CHAPTER 00
ALGEBRA REFRESHER
0.1
PURPOSE
This chapter, designed for your convenience, is a review of many pertinent topics
from algebra. For the most part, your prior exposure to this material is assumed.
Our intention is neither to develop nor to motivate any concepts; rather, we shall
provide a statement-of-fact presentation of terminology and the methods of
manipulative mathematics to which an immediate second exposure would be
beneficial. Therefore, we urge you to devote whatever time necessary to these
sections in which you need review.
0.2
AXIOMS OF THE REAL NUMBERS
Certain fundamental principles, which we assume to be true, govern the system of
real numbers. They are known as axioms (or laws) of the real number system.
Axioms of Equality
Let a, b and c be real numbers. Then
1.
a = a (reflexive)
2.
If a = b, then b = a (symmetric)
3.
If a = b and b = c, then a = c (transitive)
4.
If a = b and a + c = d, then b + c = d,
If a = b and ac = d, then bc = d (substitution)
5.
If a = b and c = d, then a + c = b + d and ac = bd
(addition and multiplication)
Although the axioms of equality may appear to be rather trivial, they are quite
powerful and vital. If we did not have them at our disposal, attempts at
developing a system of mathematics could result in futility.
Axioms of Addition
If a,b and c are real numbers then,
1.
a + b is a unique real number (closure)
2.
a + b = b + a (commutative)
3.
(a + b) + c = a + (b + c) (associative)
4.
There exists a real number 0, called “zero”, such that a + 0 = a for each
number a (additive identity)
5.
For each number a, there exists a unique real number, denoted by –a, such
that a + (-a) = 0. (additive inverse)
Axioms of Multiplication
If a, b and c are real numbers, then
1.
ab is a unique real number (closure)
2.
ab = ba (commutative)
3.
a(bc) = (ab)c (associative)
1
4.
5.
There exists a real number 1, called “one” such that
a.1 = a for each number a. (multiplicative identity)
For each non-zero number a, there exists a unique real number a-1 such
that a.a-1 = 1 (multiplicative inverse)
Division:
If a and b are real numbers and b ≠ 0, then a divide by b, written a | b ,
a
1
is defined by
 a(b 1 )  a. .
b
b
Axiom of Addition and Multiplication
This axiom relates addition and multiplication.
If a, b and c are real numbers, then
a(b+c) = ab + ac and (b+c)a = ba + ca (distributive)
Note that by the definitions of division and the distributive laws, we have
ab
1
1
1
a b
 (a  b)
 a.  b.


c
c
c
c
c c
This important result does not mean that
a
a a


bc
b c
This error is very common.
0.3
OPERATION WITH SIGNED NUMBERS
Listed below are various properties of signed numbers which you should study
thoroughly and make sure that you understand them. The ability to manipulate
signed numbers is essential to the study of mathematics.
All denominators in the following are different from zero:
1.
a–b
= a + (-b)
2.
a – (-b) = a + b
3.
– a = (-1).a
4.
a(b – c) = ab - ac
5.
–(a + b) = - a – b
6.
–(a – b) = - a + b
7.
– (-a) =
8.
a.0
9.
(-a)b
a
= (-a).0 = 0
= -(ab) = a(-b)
2
10.
(-a)(-b) =
ab
a
= a
1
11.
a
b
12.
 
a
b

a
b
a
b

13.
a
b
14.
0
a
 0 when a  0
15.
a
a
 1
when a  0
b
a   b , a  0
a
16.
17.
a
bc
18.
a
b

a 1
.

b c
1 a
.
b c
 a  c 
    
 b  c 
ac
bc
a
(b)c

a
bc
20.
a c

b d

ad  bc
bd
21.
a c

b d

ad  bc
bd
22.
a b

c c

ab
c
23.
a c
.
b d

ac
bd
19.
a
b(c)

3

where c  0
a
(b)(c)
 
a
bc
0.4
24.
a
b
c
 a
25.
a
b
c
26.
a
b
c
d
b
c

ac
b

a
c 
b
a
bc

a c

b d
a d
.
b c


ad
bc
EXPONENTS AND RADICALS
The expression
x.x.x
is abbreviated x3 and in general, for a positive integer n, xn represents the product
of n x’s. More specifically, for n > 0  we define:
1.
x n  x.x.x.....
.x

n factors
1
xn
1
x.x.x..... ..x
2.
x n
3.
x0 = 1 if x  0. 00 is not defined.


If rn = x where n > 0, then r is called an nth root of x.
The principal nth root of x is that nth root of x which is positive if x is positive,
and is negative if x is negative and n is odd.
Thus 9  3 and 3  8 = -2.
We denote the principal nth root of x by the symbol
and n is the index, and
is the radical sign.
n
x where x is the radical
We state now the basic laws of exponents and, for completeness, the laws of
radicals which are derived essentially from those of exponents:
1.
xm.xn =
xm+n
2.
x0
=
1 if x  0
3.
x-n
=
1
xn
4.
1
x n
 xn
4
5.
xm
xn
 x mn

6.
xm
xm
 1 if
x0
7.
(xm)n
=
8.
(xy)n
= xn.yn
9.
x
 
y
1
n m
x
xmn
n
xn
yn

1
11.
n
x .n y
n
x
12.
mn
n

14.
x
15.
 x
m
x


x
m
n
m
n

y
n
13.

xn
10.
n
x
y
mn
n
x
(3)
0.5
n

 x

xm
m
n
 x
Note the following pitfalls:
n
(1)
ab  n a
(2)
xy
ab

a b

2
n
a

n
b

n
b
ab
ORDER PROPERTIES
Let a,b and c be real numbers.
1.
Either a < b, b < a or a = b (trichotomy)
2.
If a < b and b < c, then a < c.
3.
If a < b, then a + c < b + c for all c.
5
4.
5.
0.6
If a < b and c > 0, then ac < bc.
If a < b and c < 0, then ac > bc.
INTERVALS
Usually there are nine types of intervals on the real number line.
Suppose that a < b.
The set of all numbers between a and b is the open interval (a,b):
(a,b) = { x: a < x < b}
a
b
The closed interval [a,b] is the open interval (a,b) together with the endpoints a
and b:
[a,b] = { x: a ≤ x ≤ b }
a
b
There are seven other types of intervals:
(a,b] = { x : a < x ≤ b}
a
b
[a,b) = { x : a ≤ x < b}
a
b
(a,∞) = { x : a < x }
a
[a, ∞) = { x: a ≤ x}
a
(- ∞, b) = { x : x < b}
b
(- ∞, b] = { x : x ≤ b}
b
6
(- ∞ , ∞) = set of all real numbers = R
0.7
ABSOLUTE VALUE
Definition:
 a; if a  0
 
 a; if a  0
Other characterization:
a
a 
a2
Geometric interpretation:

(a) 2
|a|
- distance between 0 and a
|a - c| - distance between a and c
Properties of absolute values:
1.
|a| = 0 if and only if a = 0
2.
|-a| = |a|
3.
|ab| = |a||b|
5.
a
a

b
b
|a - b| = |b - a|
6.
|a + b| ≤ |a| + |b| (triangle inequality)
7.
|a| - |b| ≤ |a - b| ≤ |a| + |b|
8.
|a – b | < c if and only if -c < a – b < c
4.
Note the pitfall:
a 2 is not necessarily equal to a, but
7
a 2 = |a|
CHAPTER 01
TRIGONOMETRY
Trigonometry is a branch of mathematics that deals with angles, triangles and the
trigonometric functions which have to do with the ratios between the sides of a right
angled triangle. It is one of the most practical branches of mathematics.
hypotenuse
opposite
θ
adjacent
The subject matter of trigonometry is based on six trigonometric functions.
We introduce the first three namely: Sine, Cosine and Tangent functions. These are
going to help us in calculating the sides and angles of a right angled triangle.
opposite
hypotenuse
(i)
sin θ =
(ii)
cos θ =
adjacent
hypotenuse
(iii)
tan θ =
opposite
adjacent
The other three additional trigonometric functions are Cosecant, Secant and
Cotangent.
(i)
cosec θ =
(ii)
sec θ
=
1
hypotenuse

sinθ
opposite
1
hypotenuse

cosθ
adjacent
8
cot θ
(iii)
=
1
adjacent

tanθ opposite
Let us consider a unit circle, i.e. a circle with radius 1 centered at the origin. Angles
are measured starting from the positive x-axis. An angle is positive if it is measured
in the counter-clockwise direction, and is negative if it is measured in the clockwise
direction. We measure angles in degrees, using the fact that a circle contains 360º.
Then we can describe any angle by comparison with the circle. Any angle in a
triangle must be between 0º and 180º. In a circle such a restriction is not there.
(0,1)
x2+y2 =1
(x,y)
0
(-1,0)
θ
(1,0)
x
θ< 90 º
(0,-1)
y
(0,1)
x2+y2=1
0
(-1,0)
θ> 180º
θ
(1,0)
(x,y)
(0,-1)
y
9
x
RADIAN MEASURE
There is another way to measure angles, the radian measure, which, in many
instances, is more useful than measurement in degrees. The radian measure of an
angle does not refer to “degrees”, it refers to distance measured along an arc of the
unit circle. This is an advantage when discussing trigonometric functions that arise in
applications that have nothing at all to do with angles.
Consider the following diagram
1
1
1
The radian is a unit of angle measurement defined such that an angle of one radian
subtended from the center of a unit circle produces an arc with arc length one.
Let R denote the radial line which makes an angle of θ with the positive x-axis. Let
(x,y) denote the point at which this radial line intersects the unit circle x2 + y2 = 1.
Then the radian measure of the angle is the length of the arc of the unit circle
from the point (1,0) to the point (x,y).
R
y
(x,y)
2
2
x +y =1
θ
0
10
(1,0)
x
Since the circumference of a circle is 2πr, where r is the radius of the circle, the
circumference of the unit circle(a circle with radius 1 unit) is 2π. Thus
360º = 2π radians.
From this follows that
360 
180 
1 radian =
=
≈ 57,3º
2

2

and
1º
=
≈ 0,0175 radians.

360 180
P
Cr
r
L
θ
x
0
y
Let Cr denote the circle of radius r centered at the origin. If OP denotes a radial line,
then OP cuts an arc from Cr of length L. Let θ be the positive angle between OP and
the positive x-axis. If θ = 360º, then L = 2πr. If θ = 180º, then L = πr. In fact

L
L
.

; or θ = 360.
360 2π r
2π r
If θ is measured in radians, then

L

2 2π r
2π L L
or
θ =
 .
2π r
r
Therefore L = rθ.
There are two basic trigonometric functions, usually written as cosθ and sinθ. Since
the equation of the circle is x2 + y2 =1, we see that cosθ and sinθ satisfy the equation
sin2 θ + cos2 θ = 1.
As θ varies, cosθ and sinθ oscillate between +1 and -1. We note that if we add 2π to
the angle θ, then we end up with the same point (x,y) on the circle.
Thus,
11
cos(θ+ 2π) = cosθ
and
sin(θ + 2π) = sinθ
In general, if n is the smallest positive integer such that f(x + n) = f(x), we say that the
function is periodic with period n. Thus the functions cosθ and sinθ are periodic with
period 2π.
We also have a number of identities:
1.
cos(- θ)
=
cosθ
2.
sin(- θ)
=
- sinθ
3.
cos(π+θ)
=
- cosθ
4.
sin(π+θ)
=
- sinθ
5.
cos(π-θ)
=
- cosθ
6.
sin(π -θ)
=
sinθ
7.


cos    
2

=
- sinθ
8.


sin    
2

=
cosθ
9.


cos    
2

=
sinθ
10.


sin    
2

=
cosθ
11.
cos2θ
= cos2θ - sin2θ
= 2cos2θ - 1
= 1 – 2sin2θ
12.
sin2θ
= 2sinθcosθ
13.
cos

2
=
12
1  cos 
2
14.
sin

2
=
1  cos 
2
15.
cos2θ
=
1  cos 2
2
16.
sin2θ
=
1  cos 2
2
17.
cosθ - cosβ
= 2 sin
18.
sinθ - sinβ
= 2 sin
19.
sinθ + sinβ
= 2 sin
20.
cosθ + cosβ
= 2 cos
21.
sin(θ + β)
= sinθ cosβ + cosβsinθ
22.
cos(θ + β)
=
cosθcos β - sinθ sinβ
23.
tan(θ + β)
=
tan   tan 
1  tan  tan 
24.
tan(π + θ)
= tanθ
25.
1 + tan2θ
= sec2θ
26.
1 + cot2θ
= cosec2θ
27.
tan

2
=
=
13
 
2
 
2
 
2
 
2
sin 
1  cos 
1  cos 
sin 
sin
cos
 
2
 
cos
cos
2
 
2
 
2
CHAPTER 02
LIMITS AND CONTINUITY
1.
TWO-SIDED LIMITS
INTRODUCTION
The concept of a limit lies at the heart of calculus and provides the foundation for both
the derivative and the integral. Hence we must develop some understanding and insight
of this concept.
We must first attempt to cultivate a feeling for the concept of a limit, that is, to convey
the idea of a limit in an intuitive manner. In lay term we can say a limit means a
boundary: a point, line or number that may not be passed or crossed.
The following two examples will clear our way towards making the limit concept precise.
Example 1
Consider the perimeter (circumference) of a circle.
Draw the inscribed triangle in the circle. The perimeter of the triangle is less than the
perimeter of the circle.
Draw the inscribed square in the circle. The perimeter of the square is less than that of
the circle.
Draw the inscribed pentagon. The perimeter will be less than that of the circle.
Fig A
As we increase the sides of a polygon, then the perimeter becomes closer and closer to
that of the circle. However, it will never be equal to that of the circle. Hence the
perimeter of a circle becomes the limit.
Example 2
Suppose we consider the function f defined by
2x 2  5x  2
f(x) =
x2
14
Observe that f(x) is defined for all values of x except when x = 2, for at that point the
denominator is zero. Hence the domain of f = { x | x ≠ 2}.
Let us now closely examine the behavior of f(x) for values of x “near 2”. The
corresponding values of f(x), when x is first less than 2, and then greater than 2, are
indicated in the table,
x<2
f(1.7) = 2.4
f(1.8) = 2.6
f(1.9) = 2.8
f(1.99) = 2.98
f(1.999) = 2.998
x>2
f(2.3) = 3.6
f(2.2) = 3.4
f(2.1) = 3.2
f(2.01) = 3.02
f(2.001) = 3.002
From an inspection of both situations, it becomes apparent that as x takes on values closer
and closer to 2, regardless of whether x approaches 2 from the left (x < 2) or from the
right (x > 2), the corresponding values of f(x) become closer and closer to 3. To express
this situation, we say that 3 is the limit of f as x approaches 2 . Symbolically we write
2x 2  5x  2
 3.
x 2
x2
lim
Actually, by taking any value of x sufficiently close to 2, we can make the number f(x) as
arbitrarily close to 3 as we wish.
We can reach the same conclusion by viewing f(x) in another way.
Since 2x2 – 5x = 2 = (2x-1)(x-2), then
f(x) 
(2x  1)(x  2)
.
x2
For x ≠ 2, we can find another form for f(x) by dividing both numerator and denominator
by the common factor (x-2); thus
f(x) = 2x-1; x ≠ 2,
x<2
f(1.7) = 2.4
f(1.8) = 2.6
f(1.9) = 2.8
f(1.99) = 2.98
f(1.999) = 2.998
x>2
f(2.3) = 3.6
f(2.2) = 3.4
f(2.1) = 3.2
f(2.01) = 3.02
f(2.001) = 3.002
By considering the table for this simplified form of f(x), we obtain the same values as in
the first table. Thus, we see that if x is sufficiently close to 2, the number (2x – 1) will be
as close to 3 as we please. The figure below illustrates this.
15
y
Fig B
f(x) = 2x-1, x≠2
3
x
0
2
Example 3
Let us determine the limit of the function f(x) = x2 as x approaches -1. From the figure
below we can readily conclude that as x approaches -1, from both the left and the right,
the corresponding functional values, x2, approach 1.
y
Fig C
f(x) = x2
1
x
0
-1
Thus,
lim x 2  1
x 1
Remark: Note that in determining this limit, we are not at all concerned with what
happens to the function when x equals -1.
16
For an arbitrary function f, our results can be generalized by saying that
lim f(x)  L
x a
(read: “the limit of f(x), as x approaches a, is L) means that f(x) will lie as close to the
number L as we please for all x sufficiently close to the number a.
Again, we are not concerned with what happens to f(x) when x = a, but only with what
happens to it when x is close to a. In our first example, the function
2x 2  5x  2
f(x) =
was not even defined for x = a = 2. We emphasize that a limit is
x2
independent of the manner in which x approaches a. The limit must be the same
whether x approaches a from the left or the right, i.e. for x < a and x > a respectively.
Here follow some theorems of limits. We state them without proofs for the moment. We
shall prove them later in the next chapter.
1.
If f(x) = c is a constant function , then lim f(x)  lim c  c
2.
If f(x) = xn, then lim x n 
x a
x a
 lim x
n
x a
x a
 an
Examples
lim 1  1
(a)
x 7
  2
lim x 5  limx
(b)
x 2
5
x 2
5
 32
If lim f(x)  L1 and lim g(x)  L 2 , where L1 and L2 are real numbers, then
x a
3.
x a
lim[f(x)  g(x)]
x a
=
lim f(x) ± lim g(x)
=
L1 ± L2
xa
xa
This property can be extended to the limit of any finite number of sums or
differences.
4.
5.
lim[f(x).g(x) ]
xa
lim[c.g(x)]
xa
=
lim f(x) . lim g(x)
=
L1 . L2
=
c. lim g(x)
=
c.L2, where c is a constant.
xa
xa
xa
17
6.
f(x)
x a g(x)
lim f(x)

lim
x a
lim g(x)
x a
L1
, provided that L2 ≠ 0
L2
=
7.
=
lim n f(x)
x a
lim f(x)
n
x a
=
n
L1
We can summarize these theorems
1. The limit of a sum is the sum of the limits.
2. The limit of a difference is the difference of the limits.
3. The limit of constant times a function is the constant times the limit of a function.
4. The limit of a product is the product of the limits.
5. The limit of a quotient is the quotient of the limits (provided that the limit of the
denominator is not zero).


