ( + ) − ( − ) . Hence, evaluate √3 + √2
− √3 − √2 .
3. Find ( + 1) + ( − 1) . Hence evaluate
√2 + 1 − √2 − 1 .
4. Use the binomial theorem to evaluate (1.06) correct to four decimal places.
5. Using binomial theorem, evaluate each of the following:
a. (96 )
b. (98 )
c. (102)
d. (101)
6. Using binomial theorem, indicate which is larger (1.1)
or 1000.
7. Using binomial theorem prove that 50 − 49 − 1 is divisible by 49 for all
positive integers .
8. Using binomial theorem, prove that 625 − 48 − 1 is divisible by 576 for all
positive integers .
9. If n is a positive integer, then prove that 81 + 20 − 1 is divisible by 100.
6.2. General and Middle Terms
Learning objectives:
To find the general and middle terms in a binomial expansion.
And
To practice the related problems.
Let be a positive integer. Let , , , … denote the terms of expansion of a
binomial
( + ) =
=
=
+
1
+
2
+⋯+
…+
Then
=
=
=
2
+
+
1
+
0
1
2
We call the ( + 1) term of the expansion the general term of the expansion. It is
denoted by
.
The general term of a binomial expansion ( + ) is given by
=
Example 1: Find the fifth term in the expansion of (2 + 3 ) .
Solution: The fifth term is given by
∙ ∙ ∙
12 (2 )
(3 ) =
(2)
(3)
=
=
∙ ∙ ∙
4
= 495(256 )(81 ) = 10,264,320
Example 2: Find the ninth term of
− ⁄
.
∙ ∙ ∙
12
Solution: =
=
− ⁄
= ∙ ∙ ∙ ∙
8
Example 3: Find the middle term in the expansion of
∙
⁄
= 495
+
⁄
.
Solution: Since there are 11 terms in all, the middle term is the sixth.
10
⁄
=
=
⁄
5
= 252 ⁄ ⁄ = 252 ⁄
Middle term:
The middle term of a binomial expansion depends upon whether the index
or odd.
If is even, the number of terms in the expansion is + , which is odd.
Therefore, the middle term is given by
=
= +
is even
In example 3, where = 10, the middle term is + 1 = 6.
If is odd, then + is even, so there will be two middle terms in the expansion:
and
For example, in the expansion of ( + ) , the two middle terms are
= 3 and
= 4.
+
Note: In expansions of the type
2
=
=
∙
=
,the middle term
2
is independent of , and is therefore a constant.
PROBLEM SET
IP1: If the 3rd term of the expansion
+
is
, then the value of
Solution:
Given T3 T21 1000
52
1
C2
x
5
x log10 x
2
1000 10 x 3 x log10 x
x 2 log10 x 3 10 2 2log10 x 3 2log x 10
Put log
= , we get
2
1000
is
2 −3=
− 3 − 2 = 0 ⟹ ( − 2)(2 + 1) = 0
⟹2
⟹
= 2 or
=
⟹
= 10 or
⟹ log
= 10
⟹
= 2 or log
= 100 or
=
=
√
P1: If the 2nd, 3rd and 4th terms in the binomial expansion ( + ) are 240, 720 and
1080 respectively, then find , and .
Solution:
Given
T2 T11 nC1 x n1 .a 240 ....( A), T3 T21 nC2 xn 2 .a 2 720....( B)
T4 T31 nC3 x n3 .a 3 1080......(C )
n ( n 1)
n
n 2 2
B C2 x .a 720 , 2! . a 3 a 6 ......(1)
n
C1 xn 1 .a
240
n
x
x ( n 1)
A
C
B
n
C3 x n3 .a 3 1080
n
C2 x n2 .a 2 720
( n 2) a 3
a
9
.
...........(2)
3
x 2
x 2( n 2)
By solving (1) and (2), we get
=
⟹ =5
(
)
(
)
From (1), we have
=
From ( ), we have
3x
C1 x 4 a 240 5 x4 240 x5 32 x 2
2
5
and = (2) ⟹ = 3
Therefore, the values of , ,
are 2, 3, 5 respectively
IP2: Find the coefficients of the middle terms of the expansion
Solution:
We have = 7 (odd). The middle terms are
The middle terms of the given expansion 3 −
T4 T31 7C3 3 x
3
=5
are 4th and 5th terms
x3 7
3 4 4 9 3
C3 1 .3 .x .x .6
6
7 3
x9
105 13
35 81x .
x
216
8
4
= 4,
−
.
4
T5 T41 7C4 3 x
7 4
x3 7
3 3 12 4
C4 .3 .x .x .6
6
35(27) 15 35 15
x
x
1296
48
P2: Find the middle term in the expansion of
+
Solution:
Here = 20. There exists only one middle term, since
Middle term =
?
is even.
+1
=
+1
= 11 term
3
T11 T10 1 C10 3
x
20 10
10
20
5x 4
20C101510 x10
IP3: If the coefficient of
+
in the expansion
is
Solution:
+
The given expansion is
Now, general term is
2 5 r
5
Tr 1 C r x
r
k
5
10 3 r r
k
Cr x
x
To find the coefficient of , we must have 10 − 3 = 1
⟹ =3
T31 5C3 k 3 x
By the hypothesis, the coefficient of is 270
!
= 270
! !
⟹
(10) = 270 ⟹
=3
P3: Find the independent term of
Solution:
The given expansion is 2
−
in
−3
C r (2 x1/2 ) 20 r (3 x 1/3 ) r
Tr 1
20
20
Cr 2
20 r r
3
20 r
x 2 ( 1) r
x
r
3
?
then find
=?
20
r
Cr (1) 2
20 r
r r
r 10 2 3
3 x
To, find the independent term, we must have
10 − − = 0 ⟹ 10 = ⟹ = ⟹ = 12
T121 20C12 ( 1)12 2 2012 3 12
,
IP4: The sum of the coefficients of
−
,
≠ , (
20
C12 28 3 12
,
being a natural number) is
in the expansion of
. Find the coefficient of
in the expansion.
Solution:
,
The coefficients of
(
,
being a natural number), are
−
in the expansion of
m
C0 ,
m
,
≠ 0,
C1 (3) and mC2 9 .
By hypothesis, we have
m
C0 3 mC1 9 mC2 559
(
)
⟹ 1−3 +
= 559
⟹ 2−6 +9
− 9 = 1118
⟹3
− 5 − 372 = 0
⟹ (3 + 31 )( − 12 ) = 0
⟹ = 12 ( being a natural number)
Now,
r
12
12 r
Tr 1 Cr x
r 12 3r
3 12
C
3
x
2
r
x
To compute the term containing
Thus, the required term is
12
, put 12 − 3 = 3 ⟹
=3
3
C3 3 x129 5940 x3 .
P4: If the coefficients of 5th, 6th, 7th terms of ( + ) are in A. P then
Solution:
The coefficients of 5th, 6th and 7th terms of (1 + ) are
n
C4 , nC5 and nC6
n
n
n
Given, C4 , C5 and C6 are in A.P.
2 nC5 nC4 nC6
=?
!
)! !
⟹ 2. (
⟹
(
⟹
)(
(
⟹
(
=(
)! . !
=(
)(
−
=
)
⟹
!
+(
)! !
=(
)
)(
)
!
)! !
)(
)! !
+(
)! . . !
+
=
⟹ 12 − 48 − 30 =
− 9 + 20
⟹
− 21 + 98 = 0
⟹ ( − 14)( − 7) = 0 ⟹
= 7 or 14
Exercises:
1. Write down and simplify
a. 5 ℎ term in (3 − 4 )
b. 6 ℎ term in
+
c. 7 ℎ term in (3 − 4 )
d. 7 ℎ term in
+
⁄
− ⁄
and simplify.
2. Find the twelfth term of
3. Find if the 17 and 18 terms of the expansion (2 +
4. Find the middle term(s) in the expansion of
)
are equal.
is
∙ ∙ ∙∙∙(
!
−2
a.
b. (3 − 2 )
c.
4 +
d. (2 + 3 )
5. Show that the middle term in the expansion of (1 + )
where is a positive integer.
6. If the
and
th term is the middle term in the expansion of
.
7. Find the term independent of
−
a.
b.
√
+ 5√
in the expansion of
−
)
2
,
, then find
4
c.
+
+
d.
−
e.
in the expansion of ( + 2 ) .
8. Find the coefficient of
9. The coefficients of three consecutive terms in the expansion of (1 +
) are in
the ratio 1: 7: 42. Find .
10. Show that the coefficient of the middle term in the expansion of (1 + 2 ) is
equal to the sum of the coefficients of two middle terms in the expansion of
(1 + )
.
(1 + 2 ) (2 −
11. Find the coefficient of in the product
) using binomial theorem.
12. If the coefficients of ( − 5) and (2 − 1) terms in the expansion of
(1 + ) are equal, then find .
6.3. Greatest Coefficient and Greatest Term
Learning objectives:
To find the greatest coefficients and greatest terms in the binomial expansion.
And
To practice the related problems.
In any binomial expansion middle term has the greatest coefficient. If there are two
middle terms, then their coefficients are equal and greatest.