6. lim[ f ( x)]n  lim f ( x) where n is a positive integer.
x a
n
x a
7. lim c  c .
x a
8. lim x  a .
x a
9. lim x n  a n , where n is a positive integer.
x a
10. lim n x  n a , where n is a positive integer.( if n is even, we assume a  0 ).
x a
11. lim n f ( x)  n lim f ( x) where n is a positive integer. .( if n is even, we assume
x a
x a
lim f ( x)  0 ).
x a
Examples
(a)
lim (x 3  x 2 ) =
x 3
lim x 3
x 3
=
33
=
36
 lim x 2
x 3
+ 32
18
(b)
(c )
(d)
(e)
lim (x 2  4x  2)
x 4
lim (x 2  1)(x 3  4)
x 1
lim [3 (x  4)]
x  2
t3  t2  2
lim
t 0
t3 1
=
lim x 2  lim 4x  lim (2)
=
4
2
=
30
x 4
x 4
+ 4(4) - 2
lim (x 2  1) lim (x 3  4)
=
lim x
=
[1 + 1][1- 4]
=
-6
=
3 lim (x  4)
=
3(-2 -4)
=
-18
x 1
x 1
2
x 1


 lim 1 lim x 3  lim 4
x 1
x 1
x 1
x  2
lim (t 3  t 2  2)
=
t 0
lim (t 3  1)
t 0
=
lim x 2  5
x 4
=
=
(f)
x 4
=
2
1
2
lim (x 2  5)
x 4
=
lim x 2  lim 5
=
16  5
=
21
x 4
x 4
In all the examples above, we could evaluate the limits by direct substitution. However,
if direct substitution gives us a denominator which approaches zero, we have to try some
“algebraic tricks” to eliminate such a denominator. Here follow some of those “tricks”:
19
(g)
(h)
x2  9
x 3 x 2  4x  3
(x  3)(x  3)
x 3 (x  3)(x  1)
= lim
lim
lim
x 0
(x  3)
x 3 (x  1)
=
lim
=
6
2
=
3
4x 2
x
In this case factorization will not be possible. We can make use of the following
technique called rationalization:
lim
x 0
4x 2
x
=
 4  x  2  4  x  2 
lim 


x 0
x

 4  x  2 
=
lim
=
lim
=
lim
=
lim
=
[ 4  x  2][ 4  x  2]
x[ 4  x  2]
x 0
x 0
x 0
x 0
(4  x)  4
x[ 4  x  2]
x
x[ 4  x  2]
1
4x 2
1
22
=
1
4
Not all limits will exist, as in the following cases:
(i)
lim
x 5
x4
9

2
x  25 0
Symbolically we write
9
  . This does not mean that ∞ (read “infinity”) is the
0
limit, but that the limit does not exist because the quotient becomes large in size
as x approaches 5.
20
(ii)
x2
= ∞
x 3 (x  3) 2
lim
This quotient becomes very large as x approaches 3, thus it becomes positively
infinite.
(iii)
x7
= -∞
x  2 (x  2) 2
lim
Since the quotient becomes large negative as x approaches 2, the expression
becomes negatively infinite.
EXERCISES
Evaluate the following limits if they exist.
1.
3.
lim (x 3  4x  1)
x  1
lim
t  1
1  2t
3t  21
x 3  3x 2  10x
x 2
x2
5.
lim
7.
x 2  5  30
lim
x 5
x 5
1

1
2
9.
lim
2 x
x2
11.
lim
1 t  1 t
t
x 2
t 0
x2 1
x 1 x  1
2.
lim
4.
lim
6.
lim
x4 1
x 1 x 2  1
4  x  15
x 1
x 2 1
8.
10.
lim
x 1
.
12.
x3 1
2x  2  2
3 t  1  2t
t 3
t (t  3)
lim
lim
x 2
6 x 2
3  x 1
13. (a) Estimate the value of
lim
x 0
x
1  3x  1
By graphing the function f (x) 
x
1  3x  1
.
(b) Make a table of values of f (x) for x close 0 and guess the value of the limit.
(c) Use the limit laws to prove that your guess is correct.
21
2.
ONE-SIDED LIMITS
Sometimes the values of a function f(x) tend to different limits as x approaches a number
c from different sides. When this happens, we call the limit of f(x) as x approaches c
from the right, the right-hand limit of f(x) at c. Similarly we call the limit of f(x) as x
approaches c from the left, the left-hand limit of f(x) at c. Limits such as these are
referred to as ONE-SIDED LIMITS.
Definition “Informal”:
Let L be a real number
(i)
Suppose that f(x) is defined near c for x > c, and that as x gets close to c, f(x)
gets close to L. Then we say L is the right-hand limit of f(x) as x approaches
c from the right and we write
lim f(x)  L
x c
(ii)
Suppose that f(x) is defined near c for x < c and that as x gets close to c, f(x)
gets close to L.
Then we say that L is the left-hand limit of f(x) as x
approaches c from the left, and we write
lim f(x)  L
x c
Theorem 2.1
The two-sided limit
lim f(x)  L
x c
if and only if lim f(x) = lim f(x)  L
x c
x c
Examples
(a)
Find lim g(t) if
t0
 2t 2  4 ; t  0
g(t) = 
2
(t  2) ; t  0
Solution:
lim g(t) = lim (2t 2  4)
t 0 
t 0
22
lim g(t)
t 0 
=
2.02 + 4
=
4
=
lim (t  2) 2
t 0 
=
(0-2)2
=
(-2)2
=
4
Therefore , lim g(t)  4 since lim g(t) = lim g(t) = 4.
t 0
(b)
Find
(i)
(ii)
if
t 0
t 0
lim h(k) and
k  1
lim h(k)
k  3
 k7
 k2

h(k)   2
k  2k
 2k  3
Solutions:
(i)
k  3
;
;  3  k  1
;
1  k  3
;
3 k
lim h(k)
k  1
lim  h(k)
k  1
Therefore,
(ii)
lim h(k) = 3, since
k  1
=
=
|-1 -2|
=
|-3|
=
3
=
lim h(k)
k   3
23
k  1
(-1)2 – 2 (-1)
=
1+2
=
3
lim h(k) =
lim h(k)
lim  (k 2  2k)
=
k  1
k  3-
lim k  2
k  1
=
lim h(k) = 3.
k  1
lim (k  7)
k  3-
=
-3 + 7
=
4
=
lim k  2
k   3
=
│-3 - 2│
=
│-5│
=
5
Therefore lim h(k) does not exist, since lim - h(k) ≠ lim  h(k)
k  3
k  3
k  3
EXERCISES
1.
3x  4
2x  3

f(x)  
 4x  1
3x  2
Let
x  1
;
; 1  x  1
;
1 x  2
;
2x
Find
(i)
2.
3.
(ii)
lim f(x)
(iii)
lim f(x)
x 2
;t  1
;t  1
;t  1
lim f (t )
t1
  2x
 x2 1

g(x)  
 2x
 x2  3

Let
Find (i)
3.
x 1
  2t  4

f (t )   4
 t 2 1

Let
Find
lim f(x)
x  1
lim g(x)
x 1
(ii)
;
x  1
; 1  x  1
;
1 x  2
;
x2
lim g(x)
x 2
CONTINUITY
In ordinary language to say a certain process is “continuous” is to say that the process
goes on without interruption and without abrupt changes or without a break. In
mathematics the word “continuous” has much the same meaning. Intuitively speaking, a
function is continuous at a point if its graph does not have a break or jump at that point.
(Note that this is NOT a definition)
24
We can show that the function f(x) = 2 is continuous by drawing the curve of f(x).
y
3
f(x) = 2
1
x
0
We cannot always rely on the drawing of graphs for continuity of functions, because
sometimes it is difficult, if not impossible, to draw some of the graphs. We need a
general mathematical definition.
Definition 2.2:
A function f(x) is continuous at c if the following three conditions are satisfied:
(i)
f ( c) is defined, where c lies in the domain of f.
(ii)
lim f(x) exists
xc
lim f(x) = f (c )
(iii)
xc
Remark:
The limit in the definition above is to be a two-sided limit.
Examples:
 x 1 ; 0  x  1

;
x 1
Determine whether the function f defined as f(x)   1
2x  2 ;
x 1

is continuous at x = 1.
1.
Solution:
(i)
f(1) = 1
lim f(x)
(ii)
x 1
lim f(x)
x 1
= lim (x  1)
x 1
=
=
=
1–1
0
lim (2x - 2)
x 1
=
2–2=0
Therefore lim f(x) = 0 since lim f(x) = lim f(x)
x 1
x1
(iii)
x 1
Since f(1) ≠ lim f(x) , we conclude that the function f is not
x1
continuous at x = 1.
25
2.
x2  4

Determine whether the function g defined as g(x)   x  2 ; x  2

; x2
 4
is continuous at x = 2.
Solution:
(i)
(ii)
g(2) = 4
Since g is defined in the same way on the left-hand side of 2 as on
the right hand side of 2, we need only find the two-sided limit
x2  4
lim g(x) = lim
x2
x 2 x  2
(x  2)(x  2)
= lim
x 2
x2
= lim (x  2)
x 2
=
=
2+2
4.
Hence lim g(x) exists.
x2
(iii)
Since lim g(x) = g(2) = 4, g is continuous at x = 2.
x2
It is also important to consider the continuity of a function over an interval [a,b]
Definition 2.3:
A function is continuous over (or in) the open interval (a,b) if f is continuous at every
point in that interval. ( a may be - ∞ and/or b may be ∞)
Definition 2.4:
The function f is continuous in the closed interval [a,b] if all the following conditions
hold:
(i)
f is continuous at every x in the interval (a,b)
(ii)
f(a) and f(b) both exist.
lim f(x)  f(a) and lim f(x)  f(b)
(iii)
x a
x b
Remark:
“f is continuous at every point in [a,b]” means
(i)
f is continuous at every x in the interval (a,b)
(ii)
f is continuous at a , which means f is right-continuous at a, and is given
by lim f(x)  f(a)
x a
(iii)
f is continuous at b, which means f is left-continuous at b, and is given by
lim f(x)  f(b) .
x b
26
Example:
Determine whether the function h defined by
x  1
 1 ;
 x3
; 1  x  1

h(x)  
 1 x ; 1  x  2
3  x 2 ;
x2
is continuous at (a) x = -1
(b) the interval [1,2]
Solution:
(a)
(i)
h(-1) = (-1)3 = -1
(ii)
lim  h(x) = lim  (1) = -1
x  1
x  1
lim h(x) =
x  1
(iii)
(b)
lim x 3 = (-1)3 = -1
x  1
lim h(x) = h(-1) = -1
x  1
Hence h is continuous at x = -1.
(i)
For every x  (1,2) , h is defined by h(x) = 1-x. Hence
lim h(x) = lim (1  x) = 1- c
x c
x c
and therefore h(c ) = 1-c = lim h(x) , which gives h continuous at every
x c
(ii)
(iii)
point c  (1,2).
Therefore h is continuous at every x in the interval (1,2)
h(1) = 13 = 1 and h(2) = 3 – 22 = -1
and h is therefore defined at x = 1 and x = 2.
lim h(x) = lim (1  x) = 1-1 = 0
x 1
x 1
Hence lim h(x) ≠ h(1) and h is not right continuous at x = 1.
x 1
lim h(x) = lim (1 - x) = 1 2 = -1
x  2
x 2
Hence lim h(x) = h(2) = -1 and h is left continuous at x = 2.
x 2
Conclusion: h is continuous over the interval (1,2].
Definition 2.5:
If the function f is not continuous at x = c, then f is said to be discontinuous at c.
The function f is discontinuous at c if one or more of the three conditions given in the
definition (Definition 2.2) of continuity fails to hold.
In most cases the function f can fail to be continuous at x = c only for one of the
following two reasons:
(i)
either f does not have a limit as x tends to c
(ii)
or f does have a limit as x tends to c, but the limit is not equal to f(c).
27
The discontinuity as a consequence of (i) is called an essential discontinuity
The discontinuity as a consequence of (ii) is called a removable discontinuity. In this
case the discontinuity can be removed by redefining f(x) at c. If the limit is L, we
define f(c ) to be L.
Example:
x2  x  6
is not continuous at x = 2, since 2 is not in the domain
x2  4
of f. Can we make it continuous at x = 2?
Solution:
For f to be continuous at x = 2, lim f(x) = f(2).
The function f(x) =
x2
x x6
x2
x 2
x2  4
(x  2)(x  3)
= lim
x 2 (x  2)(x  2)
(x  3)
= lim
x  2 (x  2)
23
=
22
5
=
4
5
5
If we define f(2) as
, i.e. f(2) = , then lim f(x) = f(2) and f will be continuous at
x2
4
4
x = 2.
2
lim f(x) = lim
The extended function
f(x) ; x  2


g(x)   5
; x2

 4
x2  x  6
; x2
 2
x

4
= 
5

; x2
4

The function g is called the continuous extension of the function f to the point x = 2.
Theorem 2.6
If the functions f and g are continuous at x = c, then
(i)
f + g is continuous at x = c
(ii)
f – g is continuous at x = c
(iii) f.g is continuous at x = c
(iv)
k.g is continuous at x = c, (k any real number)
28
(v)
f
is continuous at x = c, if g(c ) ≠ 0.
g
Proof:
Since f and g are continuous at x = c, then lim f(x) = f (c ) and lim g(x) = g(c)
xc
(i)
lim (f  g)(x)
x c
xc
= lim[f(x)  g(x)]
x c
=
lim f(x) + lim g(x)
xc
xc
= f (c)+ g(c)
= (f+g)(c)
(ii)
lim (f  g)(x)
x c
= lim[f(x)  g(x)]
x c
=
lim f(x) - lim g(x)
xc
xc
= f (c)- g(c)
= (f - g)(c)
(iii)
lim (f.g)(x)
xc
= lim[f(x).g(x)]
xc
=
lim f(x) . lim g(x)
xc
xc
= f (c). g(c)
= (f.g)(c)
(iv)
lim (k.g)(x)
xc
= lim[k.g(x)]
xc
=
k. lim g(x)
xc
= k. g(c)
= (k.g)(c)
(v)
f 
lim  (x)
x c g
 
=
=
f(x)
xc g(x)
lim f(x)
lim
x c
lim g(x)
x c
f(c)
g(c)
f 
 (c) provided g(c) ≠0.
=
g
Similar arguments with right-hand and left-hand limits establish the theorem for
continuity at end-points.
=
29
EXERCISES
Determine if the following functions are continuous at the given points and/or
intervals.
1 ; x0

1.
at x = 0.
f(x)   0 ; x  0
 1 ; x  0

2.
 x 2  2x
; x2

f(x) = 
1
; x2
2
x  6x  4 ; x  2

at x = 2
3.
 x2
; x  0


g(x) =  2
; x  0
 x  1 ; x  0


at x = 0
4.