We consider the binomial expansion of ( + ) ; let
coefficient.
n
Now,
Since
n
n
Cr
=
Cr 1
Cr
!
!(
∙
)!
(
)!(
)!
!
=
n
Cr
-------(1)
is the greatest coefficient
≥1 ⟹
+1 ≥
−
⟹
≥
-------(2)
n
If we substitute − 1 for in (1), we get
Cr 1
n
Cr
=
be the greatest
Since
n
Cr
is the greatest coefficient
≤1 ⟹
≤
− +1 ⟹
From (2) and (3), we have
≤
≤
≤
------(3)
--------(4)
When = 2 (an even number), − ≤ ≤ +
This implies = , and the greatest coefficient is the coefficient of the middle term.
When = 2 + 1 (an odd number)
≤ ≤ +1
This implies
= , +1
In the next module, we show that the coefficients of the terms in the binomial
expansion of ( + ) equidistant from the beginning and the end are equal. So, the
greatest coefficient is the coefficient of the two middle terms.
Example: In the expansion of ( + ) , there are 5 terms. The middle term is 3. The
greatest coefficient is the coefficient of the 3 term
4
= C2
=6
Example: In the expansion of ( + ) , there are 6 terms. There are two middle
terms given by
= 3 and
= 4. The two middle terms are 3 and 4 terms.
Their coefficients are greatest and equal.
=
5
C2
= 10
,
=
5
C3
= 10
Greatest Term
In a binomial expansion, greatest term means numerically greatest term. Since we
are only concerned with the numerically greatest term, the investigation will be the
same for ( − ) as for ( + ) . Therefore, in any numerical example it is
unnecessary to consider the sign of the second term of the binomial.
We write ( + ) =
Since
1+
multiplies every term in 1 +
, it will be sufficient to find the greatest
term in this latter expansion.
Let the
and ( + 1) be any two consecutive terms. The ( + 1) term is
obtained by multiplying the
term by
∙ ; that is, by
−1 .
− 1 decreases as increases; hence the ( + 1)
The factor
greater than the
term, but only until
1.
Now
−1
⟹
>1 ⟹
> +1
⟹
−1 >
> --------(5)
−1
term is not always
becomes equal to 1, or less than
If
is an integer, we denote it by ; then if = , the multiplying factor becomes
1, and the ( + 1) term is equal to the
; and these are greater than any other
term.
If
is not an integer, we denote its integral part by ; then the greatest value of
consistent with (5) is ; hence
the ( + 1) term is the greatest.
Example: If = , find the greatest term in the expansion of (1 + 4 ) .
=8 ,
Solution: Here
Now,
=
=
=4 ⟹
=
=
∵
=
, not an integer.
Therefore, its integral part = 5 and
( + 1) = 5 + 1 = 6 term is greatest term and its value is
4
T6 T51 C5
3
5
8
Example: Find the greatest term in the expansion of (3 − 2 ) when
Solution:
Given (3 − 2 ) = 3
Here
Now,
= 9,
=
= 1.
1−
=
(neglecting the sign), ⟹
=
= 4, is an integer.
=
= (∵
= 1)
∴ 4th and 5th terms are greatest terms, which are numerically equal and its value is
3
9 9
3
2
C3 36 84 8 4,89,888
3
PROBLEM SET
IP1: The greatest Binomial Coefficient in the expansion ( + )
Solution: The given expansion is ( + )
Here index = 105, which is odd
∴ The greatest coefficients are
n
Cn 1 and nCn 1
2
i.e.
105
C1051 and
105
C1051
2
2
105
C104 and
2
Notice that
2
105
C106
2
105
C52 and 105C53
105
C52 105C53 (since
n
Cr n Cn r
)
is
22
Cr
13
Cr ?
P1: If
is the greatest Binomial Coefficient in the expansion ( + )
then
Solution: The given expansion is (1 + )
Here index = 22, which is even
∴ The greatest coefficient is
n
Cn
i.e.
22
C22 22C11
2
22
22
2
Cr C11 ⟹ = 11
13! 13 12
13
Cr 13C11
78
11! 2!
2
By the hypothesis,
Now,
IP2: Find the numerically greatest term in the expansion of ( −
Solution: The given expansion is (1 − 5 )
Here = 12 , = 5 (neglecting the sign)
⟹
=
Now,
=
∵
=
= 10 , is an integer.
)
=
when
=
∴ 10th and 11th terms are the greatest terms which are numerically equal and its
10
10
C10
3
12
value is
P2: Find the numerically greatest term in the expansion
Solution:
Given 3 +
=3
+
when
= .
1+
Here
= 12,
=
⟹
=
= 10 ∵
Now,
=
=
, is not an integer.
=
Therefore, its integral part is = 1
∴ + 1 =2nd term is the greatest term and its value is
1 6
T2 312 12C1 312
10 5
IP3: Find the numerically greatest term in the expansion (
and = .
Solution: Given (3 − 4 )
= 14,
Here
⟹
Now,
=
= 2 (∵
=
=
1−
(neglecting the sign)
=8
=
= (3 )
= 3)
= 5, is an integer.
−
)
when
=
∴ 5th and 6th terms are greatest terms which are numerically equal and its value is
14
24
1
C4
2
4
14
P3: Find the numerically greatest term in the expansion of (
= and = .
Solution:
Given (2 − 3 )
= (2 )
= 12,
Here
⟹
=
=
=
∵
)
when
+
)
when
−
) when
1−
(neglecting the sign)
=1
=
Now,
−
=
=
is not an integer
Therefore, its integral part is = 9
∴ + 1 = 10th term is the greatest term and its value is
9
5
C9 12C9 23 59
2
12
12
2x
IP4: Find the numerically greatest term in the expansion of (
= and = .
Solution:
Here
= 11,
=
⟹ =
=
∵
Now,
=
=
, is not an integer.
=
,
=
Therefore, its integral part is = 5
∴ + 1 =6th term is the greatest term and its value is
(3 )
∙
11
C5
=
11
C5 (2
) (3 ) =
11
C6 (6)
(2)
P4: Find the numerically greatest term in the expansion of (
= and
Solution:
=
.
First write (3 − 5 ) = (3 − 5 )
= 17,
Here
⟹
Now,
=
=
1−
(neglecting the sign)
=
(
)
=
=
, which not an integer
Therefore, its integral part is
(3 )
= (3 )
∙
= (3 )
∙
17
C6
= 6 and its greatest term is 7th term.
= ,
=
17
=
17
C6 (5
) (3 )
C6
Exercises:
1. Find the greatest binomial coefficient(s) in the expansion of
a. (1 + )
b. (1 + )
c.
+
2. Find the numerically greatest term(s) in the expansion of
a. (2 + 3 ) when =
b. (4 + 3 )
when
c. (7 − 5 )
where
d. (1 − 3 )
when
e. (3 + 5 )
f.
g.
h.
i.
=
=
=
when
=
,
=
(3 + 7 ) when = , =
(4 − 6 ) when = 3, = 5
(3 + 7 ) when = , = 15
(3 − 4 ) when = 8, = 3
6.4. Binomial Coefficients
Learning objectives:
To derive some properties of binomial coefficients.
And
To practice the related problems.
n
The values of Cr are often referred to as binomial coefficients. This is so because of
their prominence in the binomial theorem.
Some properties of the binomial coefficients are derived below.
The binomial coefficients are defined by
n
We note that
n
Therefore,
n
Cr
n!
n!
(n (n r ))!(n r )! r !(n r )!
nC r
Cn r
C n r
n!
(n r )!r !
If
n
C x nC y , then either
=
+
or
= .
The first relation readily follows from the formulas of the left and right hand sides.
For the second relation, we note that
=
Therefore,
−
⟹
+
n
Cx nC y nCn y
=
The following relations can be easily obtained from the formula for
∙
n
=
n
Cr
Cr
∙
n
Cr .
n 1
Cr 1
r 1 n 1
C r 1
n 1
A useful combinatorial identity (called Pascal’s identity) is
n
Cr
n 1
C r 1
n 1
Cr
≤ ≤ ------(1)
Equation (1) may be proved analytically or by the following combinatorial argument.
Consider a group of objects and fix attention on some particular one of these
n 1
objects – call it object 1. Now, there are
Cr 1 groups of size that contain object
1 (since each such group is formed by selecting − 1 from the remaining − 1
objects). Also there are
n
is a total of
Cr
n 1
Cr
groups of size that do not contain object 1. As there
groups of size , equation (1) follows.
The coefficients of the terms equidistant from the beginning and the end are equal.
This is seen as follows.
n
The coefficient of the ( + 1) term from the beginning is Cr .
There are altogether ( + 1) terms and the ( + 1) term from the end has
( + 1) − ( + 1) or ( − ) terms before it. Hence the ( + 1) term from the end
is ( − + 1) term from the beginning and its coefficient is
n
n
n
Since Cr Cn r , the assertion is proved.
We put = 1 in the expansion of
n
(1 + ) =
we get
n
C0 + nC1
n
+
n
n
C2
+⋯+
n
Cn
n
2 = C0 + C1 + C2 + ⋯ + Cn
Therefore, the sum of the binomial coefficients is 2 .