g(x) = 


5.
6.
; x  2 

;2  x  4

; 4  x 
at x = -2 and the interval [2,4]
; x  1 
2x  1


;  1  x  1
f(x) =  3x
at x = - 1 and the interval [ 1,4]
2x  1
; x  1 

In each of the following exercises find the value of the unknown that will make
the given function to be continuous on  ,  .
a)
b)
c)
7.
1
x
2
x
 x2  4

f ( x)   x  2
k

ax  b

f ( x )  4
2ax  b

kx2  3
f ( x)  
kx  1
if x  2
if x  2
if x  1
if x  1
if x  1
if x  2
if x  2
Let
 A2 x 2 ,
x2
f ( x)  
(1  A) x, x  2
Find A given that f is continuous at 2 .
30
Continuous functions have two properties which are particularly important in calculus:
the Maximum-Minimum Value property and the Intermediate Value property.
Glancing at a graph of a typical continuous function, it is easy to accept the truth of these
two properties. No proofs will be given, as proofs depend on the precise definition of a
limit which will be discussed in the next chapter.
Definition 2.6:
A set of numbers S is bounded above if there exists a number M such that x ≤ M for all
x  S. Such M is called an upper bound for S. The least upper bound of S is the
smallest upper bound.
Similarly a set of numbers S is bounded below if there exists a number m such that
x ≥ m for all x  S. Such an m is called a lower bound. The greatest lower bound of S
is the largest lower bound.
Example:
Consider the set S = { x | 0 < x <1} = (0,1)
M = 2 is an upper bound, but M = 1 is the least upper bound.
M = - ½ is a lower bound, but m = 0 is the greatest lower bound.
Theorem 2.7
Let f be a continuous function at every point of the closed interval [a,b]. Then f takes on
a minimum value m and a maximum value M on [a,b]. That is, there exist numbers x1
and x2 in [a,b] such that f(x1) = m, f(x2) = M and m ≤ f(x) ≤ M for all x  [a,b].
y
M
m
0
x=a
x
x=b
31
x=a
y
x=b
M
x
0
m
REMARKS:
1.
The theorem above is in general not true on an open interval (a,b).
1
For example, f(x) =
is continuous on (0,1), but as x approaches 0 from the
x
right, f(x) becomes very large and f is therefore not bounded on that interval.
Similarly, f(x) = x has the least upper bound 1 on (0,1), but there is no x  (0,1),
such that f(x) = 1.
2.
The theorem above is generally also not true if f is not continuous in [a,b].
1
For example, f(x) =
is not bounded on the interval [-1,1] because it is not
x
continuous at x = 0.
Theorem 2.8 (Intermediate Value Theorem)
Let f be a continuous function on [a,b]. If k is any number between f(a) and f(b), there
exists a number c  (a,b) such that f(c) = k.
32
y
y = f(x)
f(b)
k
f(a)
x
0
a
c
b
EXERCISES
1.
Verify the Intermediate Value Theorem for continuous functions on the given
interval.
(i)
f(x) = x2 -5 on [2,3] ; k = 2
(ii) f(x) = x2 + x + 1 on [ -2,3] ; k = 6
(iii) f(x) = x3 -2x + 1
(iv) f(x) =
2.
on [-2,2] ; k = 1
10
on [0,1];
x 1
2
k=8
Use the Intermediate Value Theorem for continuous functions to show that the
equations have at least one real solution on the given interval.
3.
(i)
2x3 + x2 – x + 1 = 5
(ii)
x5 – 2x2 + x + 11 = 0 on [-2,0]
(iii)
4 – x = 2x on [1,2]
on [ 1,2]
Mpho is 50 centimeters tall when born and grew to a height of 175 centimeters.
Use the Intermediate Value Theorem to argue that Mpho was at some time in her
life exactly 122 centimeters tall.
33
CHAPTER 03
THEORY OF LIMITS
FORMAL DEFINITION OF A LIMIT
In the previous chapter we provided an intuitive definition of a limit, and we stated, but
did not prove, a number of useful limit theorems. At this point you should be
comfortable with the notion of a limit. The intuitive definition earlier introduced is
correct. The only problem is that the words “close to” are not precise. What do we mean
with close?
In everyday life the use of “close” is relative, but in mathematics we need a more precise
description.
The definition of a limit we give next is a precise statement of what you already know:
namely, that as x gets close to c, f(x) gets close to L.
Definition 3.1
(i)
An open interval (a,b) is the set { x  R  a < x < b}.
(ii)
A neighborhood of a point c is an open interval (a,b) such that c  (a,b).
Remark:
The Greek letters  (“delta”) and  (“epsilon”) are used extensively in this chapter. They
will almost always represent small positive quantities.
Definition 3.2
Suppose that the function f is defined near a point c (i.e. in a neighborhood of a point c).
Then
lim f(x)  L
x c
if and only if for every  > 0 there exists  > 0, such that if
0 < |x - c| < 
then
|f(x) – L| < 
Remark:
The formal definition is difficult to understand at first reading.
The inequality 0 < |x - c| that appears in the definition means “x is not c”. The
inequality |x - c| <  states that x lies in the interval ( c - , c +). The two inequalities
combined as a single statement 0 < |x - c| <  describes the open interval
( c - , c + ) from which c is deleted. This deletion is made since the function f need not
be defined at x = c.
The formal definition of a limit requires that for every open interval about L,
34
(L - , L + ), no matter how small, we can find an interval ( c - , c +) about c, of
whose function values lie within that interval about L. In the definition the number  is
the challenge and the number  is the response. No matter how small  is chosen,  can
be made small enough so that f(x) lies within a distance  from L.
The geometric significance of the precise definition of a limit is shown in the following
figure.
y
L+
 L
L-
x
0
c- cc+
Definition 3.3 (Left-Hand Limit)
Let f be a function and c  R, then lim- f(x)  L iff (read “if and only if”) for each every
x c
 > 0 there exists  > 0, such that if
c -  < x < c, then |f(x) – L| < .
Definition 3.3 (Right-Hand Limit)
Let f be a function and c R, then lim f(x)  L iff for each every  > 0 there exists  > 0,
x c
such that if
c < x < c + , then |f(x) – L| < .
Examples:
1.
Show that lim (3x  1)  5
x 2
Solution:
Given any  > 0, we must find a  > 0 such that if 0 < |x-2| <  then
|(3x-1)-5| < .
Now,
|(3x-1)-5|
= |3x-1-5|
= |3x- 6| = |3(x- 2)|
= |3 ||x-2| = 3 |x-2|
35
|(3x-1)-5|
= 3 |x-2| <  if
|x-2| < /3
Thus we can take  = /3.
And
Note the following:
(i)
The value  = /3 is not unique, that is, it is not the only value of  that
will make the - inequality holds. Any smaller positive  will also work.
(ii)
The definition does not ask for a “best”, but only one that will work.
(iii) The value of  depends on the value chosen for .
(iv)
The logical implication symbol “” is used often in a proof to indicate
that a statement follows logically from a previous statement. It does not
mean the two statements are equal, and should not be used in the place
of “=”.
2.
Show that lim x  4  1
x -3
Solution:
Given any  > 0, we must find a  > 0 such that if 0 < |x+3| < , then
| |x+4|-1| < .
Now, consider | |x+4|-1| < . From the properties of absolute values it follows that
|x+4| - |1| ≤ | |x+4|-1| < 
Hence
|x+4| - |1| < 
Or
|x+4| <  + 1
From the properties of absolute values follow:
- ( + 1) < x + 4 < 

-  -1
< x+4 < 

-  -2
< x+3 < 

-  -2
< x+3 < 

-  -2
< x+3 < 

|x + 3| <  + 2
Therefore we can choose  =  + 2
3.
+1
+1
<  +2
+2
Show that lim x 3  1
x 1
Solution:
Given any  > 0, we must find a  > 0 such that if 0 < |x-1| < , then
|x3 - 1| < .
Now, we know (x3 -1) = (x-1)(x2+x+1)
Hence, if
|x3 - 1| < 

|x-1||x2+x+1| < 
(i)
We observe that the expression |x-1| we are looking for appears on the left hand
side, but we also have the expression |x2+x+1| for which we must find a
replacement.
36
First we know from the triangle inequality that
|x2+x+1| ≤ |x2 | + |x |+ |1|
Since we want |x-1| < , we select a value for : let  = 1.
If
|x-1| < 1
Then
-1<x–1<1

0<x<2

-2 < 0 < x < 2

-2 < x < 2

|x| < 2
Therefore
|x2+x+1| ≤ |x2 | + |x |+ |1| < 22 + 2 + 1 = 7
From expression (i) follows that |x-1||x2+x+1| < 7 |x-1| < 

|x-1| < 
 = min{1, 
Therefore
7
7
}
EXERCISES
Use the formal definition of a limit to show that
lim (x  6)  5
1.
x 1
3x 

lim  4    7
5 

x  7
lim   1 
x 4 3

 3
lim (2x 2  1)  3
2.
x 5
3.
4.
x 1
lim (x 2  2x  1)  2
5.
x 1
lim 1  3x  5
6.
x 2
lim 2x  2  6
7.
x 2
lim x  1  2
8.
x 5
lim 1  5x  4
9.
10.
x 3
2 x 3  5x 2  2 x  5
lim
7
x 1
x2 1
11. Let
 1
f ( x)  
1
if x  0
if x  0
Prove that lim f ( x) does not exists.
x0
37
THEOREMS ON LIMITS
Theorem 3.4 (Uniqueness of a limit)
If lim f(x)  L and lim f(x)  M , then L = M.
x c
x c
Proof:
We use a method of proof called “Proof by contradiction”. We are going to show that if
we assume the opposite of the conclusion of the theorem, L  M, then our argument
leads to a contradiction.
Assume M ≠ L then
Since
LM
2
> 0.
lim f(x)  L , for each  =
LM
x c
0 < |x-c| < 1 , then |f(x) - L| <  =
Similarly, since
> 0 there exists a 1 > 0 such that if
2
LM
.
2
lim f(x)  M , for each  =
LM
x c
if 0 < |x-c| < 2 , then |f(x) - M| <  =
LM
2
2
> 0 there exists a 2 > 0 such that
.
Let  = min{1,2}. It follows that for each   0 , there exists   0 such that
|L - M|
= |[L – f(x)] + [f(x) – M]|
≤
|-[f(x) – L]|+|f(x)-M|
=
|f(x) - L|+|f(x)-M|
<
=
LM
2
+
LM
2
|L - M|
The conclusion |L - M| < |L - M| is a contradiction. Hence our assumption that
M ≠ L is wrong, and we conclude that L = M.
Remark:
Since the definition states “………for each  > 0 there exists a  > 0…..” we can choose
38
 > 0 as a convenient number in order to prove the theorem.
Theorem 3.5
Let a be a real number and lim f(x)  L , then lim[a f(x)]  aL
x c
x c
Proof
We want to show that for each  > 0 there exists a  > 0 such that if
0 < |x - c| <  then |af(x) - aL| < 
Let a ≠ 0. For each

> 0 there exists a  > 0 such that if 0 < |x - c| <  , then
a
|f(x) - L| <

a

a

|a||f(x) - L| < |a|

| |a|f(x) - |a|L| < 

| af(x) - aL| < 
Therefore
lim[a f(x)] =
a lim f(x) = aL
x c
xc
Theorem 3.6
If lim f(x)  L1 and lim g(x)  L 2 , then
x c
x c
lim[f(x)  g(x)]
x c
= lim f(x) + lim g(x)
xc
=
xc
L1 + L2
Proof
We want to show that for each  > 0 there exists a  > 0 such that if
0 < |x - c| <  then |f(x)+g(x) - (L1 + L2)| < 
Since lim f(x)  L1 then for each
x c

> 0 there exists 1 > 0 such that
2
0 < |x - c| < 1, implies |f(x) – L1| <

.
2
39
Also since lim g(x)  L 2 , then for each
x c
0 < |x - c| < 2, implies |g(x) – L2| <

> 0 there exists 2 > 0 such that
2

.
2
Let  = min{ 1, 2}. For each  > 0, there exists  > 0 such that
0 < |x - c| < 

|f(x)+g(x) - L1 - L2|
=
|(f(x)- L1 ) + (g(x) - L2 )|
≤
|f(x) – L1|+|g(x) – L2|
<


+
2
2
=

Therefore,
lim[f(x)  g(x)]
= lim f(x) + lim g(x)
x c
xc
=
xc
L1 + L2
Theorem 3.7
If lim f(x)  L1 and lim g(x)  L 2 , then
x c
x c
lim[f(x).g(x) ]
xc
= lim f(x) . lim g(x)
xc
=
xc
L1 . L2
Proof
We must show that for each  > 0 there exists a  > 0 such that if
0 < |x - c| <  then |f(x). g(x) - (L1 . L2)| < 
Since lim f(x)  L1 , for each 1 > 0 there exists a 1 > 0 such that
x c
0 < |x - c| < 1
 |f(x) – L1| < 1
But,
|f(x)| - |L1|
≤ |f(x) – L1 │ < 1
Hence,
|f(x)|
< 1 + |L1 |……………………………………………………..(i)
Since lim g(x)  L 2 , then for each 2 > 0 there exists 2 > 0 such that
x c
40
0 < |x - c| < 2
 |g(x) – L2| < 2

Let  = min{ 1, 2} and choose 1 =
and
2 L2
2 =

2( 1  L1 )
Then for each  > 0, there exists  > 0 such that
0 < |x - c| < 

|f(x). g(x) - (L1 . L2)|
=
|f(x)g(x) – L2f(x) + L2f(x) – L1L2|
=
|f(x)[g(x) – L2 ] + L2 [f(x) – L1]|
≤
|f(x)[ g(x) – L2 ]| + |L2 [f(x) – L1]|
=
|f(x)|| g(x) – L2 | + |L2|| (x) – L1|
<
( 1+ |L1|)2 + |L2| 1
( 1  L1 )
=

=
2
=


2( 1  L1
 L2

2 L2

2

Therefore
= lim f(x) . lim g(x)
lim[f(x).g(x) ]
xc
xc
=
xc
L1 . L2
Theorem 3.8
If lim f(x)  L1 and lim g(x)  L 2 ≠ 0, then
x c
x c
f(x)
xc g(x)
lim
lim f(x)
=
x c
=
lim g(x)
x c
L1
L2
Proof
Assume L2 ≠ 0.
1
1

x c g(x)
L2
We first show that lim
We must show that for a given  > 0, there exists  > 0 such that when
0 < |x - c|<  , then
1
1

g(x) L 2
 
41
But,
1
1

g(x) L 2
L 2  g(x)
L 2 g(x)
=
<
 …………………………..(A)
Select 1 such that 0 < |x - c| < 1 implies that |g(x) – L2| <
(Choose 1 =
L2
2
L2
2
)
Then
|L2|
=
|L2 – g(x) + g(x)|
≤
|L2 – g(x) | +|g(x)|
≤
½|L2| + |g(x)|
|g(x)| ≥
And so
½ |L2|
Then for 0 < |x - c| < 1 we have
|L2 g(x)|
=
|L2 ||g(x)|
>
|L2 | .½ |L2|
=
½ (L2)2
Taking reciprocals, we obtain
1
2
 2
L 2 g(x)
L2
Similarly, there is a 2 such that if 0 < |x - c| < 2 , then from (A)
| L2 – g(x)| <  |L2 g(x)| <  . ½ (L2)2
Now choose  = min{ 1 , 2 }. Then for 0 < |x - c| < ,
L 2  g(x)
L 2 g(x)
=
1
. L 2  g(x)
L 2 g(x)
<
2 L22
. 
L22 2
= 
1
1
.

x c g(x)
L2
which shows that lim
Finally, we obtain
f(x)
1
= lim f(x) . lim
x
c
xc g(x)
xc g(x)
lim
42
=
L1 .
=
L1
L2
1
L2
In calculus the following theorem is a technical result which is very important in proofs.
It is used to confirm the limit of a function via comparison with two other functions, of
which the limit is either known or can be easily computed.
Theorem 3.9 (The Pinching Theorem)
(Also known as the Squeeze Theorem or the Sandwich Theorem)
Let p > 0. Suppose that, for all x such that 0 < |x - c| < p, h(x) ≤f(x) ≤g(x),
lim h(x)  L and lim g(x)  L , then lim f(x)  L
x c
x c
x c
Proof
Let p > 0 be such that if 0 < |x - c| < p then h(x) ≤ f(x) ≤ g(x).
Since lim h(x)  L , for each  > 0, there exists 1 > 0 such that if 0 < |x - c| < 1
x c
then | h(x) – L | < .
Therefore
-  < h(x) - L < 

L -  < h(x) < L + 
Since lim g(x)  L , for each  > 0, there exists 2 > 0 such that if 0 < |x - c| < 2
x c
then | g(x) – L |< .
Therefore
-  < g(x) - L < 

L -  < g(x) < L + 
Let  = min{p, 1 ,2 }. For every x satisfying 0 < |x - c| <  we have
L -  < h(x) ≤ f(x) ≤ g(x) < L + 


L -  < f(x) < L + 
   f ( x)  L  

|f(x) - L| <

Therefore lim f(x)  L .
x c
43
y
g
f
L
h
x
c
INFINITE LIMITS AND LIMITS AT INFINITY
Consider the function defined by f(x) =
1
; x ≠ 0 and given by the following graph and
x2
table:
Fig. A
y=
1
x
2
x
±1
±0,5
±0,1
±0,01
±0,001
, x0
f(x)
1
4
100
10000
1000000
1
0
-1
1
Observe that as x → 0, both from the left and the right, f(x) increases without bound and
hence no limit exists. We say that as x → 0, f(x) becomes infinite and, adopting the limit
notation for convenience, we write
1
lim 2  
x 0 x
On the other hand, look at the graph of f(x) =
44
1
, x ≠ 0.
x
Fig. B
y
y=
1
,x≠0
x
x
0,01
0,001
0,0001
-0,01
-0,001
-0,0001
x
As x approaches zero from the right,
f(x)
100
1000
10000
-100
-1000
-10000
1
becomes positively infinite. As x approaches
x
1
becomes negatively infinite. Symbolically we write
x
1
1
and
lim  
lim  
x 0 x
x 0 x
1
Now let us again examine the function f(x) =
, x ≠ 0, as x becomes infinite in both a
x
positive and negative sense.
zero from the left,
x
1,000
10,000
100,000
1,000,000
f(x)
.001
.0001
.00001
.000001
x
-1,000
-10,000
-100,000
-1,000,000
f(x)
-.001
-.0001
-.00001
-.000001
From the table above we can see that as x increases without bound through positive
values, the corresponding values of f(x) approaches zero.
Like-wise, as x decreases without bound through negative values, the corresponding
values of f(x) also approach zero. Symbolically we have
1
1
and
lim  0
lim  0
x  x
x   x
We can now turn to the evaluation of the limit of a quotient of two polynomials where the
variable becomes infinite. For example consider
45
x 2  3x  5
x  x 2  x  1
It is clear that as x →∞, both numerator and denominator become infinite. However, the
form of the quotient can be changed so that we can draw a conclusion as to whether or
not a limit exists. Since x→ ∞, we are concerned only with those values of x which are
very large. Thus, we can assume x ≠ 0.
A frequently used “gimmick” is to divide both the numerator and denominator by
the power of x which is the largest in either the numerator or denominator. In our
example it is x2. Thus
3 5
x 2 (1   2 )
2
x  3x  5
x x
=
lim
lim 2
x 
x  x  x  1
1
1
x 2 (1   2 )
x x
1
1
lim 1  3 lim  5 lim 2
x 
x  x
x  x
=
1
1
lim 1  lim  lim 2
x 
x  x
x  x
1
1
As x→ ∞,
approaches 0. In fact, lim p  0 for p > 0.
x


x
x
2
x  3x  5
1  3(0)  5(0)
1
Thus, lim 2
=
=
= 1
x  x  x  1
1 0  0
1
lim
Examples
8x  2x  3
x  2x 3  3x  1
2
1.
lim
=
=
=
=
3 
8 2
x3  2  3 
x x
x 
lim 
x 
3
1 

x32  2  3 
x
x 

1
1
1
8 lim  2 lim 2  3 lim 3
x  x
x  x
x  x
1
1
lim 2  3 lim 2  lim 3
x 
x  x
x  x
8(0)  2(0)  3(0)
2  3(0)  (0)
0
= 0
2
46
2.
x2 1  3 x2 1
lim
x  4
x4 1  5 x4 1
=
=
lim
x  4
lim
x 
x 2 (1 
1
x2
)  3 x 3 ( x1 
1
x3
)
x 4 (1 
1
x4
)  5 x 5 ( 1x 
1
x5
)
x 1
1
x2
x 3
1
x