It follows that
n
If we put
C1 nC2 nCn 2n 1
= −1 in the expansion (2), we get
--------(2)
Cn r .
n
C0 − nC1 + nC2
n
C0 + nC2
+
n
−
n
C3 + nC4
−
n
n
C4
n
C5 + ⋯ = 0
n
+ ⋯ = C1 + C3 + C5 + ⋯
= (sum of all the coef icients)
= =2
The sum of the coefficients of the odd terms is equal to the sum of the coefficients of
the even terms, and each is equal to 2 .
Example 1:
10
C0 10C2 10C4 10C10 2101 29
11
C0 11C2 11C4 11C10 2111 210
10
C1 10C3 10C5 10C9 210 1 29
11
C1 11C3 11C5 11C11 2111 210
Example 2: If in the expansion of (1 + ) , the coefficient of (2 + 1) term is
equal to the coefficient of ( + 2) term, then find .
Solution:
43
C2 r 43Cr 1
Therefore, 2 + + 1 = 43 ⟹
= 14
Example 3:
A man has 6 friends; in how many ways may he invite one or more of them to
dinner?
Solution: The guests may be invited singly, in twos, threes…;
therefore the number of selection
6
C1 6C2 6C3 6C4 6C5 6C6 26 1 63
PROBLE SET
IP1: If the coefficients of ( − ) and ( − ) terms in the expansion of
( + ) are equal then the value of is
Solution:
The coefficients of ( − 5) and (2 − 1) terms in the expansion of (1 + )
34
C r 6 and
34
C2r 2
34
34
By the hypothesis, we have C r 6 C 2 r 2
Therefore, either − 6 = 2 − 2 or − 6 + (2 − 2) = 34
⟹ = −4
= 14
∴ = 14 ( ∵ is a positive integer)
are
P1: If the coefficients of ( + ) and ( − ) terms in the expansion of
( + ) are equal then =?
Solution: The (2 + 4) term of the given expansion (1 + ) is
=
(
)
=
18
C2 r 3 ∙
Thus, the coefficient of (2 + 4)
Similarly, the coefficient of ( − 2)
18
term is
is
18
C2 r 3
18
Cr 3
18
By the hypothesis, C2 r 3 C r 3
⟹ 2 + 3 = − 3 or (2 + 3) + ( − 3) = 18
⟹ = −6 or = 6
∴ = 6 ( Since is a positive integer )
IP2: Prove that
nC 3 nC 3 2 nC 3n nC 4n
i)
0
1
2
n
nC
nC
nC
nC
1
2
3
n n( n 1)
2
3
n
ii)
nC
nC
nC
nC
2
0
1
2
n 1
Solution:
n n
n
n
2
n
n
i) We know that , (1 x) C0 C1 x C2 x Cn x
n
Put = 3, we get
C0 3 nC1 32 nC2 3n nCn 4n
ii)
nC
n
n
n
nC
n
1 2 C2 3 C3 n Cn
r
r
n
n
n
n
C0
C1
C2
Cn 1 r 1 nCr 1
n
n!
( n r 1)!( r 1)! n
r
r 1
(n r )!r !
n!
n ( n 1) ( n 2) 2 1
P2: If
i.
ii.
iii.
is a positive integer, prove that
n
n
n 1
r Cr n 2
r 1
n
n
n 2
r (r 1) Cr n(n 1) 2
r2
n
r 2 nC r n(n 1) 2n 2
r 1
(n r 1)
r 1
n( n 1)
2
Solution:
n n
n
n
2
n
n
i. We have, (1 x) C0 C1 x C2 x Cn x
On differentiating both sides w.r.t , we get
n(1 x)n1 nC1 nC2 2 x nC3 3 x2 nCn nx n1
Now put = 1, we get
n 2n1 nC1 2 nC2 3 nC3 n nCn
n
n
n 1
Thus, r Cr n 2
r 1
n n
n
n
2
n
n
ii. We have, (1 x) C0 C1 x C2 x Cn x
On differentiating both sides w.r.t , we get
n(1 x)n1 nC1 nC2 2 x nC3 3 x2 nCn nx n1
Again differentiating both sides w.r.t , we get
n(n 1)(1 x)n2 2 nC2 nC3 6 x nCn n(n 1) x n2
Now put = 1, we get
n(n 1)2n2 2 nC2 6 nC3 n(n 1) nCn
n
n
n 2
Thus r ( r 1) Cr n( n 1) 2
r 2
n
n
2 n
n
iii. r Cr r ( r 1) r Cr
r 1
r 1
n
n
r ( r 1) nCr r nCr n ( n 1) 2n 2 n 2 n 1
r 1
r 1
n 2
n 2
( n 1 2) n 2
n( n 1) 2
IP3:
2nC
2
0
2
2
2
2 n C 1 2 n C 2 .......... 2 n C 2 n
?