1
x3
x 4 1
1
x4
x 5
1
x

1
x5
=
lim
=
1 0
1 0
x  4
1
1
x2
3
1
x

1
x3
1
1
x4
5
1
x

1
x5
= 1.
EXERCISES
2x 2  1
1.
lim 2
x  5x  3x
1
2.
lim
x2  4
x  x
3.
4.
5.
6.
7.
x 2 1
lim
x 
4x 2  x
x3 1
lim
x  (x  1)(x 2  x  1)
lim
x 
x
3
5x
2
2
3
2
1

lim  x  x 

x  1
lim 4x  x  1  2x
2
x 
2
x 
2
8.
x3  1
.
x    x  1 x 2  x  1
9.
a) Estimate the value of
lim

lim
x  

 x  x  1  x
2
by graphing the function f ( x)  x2  x  1  x
b) Use the table of values of f (x) to guess the value of the limit.
c) Prove that your guess is correct.
10.
Repeat question 9 for f ( x)  3x2  8x  6  3x2  3x  1 .
47
CHAPTER 04
LIMITS OF TRIGONOMETRIC FUNCTIONS
Students are advised to study Chapter 01 before continuing with this chapter,
Consider a unit circle with an angle θ measured in radians.
y
A
θ
0
C
x
B
The length L of arc AC is θ, since from Chapter 02 follows
Length of arc AC = (radius of circle)(angle in radians)
L
= r . θ = 1. θ = θ
Consider the right angled triangle AOB, then
AB
AB
sin θ =
=
= AB
AO
1
Hence
sin θ = AB < length arc AC = θ

0

lim 0
< |sinθ| < |θ|

 0

0
<
lim sin 
 0
lim sin 
 0
From the Pinching Theorem follows that
lim sin   0
 0
Similarly it can be shown that
lim cos   1
 0
48

<
lim 
 0
0
Theorem 4.1
lim
h 0
sinh
1
h
Proof:
Consider a unit circle with the triangles OAD, OAC and OBC as in the diagram.
B
y
A
h
O
D
C
x
It is clear that
Area OAC < Area of circle sector OAC < Area  OBC
(1)
But,
Area of OAC
= ½ base x height
= ½ OC.AD
AD AD
= ½ .1. sin h
[since sinh =

 AD ]
OA
1
= ½ sinh
Area of circle sector OAC
= ½ r2 h
= ½h
Area of OBC
= ½ base x height
= ½ OC. BC
= ½ 1. tanh
= ½
[since tanh =
sinh
cosh
From (1) it follows that
½ sinh
<
½h
<
½
49
sinh
cosh
BC BC

 BC ]
OC
1

sinh
<
h <

1
<
h
sinh
sinh
cosh
1
<
cosh
Taking reciprocals give
cos h <
sinh
h
< 1
sinh
 lim 1
h 0
h 0 h
h 0
Using the Pinching Theorem we can conclude that
sinh
lim
1
h 0 h

lim cosh  lim
Theorem 4.2
lim
cos   1
 0
Proof
lim
cos   1

 0

=
=
=
=
=
=
Note that
lim
1  cos 
 0

= lim
 0
0
cos   1 cos   1
.
 0

cos   1
2
cos   1
lim
 0  (cos   1)
lim
 sin 2 
 0  (cos   1)
sin 
sin 
- lim
. lim
 0 
 0 cos   1
lim
-1.0
0
 (cos   1)

= - lim
 0
Examples
1.
sin 2 2x
x 0 8x 2
lim
=
1  sin 2 2x 

lim 
2
x 0 2
 4x 
2
=
=
=
1  sin2x 
lim 

x 0 2
 2x 
sin2x
sin2x
. lim
½ lim
2x0 2x
2x0 2x
½ .1.1 = ½
50
cos   1

= -0 = 0
2.
lim
x0
sin4x
sin3x
=
=
=
=
=
3.
 sin4x 3x 4x 
lim 

x 0
 4x sin3x 3x 
4
sin4x
3x
lim
lim
3 4x0 4x 3x0 sin3x
4
sin4x
1
lim
4x

0
sin3x
3
4x
lim
3x0 3x
4 1
.1.
3 1
4
3
Determine limπ[secy  tany]
y 
2
Since the limit is not in the two known forms, we must use trigonometric
identities to transform it into one of the forms.
 1
siny 
=
lim π[secy  tany]
lim π 

y  2
y   2 cosy
cosy 

=
1  siny 
lim π 

y  2
 cosy 
(i)

), then (y +  ) = (  + y) → 0.
2
2
2


We also know that sin( + y) = cosy and cos ( + y) = - siny
2
2
Substituting this in (i), we obtain
1  cos( π2  y)
1  siny 
=
lim
lim 

(  y)0
y 
sin( π2  y)
 cosy 
[1  cos( π2  y)]
1
=
lim
(  y)0 sin( π  y)
1
2

π
- [cos( 2  y) - 1]
2  y
lim
=

(  y)0 sin( π  y)
2
2  y

- [cos( π2  y) - 1]
2  y
lim
lim
=

(  y)0 sin( π  y) (  y)0
2
2  y
[cos( 2  y )  1]
1
=
.(

1
)
lim
(   y ) 0
sin( 2  y )
( 2  y )
lim
(   y ) 0 (   y )
2
=
1.(-1).0
=
0.
If y →( -
π
2
π
2
π
2
π
2
π
2
π
2
2
2
51
EXERCISES
Find the following limits if they exist:
sin6x
1.
lim
x0 sin2x
1 - cos 2 2x
x 0
x2
2.
lim
3.
lim
4.
lim
5.
lim ( x 2  1)
6.
lim
7.
lim
3tan3t - 2sint
t0
t
1 - cosx
x0 sinx
x 0
tanx
x
sinx - cosx
x 4
cos2x
sin(x - 1)
x 1 x 2  x  2
52
CHAPTER 05
DERIVATIVES OF ORDINARY FUNCTIONS
One of the main problems of calculus is that of finding the slope of the tangent line at a
point on the graph of a function.
What line, if any, should be called tangent to the graph at the given point? You may
recall, from your study of geometry, that a tangent line, or more simply a tangent, to a
circle is a line which intersects the circle at one and only one point.
Tangent lines to a
circle
However, this notion of a tangent has some short comings when working with graphs in
general. The lines L1 and L2 both intersect the graph at one and only one point.
Intuitively, we would not want to call L2 the tangent at this point, but certainly we would
consider L1 to be one.
y
L2
L1
0
x
We would consider the line L3 to be tangent to the point (a,f(a)) even though it intersects
the graph at more than one point.
53
y
L3
(a,f(a))
x
From these examples, it becomes evident that we must abandon the notion of a tangent as
that of a line which intersects a curve at one and only one point.
If P(x1,y1) and Q(x2,y2) are any two distinct points on the curve, the line passing through
y  y1
them is called a secant line. The slope of the secant line PQ is 2
.
x 2  x1
y
y2
Q(x2,y2)
y2-y1
y1
P(x1,y1)
R(x2,y1)
x2-x1
x1
x2
x
0
If P is held fixed and Q is allowed to move along the curve towards P, the slope of the
secant line can change. Moreover, if this process is continued, the secant line may
approach a limiting position. The same limiting position may also occur if Q approaches
P from a direction to the left of P. If such a limiting position of the secant does exist,
independent of the direction, then the unique line having this limiting position is called
the tangent line to the curve at the point P.
y
Q Q’
Q”
Limiting position of tangent at P
Q1
Q2
P
0
x
54
Definition 5.1
The slope of a curve at a point P is the slope of the tangent line at P.
Since the tangent is a limiting position of secant lines, the slope is the limiting value of
the slope of the secant lines PQ as Q approaches P.
y
Q(x2,f(x2))
y2
y=f(x)
f(x2)-f(x1)
P(x1,f(x1))
y1
x2-x1=h
0
x1
x2
x
f(x 1  h)  f(x 1 )
x1  h  x1
f(x  h)  f(x 1 )
= 1
h
The slope of the tangent line at the point (x1,f(x1)) is therefore
f(x  h)  f(x1 )
lim 1
h 0
h
Slope of the line PQ =
MPQ
=
Definition 5.2
If y = f(x) defines a function of f at x, the limit
f(x  h)  f(x)
,
lim
h 0
h
if it exists, is called the derivative of f with respect to x and is denoted by f ′ (x), which
is read “f prime of x”. The process of finding the derivative of a function is called
differentiation.
Other ways of denoting the derivative of y = f(x) are
dy
d[f(x)]
(read “dee y, dee x”) ;
; Dxy ; Dxf
dx
dx
;
Dx[f(x)] ; y ′ ; f ′ .
ONE-SIDED DERIVATIVES
Definition 5.3
A function y = f(x) is differentiable on a closed interval [a,b] if
1.
it has a derivative at every interior point i.e. every point of (a,b)
55
f(a  h)  f(a)
at the endpoint a exists
h 0
h
f(b  h)  f(b)
at the endpoint b exists.
lim
h 0
h
2.
the right-hand derivative lim
3.
the left-hand derivative
In the right-hand derivative at x = a, h is positive and a+h approaches a from the right.
In the left-hand derivative at x = b, h is negative and b+h approaches b from the left.
f(a  h)  f(a)
M= lim
h 0
h
a
a+h
M= lim
h 0
b+h
b
f(b  h)  f(b)
h
x
Definition 5.4
A function has the (two-sided) derivative at a point if and only if the function’s righthand and left-hand derivatives are defined and equal at that point.
Let us consider some examples that apply the definitions.
Examples
1.
(a)
Use first principles to find the slope of the parabola y = x2 + 1 at any
point on the curve.
(b)
Find the slope of y = x2 + 1 at the point P(-3,10).
(c)
Write an equation for the tangent to the parabola at this point.
Solution:
(a)
The slope of the parabola is given by
dy
f(x  h)  f(x)
=
lim
h 0
dx
h
(x  h) 2  1  (x 2  1)
=
lim
h 0
h
2
x  2xh  h 2  1  x 2  1
=
lim
h 0
h
2
2xh  h
=
lim
h 0
h
h(2x  h)
=
lim
h 0
h
lim (2x  h)
=
h 0
=
2x + 0 = 2x
56
(b)
(c)
2.
At any point (x,y) , the result in (a) shows that the slope of the parabola is
dy
= 2x. At the point P( -3,10)
dx
dy
│x = -3 = 2(-3) = -6.
dx
Since the tangent to the parabola at P(-3,10) passes through this point, we have
from the equation
y – y1 = m(x-x1) (the point-slope from of a straight line)
that
y – 10 = -6(x-(-3)) = -6(x+3) = -6x -18
y = -6x -18 + 10
= -6x – 8
Hence, the equation of the tangent to y = x2 + 1 in the point (-3,10) is
y = -6x – 8
Find f ′ (x) if f(x) =
-1
.
2(x  3)
Solution:
f ′(x)
=
=
=
=
=
=
=
=
3.
f(x  h)  f(x)
h 0
h
1
1

2(x  h - 3) 2(x  3)
lim
h 0
h
x  3  (x  h  3)
2(x  h  3)(x  3)
lim
h 0
h
x 3 x  h 3
2(x  h  3)(x  3)
lim
h 0
h
h
lim
h 0 2h(x  3)(x  h  3)
1
lim
h 0 2(x  3)(x  h  3)
1
2(x  3)(x  3)
1
2(x  3) 2
lim
Show that f(x) = |x| is not differentiable at x = 0.
57
Solution:
f ′ (x) =
lim
=
lim
f ′ (0) =
lim
h 0
f(x  h)  f(x)
h
xh  x
h 0
h
At x = 0, we have
h 0
=
h
But
lim
h0
h
h
h
1 , h0

 1 , h  0

h
0h  0
so that
lim-
h0
h
h
= -1
and
lim
h 0
h
= 1
h
Thus
f ′ (0) = lim
h
h0
does not exist, since lim-
h
h
≠ lim
h
.
h 0 h
h
Consequently, f(x) = |x| is not differentiable at x = 0.
h0
4.
Find f ′(2) if it exists, if
f(x) =
1
x2
Solution:
Thus
f ′ (x) =
lim
f ′ (2) =
lim
h 0
f(x  h)  f(x)
h
f(2  h)  f(2)
h 0
h
1
=
lim
h 0
2h2
h

1
22
2- 4h
=
lim
h 0
4  h. 4
h
58
.
2 4h 2 4h
.
h 0
2.h. 4  h 2  4  h
=
lim
=
lim
=
lim
=
lim
=
lim
=
=
h 0
h 0
h 0
4  (4  h)
2.h. 4  h .(2  4  h )
44-h
2.h. 4  h .(2  4  h )
-h
2.h. 4  h .(2  4  h )
-1
2. 4  h .(2  4  h )
1
2.2(2  2)
1

16
h 0
5.
Show that the function
x 2 , x  0
f(x) = 
2x , x  0
has no derivative at x = 0.
Solution:
f(0  h)  f(0)
f ′ (0) =
lim
h 0
h
Using one-sided limits we shall use f  for the left-hand derivative and f  the
right-hand derivative.
f(0  h)  f(0)
lim
f  (0) =
h 0 h
=
h2  0
limh 0
h
lim- h
=
0
=
2
(using definition of f(x) for x ≤ 0)
h0
and
f  (0) =
=
=
f(0  h)  f(0)
h 0
h
2
2h - 0
lim
*(see explanation below)
h 0
h
lim 2
lim
h 0
=
2
Since f  (0)  f  (0), it follows that f ′ (0) does not exist.
Hence the function f has no derivative at x = 0.
59
* For f  (0) we use for f(0+h) the definition of f(x) for x > 0 but f(0) is still f(0) = 02.
6.
Find f ′ (3) if it exists for
 12 x 2 , x  3
f(x)
=

 3 , x  3
f(3  h)  f(3)
limf  (3) =
h 0
h
1
- 2 (3  h) 2  (3)
=
(f(3) = -3)
lim
h 0 h
- 1 (9  6h  h 2 )  3
=
lim- 2
h 0
h
2
9
- 2  3h - h2  3
=
lim
h 0 h
h 3
=
lim- (3   )
h 0
2 2h
=
-3 – 0 – (-∞)
=
∞
f(3  h)  f(3)
lim
f  (3) =
h 0
h
- 3 - (-3)
=
lim
h 0 
h
0
=
lim
h 0 h
=
0
Therefore f ′ (3) does not exist since f  (3) ≠ f  (3).
EXERCISES
1.
Use the definition of a derivative to differentiate the following:
(a)
f(x) = x2-x+1
(b)
f(x) = 2x3+1
(c)
f(t) =
(d)
r(k) =
(e)
x(f) =
1
t2
1
k 1
3
f 2
60
2.
3.
Find f ′(c ) (if it exists) from first principles:
(a)
f(x) = (3x-7)2 ,
(b)
f(x) = x +
(c)
f(x) =
(d)
f(x) =
if c = 2
2x , if c = 4
1
,
x3
1
x2
if c = -1
,
if c = 2
Find f ′(c ) (if it exists) from first principles:
(a)
x 1
 4x
f(x) =  2
for c = 1
2x  2 x  1
(b)
 x 1
f(x) = 
2
(x  1)
(c)

(x  1) 2
f(x) = 
2

(x  1)
(d)
f(x) =
x  1
x  1
x0
x0
for c = -1
for c = 0

 3x 2
x 1
for c = 1
 3

2x

1
x

1

RELATIONSHIP BETWEEN DIFFERENTIABILITY AND CONTINUITY
We have seen the function
x 2 , x  0
f(x) = 
2x , x  0
is not differentiable at x = 0. It is, however, continuous at x = 0, since
lim f(x)  lim x 2  0  lim f(x)  lim 2x  0  f(0)  0
x 0
x 0
x 0
x 0
We can clearly see it from the following graph:
61
y
f(x) = 2x
f(x) = x2
0
x
On the other hand, the function f(x) = |x| is continuous at x = 0, but it is not
differentiable at that point.
y
f(x) =│x│
0
x
Theorem 5.5
If f is a differentiable function at a point c, then f is continuous at c.
Proof
f(c  h)  f(c)
Since f is differentiable at x = c, it follows that lim
= f ′ (c) exists.
h 0
h
We want to show that lim f(x)  f(c) .
x c
Let x = c+h, then if h → 0, x → c.
Now
lim[f(c  h)  f(c)]
h 0
=
=
=
=
Hence lim f(c  h)  lim f(c)] =
h 0
and
h 0
lim f(c  h)
h 0
=
 f(c  h)  f(c) h 
lim 
. 
h 0
h
1

f(c  h)  f(c)
h
. lim
lim
h 0
h0 1
h
f ′(c).0
0
0
lim f(c) = f(c)
h0
62
or
lim f(x)
xc
=
f(c)
which shows that f is continuous at x = c.
RULES FOR DIFFERENTIATION
It should now be quite apparent that differentiating a function by direct use of first
principles (the definition), can be a tedious task. Hence we shall begin to develop rules
for simplifying the differentiation process. These rules not only give us mechanical and
efficient procedures for determining derivatives, but they also avoid the direct use of
limits.
Theorem 5.6 (Derivative of a constant function)
If f(x) = c, c a constant, then f′(x) = 0.
Proof
f(x  h)  f(x)
f ′(x) =
lim
h 0
h
f(c)  f(c)
=
lim
h 0
h
0
=
lim
h0 h
lim 0
=
h0
=
0
Example:
If f(x) = 3, then f ′ (x) = 0
Theorem 5.7 (Power Rule)
If f(x) = xn, where n is any real number, then f ′ (x) = nx n-1.
Proof
We shall prove this theorem for positive integers n.
f(x  h)  f(x)
f ′ (x) =
lim
h 0
h
(x  h) n  x n
=
lim
h 0
h
We need the Binomial Theorem which will only be covered in MATH 102. We state it
here without proof:
For n  N, x,y  R,
n
n(n  1) n 2 2 n(n  1)(n  2) n 3 3
(x+y)n = x n  x n 1 y 
x y 
x y  ............  nxy n 1  y n
1
1.2
1.2.3
Use this expansion for (x+h)n in the formula for f ′ (x).
63
x n  nx n 1 h 
f ′ (x) = lim
h 0
n(n  1) n 2 2
x h  ...........  h n  x n
2
h
n(n  1) n 2 2
x h  ...........  h n
2
lim
h 0
h
n(n  1) n 2


h nx n 1 
x h  .......  h n 1 
2

lim 
h 0
h
n(n  1) n 2


lim nx n 1 
x h.......  h n 1] 
h 0
2


n-1
nx +0+0…..+0
nx n-1
nx n 1 h 
=
=
=
=
=
Remark:
Although the theorem was only proved for n  N, the statement is true for all real
values of n.
Example:
1.
If f(x) = x7 , then f ′ (x) = 7x6
1
2
1
1 3 1
1 3
3
2.
For g(x) = x , then g ′ (x) = x
= x
3
3
Theorem 5.8
If g(x) = cf(x), then g′(x) = cf ′(x)
Proof
g(x  h)  g(x)
g′ (x) =
lim
h 0
h
cf(x  h) - cf(x)
=
lim
h 0
h
f(x  h) - f(x)
=
c lim
h 0
h
=
cf ′(x)
Example:
If f(x) = 4x -3, then f ′(x) = 4
d 3
(x ) = 4(-3x -3-1) = -12 x – 4
dx
Theorem 5.9 (Derivative of a sum)
If r(x) = f(x) + g(x), then r′(x) = f ′(x) + g′(x)
64
Proof
r′(x)
r(x  h)  r(x)
h
(f  g)(x  h)  (f  g)(x)
lim
h 0
h
[f(x  h)  g(x  h)]  [f(x)  g(x)]
lim
h 0
h
f(x  h) - f(x)  g(x  h) - g(x)
lim
h 0
h
f(x  h) - f(x) g(x  h)  g(x)

lim

h 0
h
h
f(x  h) - f(x)
g(x  h) - g(x)
lim
 lim
h 0
h

0
h
h
f ′(x) + g′(x)
=
lim
h 0
=
=
=
=
=
=
Example
3
1.
3
d
d
d
(2x 4  5x 4 ) = 2 (x 4 )  5 (x 4 ) = 2(4x3) +
dx
dx
dx
1

 3  14 
4
3 15


5 x  = 8x + x
4
4

Theorem 5.10 (Product Rule)
Let r(x) = f(x).g(x). If f ′(x) and g′(x) exist, then
r′(x) = f(x).g′(x) + g (x).f ′(x)
Proof
r(x  h)  r(x)
r ′(x) =
lim
h 0
h
f(x  h)g(x  h)  f(x)g(x)
=
lim
h 0
h
f(x  h)g(x  h) - f(x  h)g(x)  f(x  h)g(x)  f(x)g(x)
=
lim
h 0
h
g(x

h)

g(x)
f(x  h)  f(x) 

=
lim f(x  h)
 g(x)

h 0
h
h


g(x  h)  g(x)
f(x  h)  f(x)
=
lim f(x  h) lim
 lim g(x) lim
h 0
h 0
h

0
h

0
h
h
=
f(x)g′(x)+g(x)f ′(x)
Example
Find g′(x) if g(x) = (2x2-3x)(x -4 – x5)
d
d
g′(x) =
(2x2-3x) (x -4 – x5) + (x -4 – x5)
(2x2-3x)
dx
dx
=
(2x2-3x)(-4x -5 – 5x4) + (x -4 – x5)(4x-3)
=
-8x-3 -10x6 +12x -4 + 15x5 + 4x-3 -3x -4 -4x6 +3x5
=
-14x6 +18x5 +9x -4 -4x -3
65
Theorem 5.11 ( Quotient Rule)
f(x)
Let r(x) =
such that g(x) ≠0. If f ′(x) and g′(x) exist, then
g(x)
g(x) f (x)  f(x) g (x)
r′ (x) =
[g(x)] 2
Proof
r(x  h)  r(x)
r′(x) =
lim
h 0
h
f(x  h) f(x)

g(x  h) g(x)
=
lim
h 0
h
f(x  h)g(x)  f(x)g(x  h)
=
lim
h 0
h.g(x  h)g(x)
f(x  h)g(x)  f(x)g(x  h)  g(x)f(x) - g(x)f(x)
=
lim
h 0
h.g(x  h)g(x)
g(x)[f(x  h) - f(x)]  f(x)[g(x  h)  g(x)]
=
lim
h 0
h.g(x  h)g(x)
f(x  h) - f(x)
g(x  h) - g(x)
g(x)
 f(x)
h
h
=
lim
h 0
g(x  h)g(x)
f(x  h)  f(x)
g(x  h)  g(x)
g(x) lim
 f(x) lim
h 0
h 0
h
h
=
g(x) lim g(x  h)
=
g(x) f (x)  f(x) g (x)
x 0
g(x) 2
Example
Find t′(x) for the function defined as t(x) =
2x 2  3x
x3
Solution
x3
t′(x)
=
=
=
=


 
d
d 3
2x 2  3x  (2x 3  3x)
(x )
dx
dx
2
x3
x 3 (4x  3)  (2x 3  3x)3x 2
x6
4x 4  3x 3  6x 5  9x 3
x6
 6x 5  4x 4  6x 3
x6
66
=
=

x 3  6x 2  4x  6
x3x3
 6x 2  4x  3
x3

Most of the functions used in practice are composite functions, i.e. the composition of
two or more functions, like f(x) = (2x-3)2 or g(x) = 3 sin(5x 2 ) . The rule for calculating
the derivative of the composite of two or more differentiable functions, is called the
Chain Rule, and is probably the most extensively used rule of differentiation.
Theorem 5.12 (Chain Rule)
Let f and g be differentiable functions of x, then y = (fₒg)(x) = f[g(x)] is also a
differentiable function of x and
dy
f[g(x  h)]  f[g(x)]
 lim
dx h 0
h
f[g(x  h)]  f[g(x)] g(x  h)  g(x)
= lim
h 0
h
g(x  h)  g(x)
f[g(x  h)]  f[g(x)] g(x  h)  g(x)
= lim
.
h 0
g(x  h) - g(x)
h
f[g(x  h)]  f[g(x)]
g(x  h)  g(x)
= lim
. lim
h 0
h 0
g(x  h) - g(x)
h
As h → 0, then g(x+h) →g(x) because g is differentiable at x and therefore also
continuous at x. (See Theorem 5.5)
Let g = g(x+h) – g(x), then g(x+h) = g(x) + g. As h → 0, then g → 0.
Therefore
f[g(x  h)]  f[g(x)]
g(x  h)  g(x)
dy
. lim
 lim
h

0
h

0
dx
g(x  h) - g(x)
h
f[g(x)  Δg]  f[g(x)]
= lim
.g (x)
Δg0
g(x)  Δg - g(x)
f[g(x)  g]  f[g(x)]
= lim
.g (x)
Δg0
g
= f ′[g(x)].g′ (x)
Example
Differentiate y =
2x  3 3x 4
2x  5
Solution
1
dy
dx
=
1 2


4 3
d  2x  (3x ) 
1 
dx 
 (2x  5) 2 
67


1  2x  3x 
1 
2
 (2x  5) 2 
4
3
=

1
2


1  2x  3x 

.
1 
2
2
 (2x  5) 
4
3
1
1
2
4


d  2x  3x 3 
.
1 
dx 
 (2x  5) 2 
4
4
1

d 
d
3 
3
(2x  5)
2x  3x   (2x  3x ) (2x  5) 2
dx 
dx

1
2
1


2
(2x

5)




1
1
1
1
1
2
4
1

4 2

 (2x  5) 2 (2  3. 4 x 3 )  (2x  3x 3 ) 1 (2x  5) 2 d (2x  5)
3
1 2x  3x 
3
2
dx
 
.
1 
2

1
2


 (2x  5) 2 
2
(2x  5) 


1
4
1

4 2

 (2x  5) 2 (2  4x 3 )  (2x  3x 3 ) 1 (2x  5) 2 .2
3
1 2x  3x 
2
 
.
1 

2
(2x  5)
2
 (2x  5) 
=

1
2


1  2x  3x 
.
1 
2
 (2x  5) 2 
4
3
1
2
1
3
(2x  5) (2  4x ) 
4
3
(2x  3x )
1
(2x  5) 2
(2x  5)
1

1
4
(2x  5)(2  4x 3 )  (2x  3x 3 )
=
4

 2
1  2x  3x 3 
.
1 
2
2
 (2x  5) 
=
4
1
4

 2
1  2x  3x 3  (2x  5)(2  4x 3 )  (2x  3x 3 )
.
1 
3
2
 (2x  5) 2 
(2x  5) 2
=
4
4
1

 2
1  2x  3x 3  10  2x  5x 3  20x 3
.
1 
3
2
2
(2x  5) 2
 (2x  5) 

1

1
1
2
(2x  5)
(2x  5)
EXERCISES
1.
Differentiate the following functions using the appropriate rules.
(a)
f(t) = 11t5 - 6t3 +8
68
2.
3.
1
x
(b)
f(x) = x -
(c)
y = (x2 -1)(x3 – 3)
(d)
y = x(3x-2)(x4 -1)
(e)
y =
x2 1
2x  3
(f)
y =
x4  4
2x 3  1
(g)
g(y) = 2y(3y2-1)(y2+2y+3)
(h)
 4x  3 
f(x) = 

 5x  2 
(i)
f(x) =
(j)
y = 4[(3x-8)(3x2-2x+1)3]4
Find
3
x  1 x
dy
for the following. Give the final answer in terms of t.
dt
(a)
y =
1
, u = t2 -2t
2
u
(b)
y=
1  7u
, u =1 + (2t-5)2
1 u2
Find the equation of the tangent line to the graph of the function at the given
point:
(a)
(b)
(c)
2x  3 , at ( ½ , 2)
1
f(x) =
, at (1,1)
2x
f(x) = 1 – 2x2 – 3x3, at (-2, -7)
f(x) =
IMPLICIT DIFFERENTIATION
The differentiable functions we have encountered so far, were equations in which y is
9x 2
expressed in terms of x, we write it as y = f(x). Examples are y =
or y = tan(sin2x).
x 1
These equations are said to define y explicitly in terms of x. Any other kind of equation
involving y, in which we assume y is a differentiable function of x, is said to define y
implicitly in terms of x, e.g. x + 2xy + y = 1.
69
To introduce the notion of implicit differentiation, we shall consider the following
Problem:
Find the slope of the tangent line at the point ( 2 , 2 ) of the circle of radius 2 whose
centre is at the origin.
y
(x,y)= ( 2 , 2 )
-2
0
2
x
Remember the equation of the circle would be given as
x2 + y2 = 4
or
x2 + y2 - 4 = 0
(1)
dy
at this point.
dx
In the case of Equation (1), the variables are not expressed like this. We say, that Eq.(1)
is of the form F(x,y) = 0 where F(x,y) denotes a function of two variables.
The obvious thing to do is to solve Eq. (1) for y in terms of x:
x2 + y2 - 4 = 0
y2
= 4 – x2
To find the slope at the point ( 2 , 2 ) , all we need to do is evaluate
y
=  4  x2
(2)
A problem now occurs, namely that Eq.(2) gives two values of y for a particular value of
x, and hence does not express y explicitly as a function of x. We can, however,
“consider” Eq.(1) as implicitly defining y as the two different functions of x:
y =  4  x 2 and y =  4  x 2
whose graphs are given in the following figure.
y
y
( 2, 2 )
0
y=
x
0
4  x2
y =  4  x2
70
x
Since the point ( 2 , 2 ) lies on the graph of y =  4  x 2 , we should differentiate that
function:
1
dy
d
=

4  x 2 2
dx
dx
1

1
=
(4  x 2 ) 2 (2x)
2
x
=
4  x2
Then
dy
dx
=
x 2
 2
4  ( 2)
=
2
 2
42
= -1.
Thus the slope of the circle x2 + y2 - 4 = 0 at the point ( 2 , 2 ) is -1.
There are some difficulties with this method:
1.
y is not given explicitly in terms of x.
2.
Seeking to write y explicitly in terms of x, it was discovered that y is not an
explicit function of x.
3.
Some function may be very complicated or impossible to find an explicit
expression for y, e.g. x3 + y3 = 2xy.
Let us now consider a method which will avoid these difficulties.
Consider again the function F(x,y) = x2 + y2 - 4 = 0 . This function express y implicitly
in terms of x. It is assumed or implied that the equation determines y as at least one
differentiable function of x, i.e. y = f(x).
Then
x2 + [f(x)]2 – 4 = 0
dy
To find
where F(x,y) = 0, we shall treat y as a differentiable function of x and
dx
differentiate both sides of this equation with respect to x.
In particular, for the equation x2 + y2 - 4 = 0 we have
d
d
( x2 + y2 - 4 ) =
(0)
dx
dx
d 2
d 2
d
d
(0) = 0
(3)
(x ) 
(y )  (4) =
dx
dx
dx
dx
Since y is assumed to be a function of x, we have y = f(x) and y2 = [f(x)]2
d 2
d
=
(y )
[f(x)] 2
dx
dx
From the power rule and chain rule follows
d 2
d
dy
=
= 2yy′.
(y )
[f(x)] 2 = 2f(x).f ′ (x) = 2y.
dx
dx
dx
(3) can then be written as
71
and
2x + 2yy′ - 0 = 0.
dy
Solving for y′ =
from (4), gives us
dx
2yy′ = -2x
 2x
x
or
y′ =
= 2y
y
y (
Then
2, 2)
= 
2
2
(4)
(5)
= -1 , as before.
dy
is called implicit differentiation
dx
Note that Eq.(5) is not defined when y = 0. geometrically it makes sense, since the
tangent line to the circle at either (2,0) or (-2,0) is vertical and the slope is therefore not
defined for such a line.
This method of finding
Examples
Assume that y is a differentiable function of x and find
equations by implicit differentiation.
1.
x + 2xy + y = 1
d
d
(x + 2xy + y ) =
(1)
dx
dx
dy
d
dy
1 + 2x
+ 2y (x) +
= 0
dx
dx
dx
dy
(2x + 1)
= -1 – 2y
dx
dy
 1  2y
=
dx
1  2x
x y 4
2.
1
1
Since x 2  y 2  4 , then
1
1
Dx ( x 2  y 2 ) = Dx(4)
1
1
1 2 1 2 d
x  y
(y) = 0
2
2
dx
1
1
y =
2 x
2 y
y′
= 
y
x
, if x  0.
72
dy
from the following
dx
Using the original equation,
y  4 - x , we obtain
y′ = 
4 x
x
x3 = (y-x2)2
Dx( x3 ) = Dx(y-x2)2
3x2 = 2(y-x2)Dx(y-x2)
= 2(y-x2)(y′ - 2x)
2
3x
= 2(yy′ -x2 y′ - 2xy + 2x3)
= 2yy′ -2x2 y′ - 4xy + 4x3
= y′(2y - 2x2 ) - 4xy + 4x3
2
Then y′ (2y - 2x ) = 4xy - 4x3 +3x2
3x 2  4x 3  4xy
y′ =
2(y  x 2 )
If we want to find the slope of the curve x3 = (y-x2)2 at the point (1,2), we have
3(1) 2  4(1) 3  4(1)(2)
7
=
y (1,2) 
2
2
2[2  (1) ]
3.
EXERCISES
dy
Find
for the following equations.
dx
1.
xy + y2 = 1
2.
x2y + xy2 = 10
3.
xy2 + xy = 0
4.
x2+5xy-3y3-5 = 0
5.
2y2 + xy = x2 + 3
xy  2x  y
6.
7.
x  y  xy  6
HIGHER ORDER DERIVATIVES
Since the derivative of a function y = f(x) is itself a function, it makes sense to expect that
f ′ (x) may also be a differentiable function. When this is done, the result may also be
differentiated. Continuing in this manner, we obtain higher-order derivatives.
If y = f(x), then f ′ (x) is called the first derivative of f with respect to x. The derivative
of f ′ (x), denoted by f (x), is called the second derivative of f with respect to x, etc.
Some of the various ways in which higher-order derivatives of the function y = f(x) may
be denoted are indicated below.
73
First
y′ ;
Second
y″ ;
Third
y″′ ;
Fourth
dy
d
;
[f(x)]
dx
dx
d2y
d2
f″(x);
D2 x y ;
;
[f(x)]
dx 2
dx 2
d3y
d3
f ′″(x);
D3 x y ;
;
[f(x)]
dx 3
dx 3
d4y
d4
(4)
4
f (x);
D xy ;
;
[f(x)]
dx 4
dx 4
f ′(x);
y(4) ;
Dxy;
Note that when the prime notation is replaced by digits to indicate the higher derivative,
the digit is put in brackets, to distinguish between a derivative and a power of a function,
e.g.
d4
4
4
(4)
f (x) = [f(x)] while f (x) =
[f(x)]
dx 4
EXAMPLES
1.
Find y″ if y =
y′ =
y″ =
2.
8x4 + 9x3 -24x2 +6x -10
32x3 +27x2 - 48x +6
96x2 + 54x – 48
d2y
Find
if x + 2xy + y = 1
dx 2
d
d
(x + 2xy + y ) =
(1)
dx
dx
dy
d
dy
1 + 2x
+ 2y (x) +
= 0
dx
dx
dx
dy
(2x + 1)
= -1 – 2y
dx
dy
 1  2y
=
dx
1  2x
d y
dx 2
d
d
(1  2y)  (1  2y) (1  2x)
dx
dx
2
(1  2x)
dy
(1  2x)( 2 )  (1  2y)(2)
dx
(1  2x) 2
(1  2x)
2
=
=
Since
dy
dx
=
 1  2y
, we can write
1  2x
74
2
d y
dx 2
 - 1 - 2y 
(1  2x)( 2
)  2  4y
1  2x 