Solution:
We have
2n
1 x
2 nC0 2nC1 x ......... 2nC2n x2n
---------- (1)
x 12n 2nC1x 2n 2nC1x 2n1 ......... 2nC2n ------- (2)
Multiplying (1) & (2) we get
x 1
2
2n
2nC0 x2n 2nC1 x2n1 2nC2 x 2n2 ........ 2nC2n
2nC0 2nC1 x 2nC2 x2 ........ 2nC2n x 2n
By comparing the
coefficients on both sides, we get
2
2
2
2
2n
C0 2nC1 2nC2 2nC2 n
P3: Find,
+
+
+
Solution: We have (1 + ) =
By integrating, we get
(
)
=
+
+
+……. +
+
+
2n
Cn 1
n
=?
+ ⋯+
+⋯+
Applying the limits 0 to 3, we get
(
)
=
−
(
)
IP4: If
∙
+
=3
=
+
+
+
denotes
∙
+
n
+
+⋯+
+
+⋯+
+⋯+
C r , then prove that
+
∙
+∙∙∙ +(
+ )
=(
+ )∙
2n
Cn .
Solution: Let = 3 ∙
+7∙
+ 11 ∙
+∙∙∙ +(4 + 3)
------(1)
On writing the terms of the . . . of (1), in the reverse order, we get
= (4 + 3 )
+ (4 − 1)
+ (4 − 5)
+∙∙∙ +3
= (4 + 3)
+ (4 − 1)
+ (4 − 5)
+∙∙∙ +3
--(2)
(∵
= ,
0≤ ≤ )
On adding (1) & (2), we get
2 = (4 + 6)
+ (4 + 6 )
+ (4 + 6)
+∙∙∙ +(4 + 6)
= (4 + 6 )
+
+
+∙∙∙ +
2n
2n
= (4 + 6) ∙ Cn
∵
+
+
+∙∙∙ +
= Cn
∴
= (2 + 3) ∙
P4: If
denotes
2n
Cn
n
Cr
, then prove that
+
+
+⋯+
and deduce the following:
=
2n
Cn r
1. C 02 C12 C 22 ....... C n2 2n C n
2. C0C1 C1C2 C 2C 3 ....... C n1C n 2 nC n1
Solution: We have
(1 + ) =
+
+
+ ⋯+
------ (1)
=
+ + +⋯+
Multiplying (1) & (2) we get
(1 + )
=
+ + +⋯ +
------ (2)
and 1 +
( +
+
+⋯+
The coefficient of on
of (3) is
+
+
+⋯+
The coefficient of
) ------(3)
on
= The coefficient of
= The coefficient of
in
(
)
in (1 + )
2nCn r
Comparing the coefficient of
on both sides of (3), we get
2nCn r
+
+
Deduction 1: Put
+⋯+
= 0 in (4), we get
+
Deduction 2: Put
+
+⋯+
= 1 in (4), we get
+
+
+⋯+
2n
-------(4)
Cn
2nCn1
Exercises
1. Prove that + 2 ∙ + 2 ∙ +∙∙∙ +2 ∙
=3 .
2. Show that
i. 2 ∙ + 5 ∙ + 8 ∙ +∙∙∙ +(3 + 2) = (3 + 4)2 .
ii. + 3 ∙ + 5 ∙ +∙∙∙ +(2 + 1) = (2 + 2)2 .
iii. 2 ∙ + 7 ∙ + 12 ∙ +∙∙∙ +(5 + 2) = (5 + 4)2 .
3. Show that − 4 + 7 − 10 + ⋯ = 0.
4. If (1 + ) =
+
+
+ ⋯+
,
then find the value of + 2 + 3 + 4 + ⋯ + ( + 1) .
5. If (1 + ) =
+
+
+ ⋯+
,
then find the value of +
+
+⋯+
.
6. If (1 + ) =
+
+
then find the value of +
7. Find the sum of the following
+ ⋯+
,
+ +⋯+(
)
.
4∙
i.
+7∙
+ 10 ∙
+∙∙∙ +(3 + 4)
15
ii.
15
15
15
C1
C1
C3
C15
2
3
15
15
15
15
15
C0
C2
C2
C14
iii.
iv.
v.