=
(1  2x) 2
 2(-1 - 2y)  2  4y
=
(1  2x) 2
4  8y
=
(1  2x) 2
EXERCISES
Find the indicated derivative using appropriate methods
1.
x3 + y3 =1
;
y″
d2y
dx 2
2.
x  y 1 ;
3.
x2 + 6xy + y2 = 8 ;
4.
y=
5.
y =
6.
f(x) =
5x  1 ;
1 x
1 x
1
3x 2
dx 3
;
;
d2y
dx 2
d3y
y″′
f (4) (x);
75
CHAPTER 06
DERIVATIVES OF SPECIAL FUNCTIONS
1.
TRIGONOMETRIC FUNCTIONS
In this section we shall compute the derivatives of trigonometric functions. Throughout,
we shall assume that all angles are in radian measure. The material in this section
depends on the knowledge of basic Geometry, see Chapter 01.
1.
Let f(x) = sin x, then f ′(x) = cos x.
Proof
f(x  h)  f(x)
f ′(x) =
lim
h 0
h
sin(x  h)  sin x
=
lim
h 0
h
sin x cosh  cos x sinh  sin x
=
lim
h 0
h
sin x (cosh - 1)  cos x sinh
=
lim
h 0
h
 sinx(cosh  1) cosxsinh 
=
lim 


h 0
h
h


cosh  1
sinh
=
sinx lim
+cosx lim
h 0
h0 h
h
=
sinx.0 + cosx.1
=
cosx
2.
Let f(x) = cos x, then f ′(x) = - sin x.
Proof
f(x  h)  f(x)
f ′(x) =
lim
h 0
h
cos(x  h)  cos x
=
lim
h 0
h
cosxcosh  sinxsinh  cosx
=
lim
h 0
h
cosx(cosh  1)  sin x sinh
=
lim
h 0
h
(cosh  1)
sinh
=
cosx lim
 sinx lim
h 0
h 0 h
h
=
cosx.0 - sinx.1
=
- sinx
76
3.
If f(x) = tan x , then f ′(x) = sec2x
Proof
sinx
Since f(x) = tanx =
, we can use the quotient rule and the result in (1) to
cosx
obtain
d  sinx 
f ′(x) =
dx  cosx 
d
d
cosx (sinx)  sinx (cosx)
dx
dx
=
2
(cosx)
=
=
=
4.
cos x cosx  sinx(-sinx )
cos 2 x
cos 2 x  sin 2 x
cos 2 x
1
= sec2x
2
cos x
If f(x) = cosec x , then f ′(x) = -cosecxcotx
Proof
1
Since f(x) = cosec x =
, we can use the quotient rule and the result in (1)
sin x
and (2) to obtain
d  1 
f ′(x) =


dx  sin x 
sin x
=
=
=
=
=
d
d
(1)  1 (sin x)
dx
dx
(sin x) 2
sin x .0  1(cos x)
sin 2 x
- cos x
sin 2 x
1  cosx
sin x sinx
-cot x.cosecx
77
5.
If f(x) = sec x , then f ′(x) = sec x tan x
Proof
1
Since f(x) = sec x =
it follows that
cos x
f ′(x)
=
d  1 


dx  cos x 
cos x
=
=
=
=
=
6.
d
d
(1)  1 (cos x)
dx
dx
(cos x) 2
cos x .0  1(-sin x)
cos 2 x
sin x
cos 2 x
1 sin x
cos x cosx
sec x tan x
If f(x) = cot x , then f ′(x) = -cosec2x
Proof
cos x
Since f(x) = cot x =
,
sin x
d  cos x 
f ′(x) =


dx  sin x 
sin x
=
=
=
=
d
d
(cosx)  cosx (sinx)
dx
dx
2
(sin x)
sinx(-sinx )  cosx(cosx)
sin 2 x
- (cos 2 x  sin 2 x)
sin 2 x
-1
= -cosec2x
2
sin x
Examples
1.
Differentiate the following:
(a)
f(x) = cot(5x – 3)
(b)
f(x) = sec4xtanx
(c)
f(x) = sinx  tanx
(d)
x3 = y4 + x2siny +1
78
Solutions:
(a)
f ′(x)
d
(5x-3)
dx
= -5cosec2(5x-3)
(b)
f ′(x)
=
= -cosec2(5x-3).
=
=
=
(c)
(d)
f ′(x)
d
d
(tanx) + tanx
(sec4x)
dx
dx
d
sec4x sec2x + tanx .(4sec3x) (secx)
dx
sec4x sec2x + 4tanx. sec3x .secx.tanx
sec6x + 4tan2x sec4x
sec4x
d
(sinx + tanx)
dx
= ½ (sinx + tanx) -½ (cosx +sec2x)
cosx  sec 2 x
=
2 sinx  tanx
= ½ (sinx + tanx) -½
d
d 4
( x3 ) =
(y + x2siny +1)
dx
dx
d
d
d
d
3x2 = 4y3
(y) + x2 (siny) + siny (x2) +
(1)
dx
dx
dx
dx
dy
dy
= 4y3
+ x2.cosy
+ siny.2x + 0
dx
dx
dy
dy
4y3
+ x2.cosy
= 3x2 – 2x.siny
dx
dx
dy
(4y3 + x2cosy)
= 3x2 – 2x.siny
dx
3x 2  2xsiny
dy
=
dx
4y 3  x 2 cosy
EXERCISES
1.
Find f′(x) if
(a)
f(x) = 5x7 +4sinx -7cosx
(b)
f(x) = 8x5 – 5sin(3x2)
(c)
f(x) =
(d)
f(x) = cot3xcosec5x
cosecx
79
2.
(e)
 x2  2 

y  cos 2 
 x2  2 


(f)
3xsinx
y = 3
tan2x
(g)
1
3
2
y = sin ( x  1) 3
2
(h)
y=
(i)
y =
1
sin 2 2t  cos 2 3t
xsin x  2x
LOGARITHMIC FUNCTIONS
Definition 6.1:
The logarithmic function to the base b, denoted by Logb, is
Logb = [ (x,y) |x = by]
The exponent y which satisfies x = by is called the Logarithm (or Log) of x to the base
b and we write y = Logbx, where x is a positive real number.
Properties of the Logarithmic Function
1.
Logb(mn) = Logbm + Logbn
m
n
2.
Log b
3.
Log bmn
4.
Log bx
5.
Log b1
= 0 since b0 = 1
6.
Log bb
= 1 since b1 = b
7.
Log b
8.
Log bx
1
m
= Logbm - Logbn
= nLogbm
= y if and only if x = by
= - Logbm
=
Log a x
Log a b
80
9.
Log bx
=
1
Log x b
Logarithms to the base 10 (ten), called Common Logarithms, are used for
computational purposes. The subscript 10 is generally omitted from the notation. If there
is no subscript, we assume a subscript of 10. Thus
Log100 means log10100.
Which is the best of all possible bases b to use? In theory you can use ANY base for
logarithms, and the rules of logarithms work for any base. The only reason why base10
was used as base for Common Logarithms, was because it seemed reasonable at the time,
like the way the counting number base was chosen to be 10. In calculus the base e is
preferred, where e = 2.718281828459045235360….., an irrational number.
When we want to find the derivative of f(x) = log bx, we shall examine the limit
log b (x  h)  log b (x)
f ′ (x) = lim
.
h 0
h
In this process we shall encounter a limit which we have not discussed before, namely
1
h
lim (1  h) .
h 0
In order to obtain the derivative of the logarithm functions, we shall assume that the limit
exists and that
lim(1  h)
1
h
h 0
= e
(A)
in honor of the eighteenth century mathematician Leonhard Euler (pronounced “oiler”).
Another consequence of (A) is
1
lim(1  ) n  e
n 0
n
n an integer.
Theorem 6.2
1
log b e .
x
d
1
1
f′(x) =
log e x = log e e =
dx
x
x
Proof
If f(x) = logbx, then f ′(x) =
f ′(x)
=
=
=
If b = e, then for f(x) = log ex we get
f(x  h)  f(x)
h 0
h
log b (x  h)  log b x
lim
h 0
h
xh
log b
x
lim
h 0
h
lim
81
(B)
=
=
1
xh
lim log b
h 0 h
x
1x
 h
lim
log b 1  
h 0 x h
 x
x
=
1
 h h
lim log b 1  
h 0 x
 x
=
1
 h h
lim log b 1  
x h 0
 x
x
Let k =
h
.
x
If h → 0, then
h
→ 0.
x
Hence
x
f ′(x)
=
=
=
1
 h h
lim log b 1  
x h 0
 x
1
1
log b lim 1  k  k
k 0
x
1
log b e
x
If b = e, f(x) = log e x is called the natural logarithm.
1
1
1
In this case
f ′(x) = log e e = .1 = .
x
x
x
The natural logarithm can also be written as ln x. Described in simple terms,
log e x = ln(x) is the power to which the number e = 2.718281828….. would have to be
raised to equal x. For example,
log e e 2 = 2logee = 2 and log e 1 = 0 since e0 = 1.
Examples
1.
Find f ′(x) if f(x) = log27x
Using property (8) we can write
log e 7x
f(x) = log27x =
log e 2
1
Then
f ′(x) =
.D x ln(7x)
ln2
1 1
=
. D x (7x)
ln2 7x
1 7
=
.
ln2 7x
1
=
x.ln2
82
2.
Differentiate f(x) = ln[(x2+1)2(x-1)5x3 ]
Using the logarithm laws we can write
f(x) = ln(x2+1)2 +ln(x-1)5 +lnx3
= 2ln(x2+1) +5ln(x-1)+3lnx
Then
f ′(x) = 2Dxln(x2+1) +5Dxln(x-1)+3Dxlnx
1
1
1
= 2 2
D x (x 2  1)  5
D x (x  1)  3 D x (x)
x 1
x
x 1
2.2x
5.1 3.1
= 2


x 1 x 1 x
4x
5
3
= 2


x 1 x 1 x
3.
Find Dx(f(x)) if f(x) = ln
f(x) =
=
Then
Dx(f(x)) =
=
3.
x 4 1
x4 1
x4 1
½ ln 4
x 1
½ ln(x4-1) – ½ ln(x4+1)
1 4x 3
1 4x 3

2 x 4 1 2 x 4 1
2x 3
2x 3

x 4 1 x 4 1
EXPONENTIAL FUNCTION
Definition 6.3:
The function defined by
f = { (x,y) |y = bx, b > 0, b ≠ 1}
where the exponent x is any real number, is called an exponential function.
In the previous section we have encountered the irrational number e = 2.718…..
as the base to the natural logarithm. Since we know that if x = logey, then y = ex,
it makes sense to also consider the function f(x) = ex.
Definition 6.4:
The function f(x) = ex = exp(x), x any real number, is called the exponential function
with base e and exponent x.
From the graph of the function f(x) = ln(x) (See the figure below) we can see that
f(x) = ln(x) is a 1-1 function, and has therefore an inverse, i.e. (f(x))-1 = (ln(x))-1 exists.
This inverse of the natural logarithm function is exactly the exponential function to
83
base e.
Definition 6.5:
For every real number x, y = ex if and only if x = ln(y).
The graph of ex may be obtained by reflecting the graph of y = ln(x) across the line y = x.
Graph of the natural logarithm function and the exponential function
y
f(x) = ln(x)
1
x
y=e
0
1
y=x
Properties of the exponential function
1.
ex > 0 for all real numbers x
2.
e x+y = ex.ey
ex
3.
e x-y = y
e
1
4.
e –x =
ex
cx
5.
e = (e x)c
Equations involving lnx and ex
Since y = ex and y = lnx are inverses of each other,
e lnx = x for all x > 0
and
ln (ex) = x for all x
84
x
In the proof for the derivative of f(x) = ex , we shall make use of the expansion of the
function as a series. This will be discussed in later modules with you, and we just state
the result here without proof
ex =

xn
x x2
x3
x4

1




 ..........

1 1.2 1.2.3 1.2.3.4
n 0 n!
│x│ < ∞.
Theorem 6.6
If f(x) = ex, then f ′ (x) = ex
Proof
f ′(x)
f(x  h)  f(x)
h 0
h
=
lim
=
lim
=
lim
=
lim
=
e x lim
=
=
e x h  e x
h 0
h
exeh  ex
h 0
h
e x (e h  1)
h 0
h
eh 1
h 0
h
h h2
h3
h4
1 


 ..........  1
1 1.2 1.2.3 1.2.3.4
e x lim
h 0
h
 h h2 h3

h 1  

 ...........
2 6 24

e x lim 
h 0
h
=
 h h2 h3

e x lim 1  

 .........
h 0
 2 6 24

=
ex.1
=
ex
The functions ax and a f(x) and their derivatives
1.
If a is a positive number and x any real number, then
85
y = ax = exlna
and
2.
y′ = lna . exlna = ax lna.
If a > 0 and f(x) is any differentiable function of x, then
y = af(x) = e f(x)lna
y′ = lna.f ′ (x). e f(x)lna = lna.f ′ (x). a f(x)
and
Examples:
2x
1.
If y = e 3 tan4x , then
2x
2x
dy d
d

(e 3 ).tan4x  e 3
tan4x
dx dx
dx
2x
=
2.
3.
If f(x) =
ex 1
, then
ex 1
2 3
e tan4x  4sec 2 4xe
3
f ′ (x) =
(e x  1)e x  (e x  1)e x
(e x  1) 2
=
e 2x  e x  e 2x  e x
(e x  1) 2
=
2e x
(e x  1) 2
Find the derivative of 3 sinx
Since we can write f(x) = 3sinx = e sinx.ln3
It follows
f ′(x) = ln3.(cosx) . e sinx.ln3
= 3sinx .ln3.cosx
EXERCISES
1.
Find f ′ (x) if
(e)
(f)
(g)
f(x) = ln 4 1 e x
f(x) = (e 2x – e -2x)2
f(x) = ln(sinx.sin2x)
2
f(x) = ln
2  3x
2
f(x) = e (x ) e  x
f(x) = log10 (x3 + 3x -5)
f(x) = ln(x2+4x)5(tan4(2x))
(h)
(i)
f(x) = e sin(x cos
f(x) = 2 secx
(a)
(b)
(c)
(d)
2
2x
3
2x)
86
(j)
2.
f(x) =
2 2x .3 3x
Use implicit differentiation to find
x.
(a)
(b)
(c)
(d)
dy
. Assume y is a differentiable function of
dx
ylnx = xey
ln(y2-1) – ln(y-1) = sinx
e (x ) e (2x1)  e y
ln(y-2) = ln(sinx) – x
2
4.
LOGARITHMIC DIFFERENTIATION
The derivative of a function given by a complicated equation can sometimes be
calculated more quickly if we take the logarithm on both sides of the equation before
differentiating. This process is called logarithmic differentiation.
We usually use logarithmic differentiation in the following cases:
1.
If the logarithm of a function is simpler as the function itself.
2.
If it is of the form y = [f(x)] g(x)
Example
3
dy
cos2x
Find
if y  4
dx
x tan3x
Solution:
Take the natural logarithm on both sides of the equation
 3 cos2x 
ln(y) =
ln  4

 x tan3x 
=
ln 3 cos2x  ln(x 4 .tan3x)
1
=
ln(cos2x)  [lnx 4  ln(tan3x) ]
3
1
=
ln(cos2x)  4lnx  ln(tan3x)
3
Take the derivative on both sides, using implicit differentiation of the left.
1 1
1
1
1 dy
=
.(sin2x).2  4 
sec 2 3x.3
3 cos2x
x tan3x
y dx
=
Solve now for
dy
dx
dy
=
dx
 2sin2x 4 3sec 2 3x
 
3cos2x x
3
  2sin2x 4 3sec 2 3x 
 
y 

3 
 3cos2x x
87
cos2x   2sin2x 4 3sec 2 3x 
 

.
3 
x 4 tan3x  3cos2x x
3
=
1
4
(x  1) (3x  1) (x 3  2x) 6
dy
3
Find
if y 
dx
(x  3) 3 5 2x  4
Take natural logarithm on both sides
2
2
2.
lny
2
3
 ln
4
1
4
(x  1) (3x  1) (x 3  2x) 6
2
4
(x  3) 3 5 2x  4
2
1
1
lny  4ln(x 2  1)  ln(3x  1)  6ln(x 3  2x)  3ln(x  3)  ln(2x  4)
3
4
5
Differentiate left and right with respect to x
2 1 dy
1
1 1
1
1
1 1
4 2
.2x 
.3  6 3
.(3x 2  2)  3

.2
3 y dx
4 3x  1
x  3 5 2x  4
x 1
x  2x
2 dy
8x
3
18x 2  12
3
2
 2

 3


3y dx x  1 4(3x  1)
x  3 5(2x  4)
x  2x
dy
Solving for
, gives us
dx
3
18x 2  12
3
2
dy
 3y  8x
=  [ 2
]

 3


dx
x  3 5(2x  4)
x  2x
 2  x  1 4(3x  1)
=
1
4
3 (x  1) (3x  1) (x 3  2x) 6
8x
3
18x 2  12
3
2
[
][
].