∙
2 ∙
3∙
+ ∙ + ∙ +∙∙∙ +
∙
+ 3 ∙ + 4 ∙ +∙∙∙ +( + 2) ∙
+ 6 ∙ + 12 ∙ +∙∙∙ +3 ∙ 2 ∙
6.5. Multinomial Coefficients
Learning objectives:
To generalize the Binomial theorem to Multinomial theorem.
To generalize the concept of Binomial coefficients to Multinomial coefficients.
AND
To practice the related problems.
We consider the following problem: A set of distinct items is to be divided into
r
distinct groups of respective sizes
,
,
, …,
, where ni n . We wish to
i 1
know how many different divisions are possible.
n
We note that there are Cn possible choices for the first group; for each choice of
1
n n1
the first group there are
Cn2 possible choices for the second group; for each
n n1 n2
Cn3 possible choices for the third
choice of the first two groups there are
group; and so on. Hence it follows from the generalized version of the basic counting
principle that there are
nC n n1 C n n1 n2 C n n1 n2 nr1 C
n1
n2
n3
nr
n n1 ! n n1 n2 nr 1 !
n!
0! nr !
n n1 !n1! n n1 n2 !n2 !
n!
n1 ! n2 ! nr !
possible divisions.
We use the following notation:
If
+
…+
= , we define
,
,…,
=
,
!
!
!…
,…,
!
by
Thus
represents the number of possible divisions of
, ,…,
into distinct groups of respective sizes , , … , .
distinct objects
Example 1:
A police department in a small city consists of 10 officers. If the department policy is
to have 5 of the officers patrolling the streets, 2 of the officers working full time at
the station, and 3 of the officers on reserve at the station, how many different
divisions of the 10 officers into the three groups are possible?
Solution:
!
There are
= 2520 possible divisions.
! ! !
Example 2:
Ten children are to be divided into an team and a team of 5 each. The team
will play in one league and the team in another. How many different divisions are
possible?
Solution:
!
There are ! ! = 252 possible divisions.
Example 3:
In order to play a game of basket ball, 10 children at a playground divide themselves
into two teams of 5 each. How many different divisions are possible?
Solution:
We note that this example is different from previous example because now the order
of the two teams is irrelevant. That is, there is no and team but just a division
consisting of 2 groups of 5 each. Hence the desired answer is
!
÷2=
= 126
! !
The following theorem, known as multinomial theorem, generalizes the binomial
theorem.
(
+
) =
+⋯+
,
(
,
…
, …,
):
,…,
…
That is, the sum is over all nonnegative integer-valued vectors (
that + … +
= .
The numbers
,
For the case = 2,
, …,
= ,
,
are known as multinomial coefficients.
=
− , the equation
!
= ! !…
, ,…,
reduces to the binomial coefficient
!
= ( )! !
!
-------(1)
,…,
) such
Furthermore, we note that the multinomial coefficient in equation (1) is identical to
the expression for the number of distinguishable permutations of objects, of
which are identical and type (for = 1,2, … , and
+ …+
= ):
!
= ! !… !
A careful look should convince us that the two expressions must be identical.
Example 4:
(
+
+
) =
2
1,1,0
2
2,0,0
+
+
=
+
+
2
1,0,1
+2
2
0,2,0
2
0,0,2
+
+
+2
2
0,1,1
+2
+
PROBLEM SET
IP1: Find the number of ways that
apples can be divided among children if the
youngest child receives apples and each of the other apples.
Solution:
We wish to find partitions of the 12 apples into 4 cells containing 6,2,2,2 apples
respectively. So, that
!
× × × × ×
∴ ! ! ! !=
= 83160
P1: In how many ways can the 12 students in a class take 4 different tests if 3
students are to take each test?
Solution:
We have to divide (partition) 12 students for 4 different tests if 3 students are to take
each test.
!
∴
= 3,69,600 Partitions
! ! ! !
IP2: Find the coefficient of x2009 in the expansion of (1 x)2008 (1 x x2 )2007 .
Solution:
1 x
2008
2 2007
1 x x
2007
3 2007
2007
2007
1 x 3
x 1 x 3
3 2007
x
1 x
1 x 1 x
All the terms in the expansion of
1 x 1 x 1 x x 2
are of the form
3r
and all the terms
in the expansion of
form
P2:
x3 r 2 .
x 1 x3
2007
x3r 1, where as x 2009 is of the
are of the form
Thus, the desired coefficient is 0.
Find the coefficient of
Solution:
x10 in the expansion of (1 x 2 x 3 )8 .
We rewrite the given expression as [1 x2 (1 x)]8 and expand By using the
binomial theorem, we have
[1 x2 (1 x)]8 8C0 8C1 x 2 (1 x) 8C2 x 4 (1 x)2 8C3 x6 (1 x)3
8C4 x 8 (1 x)4 8C5 x10 (1 x)5 ...........