2
3
35
2
x  3 5(2x  4)
x  1 4(3x  1)
x  2x
(x  3) 2x  4
2
3.
4
dy
if y = x sinx
dx
Take logarithms on the left and right
lny = ln( x sinx) = sin(x).ln(x)
Differentiate left and right
1
1 dy
= sin(x). + ln(x)(cosx)
x
y dx
sinx
=
 ln(x)cos(x )
x
dy
Solving for
dx
dy
sinx
= y[
 ln(x)cos(x ) ]
dx
x
Find
88
= x sinx. [
sinx
 ln(x)cos(x ) ]
x
EXERCISES
1.
Find f ′ (x) for the following using logarithmic differentiation
(a)
f(x) = 2 sin5x
3
2
(b)
f(x) = e x x x
(c)
f(x) = x x
(d)
2.
5.
Find
f(x) =
x 10 x 2  5
3
8x 2  2
dy
using logarithmic differentiation
dx
(a)
y = ax
(b)
y=
x2 1
x 1
(c)
y=
(x 3  1) 4 sin 2 x
3x
(d)
2
2
y = x 5 (x 2  8) 4 e x 2
INVERSE TRIGONOMETRIC FUNCTIONS
Let y = sin(x). Can we solve x as a function of y? With our current knowledge it is not
possible.
Suppose that sin(x) = ½ , then x would be

5
13
, or
, or
, and so on. In fact, if y
6
6
6
is any number in the interval [ -1, 1], there are an infinite number of values of x for which
sin(x) = y. This is illustrated in the following figure:
89
y
1

x1 x2

2
y
0 x3
 x4
x5
x
2
sin(xi) = y.
-1
  
We can eliminate this problem by restricting x to lie in a certain interval, say  ,  .
 2 2
In this interval, the function f(x) = sin(x) is a 1-1 function and will have an inverse.
Thus, as in the figure above, for each value of y in [-1,1], there is a unique value of x in
  
 2 , 2  such that sinx = y.


Definition 6.7:
The inverse sine function is the function that assigns to each number x in [-1,1] the
  
unique number y in  ,  such that x = siny. We write y = sin-1x or y = arcsin(x).
 2 2
Theorem 6.8
In the open interval (-1,1), y = sin-1x = arcsin(x) is differentiable and,
1
dy
=
,
-1 < x < 1
dx
1 x2
Proof
Let y = sin-1x with x  (-1,1). Then x = sin y and
d
d
[siny] =
(x)
dx
dx
dy

cosy.
= 1
dx
dy
1

=
dx
cosy
1
=
1  sin 2 y
1
=
-1 < x < 1
1 x2
90
Definition 6.9:
The inverse cosine function is the function that assigns to each number x in [-1,1] the
unique number y in [0, ] such that x = cos(y). We write y = cos-1x or y = arccos(x)
y
1
-π


0
2

π
x
2
y = sin(x)
-1
Theorem 6.10
In the open interval (-1,1), f(x) = cos-1x = arccos(x) is differentiable and,
-1
f′(x) =
,
-1 < x < 1
1 x2
Proof
Let y = cos-1x with x ϵ (-1,1). Then x = cos y and
d
d
[cosy] =
(x)
dx
dx
dy

-siny.
= 1
dx
dy
-1

=
dx
siny
-1
=
1  cos 2 y
-1
=
-1 < x < 1
1 x2
Definition 6.11:
The inverse tangent function is the function that assigns to each real number x the
  
unique number y in   ,  such that x = tany. We write y = tan-1x or y = arctanx.
 2 2
91
y



2
2
x
y = tanx
Theorem 6.12
In the open interval (- ∞,∞), f(x) = tan-1x = arctan(x) is differentiable and,
1
f ′ (x) =
,
-∞<x<∞
1 x2
Proof
If y = tan-1x, then x = tany, and
d
d
[tany] =
(x)
dx
dx
dy

sec2y.
= 1
dx
dy
1

=
dx
sec 2 y
1
=
1  tan 2 y
1
=
;
-∞<x<∞
1 x2
Similarly the following can be proved:
Theorem 6.13
1.
2.
3.
dy
1
;

dx 1  x 2
dy
1

If y = cosec-1x, then
;
dx x x 2  1
If y = cot -1x, then
If y = sec-1x, then
dy
1

;
dx x x 2  1
92
-∞<x<∞
|x| >1
|x| >1
Example
1.
Find
dy
if y = tan-1 x  1
dx
Solution
Let u = x  1 = (x +1)½
du
Then
= ½ (x+1) -½
dx
dy
du
Hence
= dy .
dx
du dx
= d tan 1 u . d (x +1)½
du
dx
1

1 1
2
=
.
(x

1)
1 u2 2
1
1
=
.
1  ( x  1) 2 2 x  1
1
=
2(2  x) x  1
2.
x 2  4  2sec 1
Find f ′(x) if f(x) =
x
2
1
Solution:

1 2
d
(x  4) 2 . (x 2  4)  2
f ′(x) =
x
2
dx
2
1
=
=
3.

1 2
2
1
(x  4) 2 .(2x) 
x x 2 4 2
2
4
2
x
4

x2  4 x x2  4
dy
2
x
if y = cot 1  tan 1
dx
x
2
Solution:
1
d 2
1
d x
dy
=
 
 
2
2
dx
 2  dx  x 
 x  dx  2 
1  
1  
x
2
2
1
 1(2x )
=
 2 2
4
x
1 2
1
x
4
Find
93
1
x2 
2
d x
 
 1 dx  2 
=
=
=
2
1

 x2  4 
 4  x2


x 2 
2
2

 x 
 4
2
2
 2
2
x 4 x 4
4
2
x 4



EXERCISES
Find the derivatives of the following functions. Simplify where possible.
1. y = tan - 1 x
2. G(x) = ln(tan-1x)
3. f(x) =
1 1
sin (3x  4x 2 )
3
4. y = tan-1(e-3x)
5. y = sin[sec-1(lnx)]
6. H(x) = (arcsinx)lnx
7.
 cosx 
y = sin-1 

 1  sinx 
94
CHAPTER 07
L’HôPITAL’S RULE
In the late seventeenth century, John Bernoulli discovered a rule for calculating limits of
fractions whose numerators and denominators both approach zero. The rule is known
today as L’HôPITAL’S Rule. The rule gives fast results even when other methods are
unavailable or slow.
(i)
The Indeterminate Form
0
0
f(x)
A limit of the form lim
, where lim f(x) = 0 and lim g(x) = 0 has the indeterminate
xa
xa
xa g(x)
form
0
at x = a. It is called indeterminate because such a limit might be
0
 any real number,
 ∞ or - ∞ , or
 it might not exist.
In earlier chapters we used a few techniques for finding such limits. L’HôPITAL’S
Rule is yet another powerful technique.
Definition 7.1:
If lim f(x) = 0 and lim g(x) = 0 and both f ′(x) and g′(x) exist, then
xa
xa
f(x)
f (x)
= lim
x

a
g(x)
g (x)
provided the limit on the right exists or is ∞ or - ∞ .
lim
xa
0
f(x)
is of the indeterminate form as x → a, then the limit of
0
g(x)
f(x)
f (x)
as x → a is equal to the limit of
as x → a, which is obtained by
g(x)
g (x)
differentiating both the numerator and denominator of the original quotient.
This rule says that if
PITFALLS
1.
2.
f(x)
, do not differentiate it as quotient
g(x)
with the quotient rule. Rather, differentiate the numerator and denominator
f (x)
independently, obtaining
.
g (x)
Do not apply L’HôPITAL’S Rule to forms which are not considered
indeterminate.
In applying L’HôPITAL’S Rule to
95
Examples
1.
x 2  3x  10
x 2
x2  4
Solution:
Both the numerator and the denominator approach 0 as x approaches 2, so this is
Find
lim
0
the indeterminate form . By L’HôPITAL’S Rule
0
x 2  3x  10
lim
x 2
x2  4
d (x 2  3x  10)
dx
= lim
d (x 2  4)
x 2
dx
2x  3
x 2
2x
43
=
4
7
=
4
= lim
2.
Evaluate
ln(x  3)
x 4
x4
lim
Solution:
Both the numerator and the denominator approaches 0 as x approaches 4, so this
is the indeterminate form
0
and we can use L’HôPITAL’S Rule
0
ln(x  3)
=
lim
x 4
x4
=
=
d [ln(x  3)]
dx
lim d
x 4
(x  4)
dx
1
x -3
lim
x 4
1
1
1
= 1
If, after applying L’HôPITAL’S Rule , another indeterminate form occurs, the rule
can be applied again.
2x 3  3x 2  12x  20
lim
3.
Find
x 2 2x 3  9x 2  12x  4
Solution:
2x 3  3x 2  12x  20
0
lim
(Form )
x 2 2x 3  9x 2  12x  4
0
d (2x 3  3x 2  12x  20)
lim dxd
=
x 2
(2x 3  9x 2  12x  4)
dx
=
6x 2  6x - 12
x 2 6 x 2  18x  12
lim
96
6x 2  6x - 12
is still of the indeterminate form
x 2 6 x 2  18x  12
L’HôPITAL’S Rule again.
2x 3  3x 2  12x  20
0
Hence
(Form )
lim
x 2 2x 3  9x 2  12x  4
0
2
6x  6x - 12
0
=
(Form )
lim 2
x 2 6 x  18x  12
0
d (6x 2  6x - 12)
=
lim ddx 2
x 2
(6x  18x  12)
dx
0
But lim
=
=
=
4.
Evaluate
, hence we can apply
0
12x - 6
x 2 12x  18
18
6
3
lim
sin x  x
x 0
x2
lim
Solution:
lim
x 0
=
lim
=
lim
x 0
sin x  x
(Form
x2
d (sin x  x)
dx
)
0
d (x 2 )
dx
cos x  1
x 0
2x
(Still the form
0
)
0
=
d (cos x  1)
lim dx d
x 0
(2x)
dx
=
lim
=
0
2
x0
0
- sin x
2
= 0
(ii)
Indeterminate Form 

If f(x) and g(x) each become infinite in absolute value (that is, f(x) → ± ∞ and
f(x)
g(x) → ± ∞ as x → a, then lim
is said to be in indeterminate form 
.
xa g(x)
L’HôPITAL’S Rule is also applicable in this case.
97
Examples
1.
Find
lim
x 
2x  6
3x  4
Solution:
Since lim (2x  6)  
and lim (3x  4)   the given quotient is of the form
x 
x 
 . Applying L’HôPITAL’S Rule ,

d (2x  6)
2x  6
=
lim dx
lim
x  d (3x  4)
x  3x  4
dx
=
lim
x 
2
3
2
=
3
2
2.
Evaluate
lim
x 
x
ex
Solution:
Since lim x 2  
and lim e x   the given quotient is of the form 
.
x 
x 
Applying L’HôPITAL’S Rule ,
x2
lim x
x  e
=
d (x 2 )
dx
lim
x  d (e x )
dx
=
2x
x  e x
=
(2x)
lim dx
x  d (e x )
dx
lim
( of form 
)
d
=
lim
x 
2

0
=
=
3.
Evaluate
2
ex
lnx
lim 1
x 0 
x
Solution:
ln x
lim 1
x 0 
(The form 
)
x
d
=
(ln x)
lim dx
x 0
d 1
(
)
dx x
98
=
lim
x 0
1
x
- 12 x
3
2
1
lim (2x 2 )
=
x 0
=
=
-2 0
0
(iii) The Indeterminate Form 0.∞
If lim f(x) = 0 and lim g(x) = ± ∞ , then the product f(x).g(x) is said to be in the
xa
xa
indeterminate form 0. ∞ as x → a. By transforming the product to a form of a quotient,
such as
f(x)
g(x)
or
1
1
f(x)
g(x)
we obtain the indeterminate form
0
0
Examples
1.
Evaluate
or 
 and L’HôPITAL’S Rule may be applied.
lim x ln x
x 0 
Solution:
Since lim x  0 and lim lnx   , this is the indeterminate form 0.∞. Rewriting,
x 0
x 0
ln x
lim x ln x = lim 1
x 0
x 0
x
This is now in the indeterminate form 
 , and we can apply L’HôPITAL’S Rule
ln x
lim x ln x = lim 1
x 0
x 0
=
=
x
d (ln x)
dx
lim d
x 0 
(1)
dx x
1
lim x
x 0   1
x2
= lim ( x)
x 0
= 0
2.
Find
lim (π  2x)sec x
x π
2
Solution:
Since this is the indeterminate form 0.∞, we rewrite it as
99
limπ (π  2x)sec x = limπ
x
x
2
 - 2x
cos x
2
0
Now that it is in the indeterminate form
, we can use L’HôPITAL’S Rule.
0
limπ (π  2x)sec x = limπ
x
x
2
2
 - 2x
cos x
d ( - 2x)
= limπ dxd
x
2 dx (cos x)
-2
x - sin x
2
=
limπ
2
1
2
=
=
(iv)
The Indeterminate Form (∞ - ∞)
If lim f(x) = ∞ and lim g(x) = ∞ , then f(x) - g(x) is said to be in the indeterminate form
xa
xa
∞ - ∞ as x → a. Rewriting f(x) – g(x) as, for example, a quotient, may lead to a
suitable form
0
0
or 
 to which L’HôPITAL’S Rule is applicable.
Note that the forms : ∞ + (- ∞); - ∞ + ∞ and - ∞ - (- ∞) are also classified under the
label ∞ - ∞
Examples
1
 1
Evaluate lim  x
 
x 0  e  1
x
Solution:
The given expression is of the form ∞ - ∞. Combining the fractions in the expression we
obtain
x  ex 1
1
1
=

x(e x  1)
ex 1 x
1.
which takes on the form
0
as x → 0+.
0
Applying L’HôPITAL’S Rule
1
 1
lim  x
 
x 0  e  1
x
=
=
lim
x 0
x  ex 1
(form
x(e x  1)
d (x  e x  1)
dx
lim d
x 0 
x(e x  1)
dx
100
0
0
)
=
lim
x 0
=
)
0
lim d
(xe x  e x  1)
dx
 ex
lim
x 0  xe x  2e x
 e0
0.e 0  2e 0
1
= -½
2
=
=
Evaluate
0
x 0 
=
2.
1  ex
(form
xe x  e x  1
d (1  e x )
dx
1
1 

lim  
x 0 x
sin
x


Solution:
This is the indeterminate form ∞ - ∞, so try to write it in a form which allows the use of
L’HôPITAL’S Rule.
1
 sin x - x 
1 
0