8
8
4
10
8
10
5
The two terms which contain x are C4 x (1 x) and C5 x (1 x ) .
Thus, the coefficient of
x10 in the given expansion
8C4
x 2 in the expansion of(1 −
[Coefficient of
8C4 6 8C5
in (
x1 x2 x3
n
−
n1 ,n2 ,n3 0
n1 n2 n3 n
Taking
Now,
=2 ,
= −3 ,
2 x 3 y 5 z 8
=5
n1 ,n2 ,n3 0
n1 n2 n3 8
∴ The coefficient of
5
8!
8!
(70) (6) 56 476
6
4! 4!
3! 5!
IP3:
Find the coefficient of
Solution:
We have
) ] + 8C
+
) .
n
n1 n2 n3
n ,n ,n x1 x2 x3
1 2 3
and
= 8, we get
8
n1
n2
n3
2
x
3
y
5
z
n ,n ,n
1 2 3
8
(2) (−3) (5)
3, 3, 2
!
= ! ! ! (2) (−3) (5)
=
= 560 (2) (−3) (5)
P3:
Find coefficient of
x
7
in the expansion of
1 2x x 3
6
Solution:
3 6
1 2 x x
p ,q ,r 0
p q r 6
p ,q ,r 0
p q r 6
6!
1 p 2 x q x3
p! q! r!
6!
2 q xq 3r
p! q! r!
x 7 , we have to take
!
7
Thus, coefficient of x is = ! ! ! (−2)
For coefficient of
= 1,
r
--------(1)
= 4, = 1 and
= 3, = 1,
= 2.
!
+ ! ! ! (−2)
= 480 − 120 = 360
IP4:
15
If ( +
+
+
) =
ak x
k
7
, then find
k 0
Solution:
(1 + +
+ ) =
+
+
Put = 1 in ( ), we get
+ + + + + ⋯+
+ + + + + ⋯+
Put = −1 in ( ) , we get
− + − + − ⋯−
− + − + − ⋯−
By adding (1) and (2), we get
2( + + ⋯ +
)=4
7
+
+⋯+
a2 k
k 0
a2k
.
k0
+⋯+
------( )
= (1 + 1 + 1 + 1)
=4
------(1)
= (1 − 1 + 1 − 1)
=0
------(2)
45
512
2
P4:
) =
a. If ( +
−
+
+
+⋯+
, then find the value of
)
+
+
+ ⋯+
)
+
+
+⋯+
b. If ( + + + ) =
+
+
+⋯+
then find the value of
)
+
+
+⋯+
)
+
+
+⋯+
Solution:
a. We have
+
+
+⋯+
= (3 + 7 − 9 ) ------(1)
= 1 in (1), we get
+ + + +⋯+
Put = −1 in (1), we get
− + − +⋯+
(2) + (3) ⟹ 2( + + ⋯ +
Put
⟹
+
+
=1
+ ⋯+
------(2)
= (−13)
------(3)
) = 1 + (−13)
=
(
)
b. We have
+
+
+⋯+
= (1 + +
+ ) -----(1)
Put = 1 in (1), we get
+ + +⋯+
=4
------(2)
Put = −1 in (1), we get
− + − + ⋯−
=0
------(3)
(2) − (3) ⟹ 2( + + + ⋯ +
)=4
⟹
+ + + ⋯+
=2
Exercises:
1. Find the number of ways that 9 toys can be divided between 4 children if the
youngest is to receive 3 toys and each of the others 2 toys
2. There are 12 students in a class. Find the number of ways that 12 students can
take 3 different tests if 4 students are to take each test.
3. Find the number of ways that 12 students can be partitioned into 3 teams so
that each team contains 4 students.
4. Find the number of ways in which 15 recruits can be drafted into three different
regiments, five into each.
5. Find the number of ways in which 15 recruits can be divided into three equal
groups.
8
36
6. Find the Coefficient of x in the expansion of (1 2 x x ) .
7. Find the coefficient of
in the expansion of ( + + +
) .
8. Find the coefficient of
in the expansion of ( + + + + ) .
9. Find the sum of the coefficients in the (1 + − 3 ) .
10.If (1 + 3 − 2 ) =
+
+
+⋯+
, then show that
a.
+ + +⋯+
=2
b.
− + − …+
=4
11.If (1 + + ) =
+
+
+ ⋯+
, then show that
a.
+ + +⋯+
=3
b.
+ + +⋯+
=
c.
d.
+
+
+
+
+⋯+
=
+ +⋯ = 3
+