 (form )
=
lim  
lim 
x 0 x
x 0
sin x 
0

 xsin x 
=
(sin x  x)
lim dxd
x 0
(xsin x)
dx
=
lim
d
x 0
cos x  1
(form
x cos x  sin x
=
d (cos x  1)
lim d dx
x 0
(x cos x  sin x)
dx
=
lim
=
0
2
0
=
x0
- sin x
2cos x - xsinx
EXERCISES
Use L’HôPITAL’S Rule to determine the following limits:
3  x3
1.
lim
x 1 5  5x
x2
lim x
2.
x  e
101
0
0
)
(5x  1) 3
x  (7  2x) 3
3.
lim
4.
lim
5.
lim cosec x. ln(1  sin x)
cos 2 x
π
x  sin x  1
2
6.
7.
8.
9.
10.
x 0
 sinx 1 
lim  3  2 
x 0
x 
 x
6  6x  3x 2  6e x
lim
x 0
x  sinx
2
6x  5x  7
lim
x 
4x 2  2x
5
 1

lim 
 2
x  x  1
x  3x  4 

 1

lim x e x  1
x 


102
CHAPTER 08
APPLICATION OF DIFFERENTIATION
1.
THE MEAN-VALUE THEOREM
The Mean-Value Theorem is one of the most important and applicable theorems of
Mathematics. We can use it to prove some important facts about differentiation. We can
also use it to obtain information that is useful in graphing a wide variety of functions.
Before stating and proving the Mean-Value Theorem, we need one preliminary result.
Theorem 8.1 (Rolle’s Theorem)
Let f : R → R be a function which is continuous at every point of the closed interval [a,b]
and differentiable at every point of the open interval (a,b). If f(a) = f(b) = 0, then there
exists at least one number c in (a,b) at which f ′(c) = 0.
f ′ (c) = 0
y
a
0
c
b
x
Proof
Assume f is a continuous function on [a,b] and differentiable on (a,b). Assume also that
f(a) = f(b) = 0. Since f is continuous on [a,b], it follows from Theorem 2.7 (Chapter 02)
that f assumes a maximum value M and a minimum value m in [a,b].
If f(x) = 0 for all x ϵ [a,b], then f ′(x) = 0 for all x in in (a,b), In particular, if c ϵ (a,b), we
have f ′ (c) = 0.
If f is not the zero function, then f(x) must become positive or negative in (a,b).
Suppose f takes on some positive values in (a,b).
Then there exists a number c in (a,b) such that f(c) = M is the maximum value of f(x) on
[a,b]. M > 0, so c is not equal to a or b, since f(a) = f(b) = 0. Therefore, since c  (a,b)
and f is differentiable in (a,b), f ′(c) exists where
f(c  h)  f(c)
f ′(c) =
.
lim
h 0
h
f(c  h)  f(c)
It follows that
 f (c) <  where  > 0.
h
103
f(c  h)  f (c)
 f  (c) = h.
h
f(c  h)  f(c)
And therefore
= f ′ (c) + h.
h
Since f takes its maximum value at x = c, f(c+h) ≤ f(c).
This implies that f(c+h) - f(c) ≤ 0.
Hence
If h > 0, then
f(c  h)  f(c)
≤ 0,
h
f(c  h)  f(c)
which gives
= f ′ (c) + h ≤ 0.
h
If h → 0, then h → 0 and we will have f ′ (c) ≤ 0.
f(c  h)  f(c)
≥ 0,
h
f(c  h)  f(c)
which gives
= f ′(c) + h ≥ 0.
h
If h → 0, then h → 0 and we will have f ′(c) ≥ 0.
f(c  h)  f(c)
Therefore
= f ′(c) = 0.
lim
h 0
h
If h < 0, then
Theorem 8.2 (The Mean-Value Theorem)
Let f : R → R be a function which is continuous at every point of the closed interval [a,b]
and differentiable at every point of the open interval (a,b). Then there exists at least one
number c in (a,b) such that
f(b)  f(a)
.
f  (c) 
ba
Before proving the theorem, let us describe the geometric implication of the theorem.
The expression
f(b)  f(a)
ba
represents the slope of the secant line joining the points (a,f(a)) and (b,f(b)).
We know that f ′(x) is the slope of the tangent line to the curve at any point on the curve.
The Mean-Value Theorem then states that there always exists a number c between a and
b such that the slope of the tangent line at the point (c,f(c)) is the same as the slope of the
secant line between (a,f(a)) and (b,f(b)), i.e. the secant and the tangent lines are parallel.
104
y
(b,f(b))
y
y = f(x)
(c,f(c))
(c,f(c))
y = f(x)
(b,f(b))
(a,f(a))
(a,f(a))
0
a
c
b
a
x
c
0
b
x
Proof
Introduce the function
 f(b)  f(a)

g(x) = f(x) - 
(x  a)  f(a)  .
 ba

The expression
 f(b)  f(a)

y = 
(x  a)  f(a) 
 ba

is the equation of a straight line (the secant line discussed above) that passes through the
points (a,f(a)) and (b,f(b)). [Check: when x = a, y = f(a) and when x = b, y = f(b).]
Since f is continuous on [a,b] and differentiable on (a,b), g is also continuous on [a,b] and
differentiable on (a,b).
Moreover,
 f(b)  f(a)

g(a) = f(a) - 
(a  a)  f(a)
 ba

= f(a) - f(a)
= 0
and
 f(b)  f(a)

g(b) = f(b) - 
(b  a)  f(a)
 ba

= f(b) – f(b)
= 0.
Therefore Rolle’s Theorem applies to the function g.
Hence there exists a c  (a,b) such that g′ (c) = 0.
But
 f(b)  f(a) 
g′ (x) = f ′ (x) - 

 ba 
so that , since g ′(c) = 0, then
 f(b)  f(a) 
g ′(c) = f ′(c) - 
 =0
 ba 
and therefore
105
f ′(c) =
f(b)  f(a)
.
ba
Examples
1.
Exhibit the validity of Rolle’s Theorem on the given intervals:
(a)
f(x) = x2 - 4x + 3 ; [ 1,3]
x2 1
(b)
f(x) =
;
[-1,1]
x2
Solutions:
(a)
f(x)
=
x2 - 4x + 3
f(1)
=
12 – 4(1) + 3
=
1–4+3
=
0
f(3)
=
32 – 4(3) + 3
=
9 -12 + 3
=
0
Since f is continuous on [1,3] and differentiable on (1,3), and f(1) = f(3) = 0, it satisfies
the conditions of Rolle’s Theorem. Hence there is a point c in (1,3) such that f ′(c) = 0.
Now,
f ′(x) = 2x - 4
Hence
2x – 4 = 0
x = 2.

Therefore, c = 2  (1,3).
(b)
x2 1
Since f(x) =
is not defined for x = 0  [-1,1], f(x) is not continuous on
x2
[-1,1]. Therefore Rolle’s Theorem does not apply.
2.
Exhibit the validity of the Mean-Value Theorem on
f(x) = x3 – 3x2 + 1 ;
[1,4]
Solution
f is a continuous function on [1,4] and differentiable on (1,4). Further
f(4)  f(1)
17  1

 6,
4 1
3
and
f ′(x) = 3x2 – 6x = 6.
Therefore
3(x2 -2x -2) = 0, which implies that
And
x2 -2x -2 = 0.
Using the quadratic formula
x
=
=
=
 (2)  (2) 2  4(1)(2)
2(1)
2  12
2
1 3 .
106
Therefore,
c =
1 3
 (1,4).
EXERCISES
1.
Determine if the given function satisfies the hypotheses of Rolle’s Theorem on
the indicated interval. If so, find the value(s) of c which satisfies the conclusion
of the theorem:
(a)
f(x) = x2 – 6x + 5,
[1,5]
3
2
(b)
f(x) = x – 5x + 4x
[0,4]
2

2x  1
2 2
(c)
f(x) =
,


2
2
2 
1 x

2
2.
1
(d)
f(x) =
x 3  3x 3  2
(e)
f(x) =
ln(sin x – ½ )
[1,8]
  5 
6 , 6 


Determine if the given function satisfies the hypotheses of the Mean-Value
Theorem on the indicated interval. If so, find the value(s) of c which satisfies the
conclusion of the theorem:
(a)
f(x) = x3 + x + 2
[2,5]
1
(b)
f(x) =
[-2, -½ ]
x
(b)
f(x) =
[2,6]
4x  1
(c)
f(x) = 3 x  4x
[1,4]
x 1
(d)
f(x) =
[-2,-1]
x 1
 
(e)
f(x) = x + sin x
 2 , 


107
2.
ELEMENTARY CURVE SKETCHING
Consideration of arbitrary selected points on a curve is, in most cases, insufficient
for proper determination of a curve’s shape.
.
B
D
y
f(x)
F
A
E
C
0
G
x
If we look at the shape of the curve from point E to F we realize that the curve is
going down, in other words it is decreasing from E to F. On the other hand, from point F
to G, the curve is going up. We say the function f increases from F to G.
If we draw any tangent line between points E and F, it will be observed that the slope of
the tangent line is negative, whereas the slope of the tangent line between F and G is
positive.
Points A B, C and D are turning points of the curve. Notice that at these points the
derivative of f(x) is equal to zero, because the tangent line at these points is horizontal.
These turning points can be locally or relatively minimum or they can be absolutely or
globally minimum, depending on the nature of the function. They can also be locally or
relatively maximum, or globally or absolutely maximum depending on the nature of the
function.
Theorem 8.3
A function f is said to be an increasing [decreasing] function on the interval (a,b) if and
only if for any two points x1, x2 in (a,b) such that if x2 > x1, then f(x2) > f(x1)
[f(x2) < f(x1)].
Remark
If f ′(x) > 0 on an interval (a,b), then f is an increasing function on (a,b). If f ′(x) < 0 on
(a,b), then f is a decreasing function on (a,b).
Definition 8.4
A function has an absolute maximum [minimum] at the point x0 if f(x0) ≥ f(x)
[f(x0) ≤ f(x)] for all x in the domain of the function f .
Definition 8.5
A function f has a relative maximum [minimum] at the point x0 if there is an interval
around x0 on which f has an absolute maximum [minimum] when x = x0.
108
Remark
1.
If f has a relative maximum or minimum at x = c and if f ′(c) exists, then
f ′(c) = 0.
2.
If c  Df and f ′(x) changes from positive to negative values as x increases
through c, then f has a relative maximum when x = c. If f ′(x) changes from
negative to positive values as x increases through c, then f has a relative minimum
when
x = c.
Definition 8.6
If f ′(c) = 0 or f ′(c) is not defined, c is called a critical value of f. If c is a critical value
and c  Df , then (c, f(c)) is called a critical point.
First Derivative Test for relative maxima/minima of y = f(x).
1.
2.
3.
4.
Find f ′(x).
Determine all the critical values.
In the intervals suggested by the critical values, determine whether f is
increasing (f ′(x) > 0) or decreasing (f ′(x) < 0).
For each critical value c  Df, determine how f ′(x) changes sign as x changes
through x = c.
 If f ′(x) changes from (+) to (-) there is a relative maximum at x = c
 If f ′(x) changes from (-) to (+) there is a relative minimum at x = c
 If f ′(x) does not change sign, there is no relative maximum or minimum.
Definition 8.7
1.
A function f is said to be concave up on an interval (a,b) if f ′ is an increasing
function on (a,b), that is, if f ″ (x) > 0 for all x  (a,b).
2.
A function f is said to be concave down on an interval (a,b) if f ′ is a
decreasing function on (a,b), that is, if f ″ (x) < 0 for all x  (a,b).
Note the following: concavity of a function relates to whether the derivative f ′ is
increasing or decreasing (not the function f).
Second Derivative Test for relative maxima/minima of y = f(x).
Suppose f ′(c) = 0 and c  Df, then
1.
If f ″ (x) < 0, there is a relative maximum when x = c;
2.
If f ″ (x) > 0, there is a relative minimum when x = c;
3.
If f ″ (x) = 0, the test fails. Use the First Derivative Test in this case.
Definition 8.8
A point on the graph in which f ″(x) = 0 or where f ″(x) is not defined is called a point of
inflection. At such a point the sign of f ″(x) must go from (-) to (+) or from (+) to (-),
which means concavity changes from downward to upward, or vice versa.
109
Example
Sketch the graph of y = x3- 3x2 + 2.
Solution:
Step 1
Find the intercepts with the axes.
If x = 0 :
y=2
If y = 0: (x-1)(x2-2x-2) = 0
or (x-1)(x-(1+ 3 ))(1-(1- 3 )) = 0.
Hence the intercepts occur at
(0,2); (1,0); (1+ 3 , 0) and (1- 3 , 0).
Step 2
Find the critical points (turning points):
f ′(x) = 3x2 -6x
Setting f ′(x) = 3x2 -6x = 3x(x - 2) = 0
gives
x = 0 and x = 2.
The turning points are therefore
(0,f(0)) = (0,2) and (2, f(2)) = (2, -2).
Step 3
Find the relative maxima and relative minima.
Using the First Derivative Test we consider the three subintervals suggested by the
critical points: (- ∞ ,0); (0, 2) ; (2, ∞)
Choose an arbitrary point on each of the intervals and work out the sign of f ′(x) in these
intervals :
f ′(x) = 3x2 – 6x
x<0
(e.g choose x = -1)
f ′(-1) = 9 > 0 (+)
0<x<2
(e.g choose x = 1)
f ′(1) = -3 < 0 (-)
x>2
(e.g choose x = 3)
f ′(3) = 9 > 0 (+)
From the table above we can make the following conclusions:
1. On the interval (- ∞ , 0), f ′(x) > 0 which indicates that f is increasing on this
interval.
2. On the interval (0,2), f ′(x) < 0 which indicates that f is decreasing on this
interval.
3. On the interval (2, ∞), f ′(x) > 0 which indicates that f is increasing on this
interval.
4. The sign of f ′(x) changes from positive to negative at x = 0. Hence f(x) has a
relative maximum at the point (0,2).
5. The sign of f ′(x) changes from negative to positive at x = 2, and thus f(x) has a
relative minimum at the point (2,-2).
Step 4
Find the concavity of the curve.
f ″(x) = 6x – 6 = 6(x - 1).
Then
f ″(x) = 6(x-1) = 0 if x = 1.
Thus f has a point of inflection at (1,0). The point of inflection suggests 2 subintervals,
(- ∞, 1) and (1, ∞ )
110
Choose an arbitrary point on each of the intervals and work out the sign of f ″ (x) in these
intervals :
f ″(x) = 6(x - 1)
x < 1(e.g choose x = 0)
f ″(0) = -6 < 0
x > 1(e.g choose x =2)
f ″ (2) = 6 > 0
From the table above we can make the following conclusions:
1.
f ″(x) < 0 on the interval (- ∞, 1) , thus f is concave down.
2.
f ″(x) > 0 on the interval (1, ∞) , thus f is concave up.
By the Second Derivative Test we also note that f ″(0) = - 6 < 0, thus f has a relative
maximum at x = 0. Further f ″(2) = 6(2) - 6 = 6 > 0, thus f has a relative minimum at
x = 2.
Step 5
We can now sketch the graph of the function.
y
Relative
maximum
(0,2
)
0
Point of
inflection
(1,0
)
(2,-2)
3.
f(x) = 3x2-6x
x
Relative
minimum
Curve sketching of a quotient of two polynomial functions
We complete our discussion of curve sketching by describing what happens either as
x → ± ∞ , or when f(x) → ± ∞ as x approaches a finite value.
In geometry, an asymptote of a curve is a way of describing its behavior far away from
the origin by comparing it to another curve. Informally, it means that as a point P on the
graph of a function y = f(x) moves increasingly further from the origin, it may happen
that the distance between P and some fixed curve tends to zero. In other words, the curve
“approaches” the asymptote as it gets farther away from the origin
111
An important case is when the asymptote is a straight line, called a linear asymptote.
Definition 8.9
The line x = a is a vertical asymptote of the graph of the function f if and only if
lim f(x)   or  
x a
or
lim f(x)   or   .
x a 
Note: for a function which is a quotient of two polynomials, a vertical asymptote
will occur at a point at which the denominator turns to 0.
Definition 8.10
The line y = b is a horizontal asymptote of the graph of f if and only if
lim f(x)  b
x 
or
lim f(x)  b
x 
.
Examples:
1.
The x-axis (y = 0) is a horizontal asymptote for the graph of the function y = ex
and the y-axis (x = 0) is a vertical asymptote for the graph of y = lnx. (See the sketches
on page 84 of this guide).
1
2.
Draw the graph of y =
 3 and indicate clearly the asymptotes.
x2
Solution:
Step 1:
1
If x = 0, y =
 3   12  3  2 ½ .
2
1
1
If y = 0,
 3 = 0, which gives
 3 , for x ≠ 2.
x2
x2
 x - 2 = - 13

x = 1 13
Hence, the intercepts with the axes are (0, 2½) and (1⅓, 0).
Step 2:
Critical points:
dy d
 [(x  2) 1  3]
dx dx
= -(x-2)-2 + 0
-1
=
≠ 0 for any value of x  R .
(x  2) 2
Hence there are no critical points.
112
Step 3
-1
dy
< 0 for all x  R,

dx (x  2) 2
From
it follows that y =
1
 3 is a decreasing function for all x  R, x ≠ 2.
x2
Step 4
We consider the regions of concavity
d  1 
d2y



2
dx  (x  2) 2 
dx
2
=
.
(x  2) 3
1.
2.
3.
d2y
≠ 0 for all x  R, except x = 2, where it is not defined.
dx 2
Thus the only point of inflection is x = 2.
d2y
For x < 2,
< 0, which means y = f(x) is concave down on ( - ∞ , 2).
dx 2
d2y
For x > 2,
> 0 and y = f(x) is concave up on (2,∞).
dx 2
We observe that
Step 5
1
 3 is not defined for x = 2. This means that
x2
1
 1

 lim 3
lim 
 3  = lim
x 2 x  2
x 2
x 2  x  2

= ∞ +3
= ∞
1
1


and
 lim 3
lim 
 3  = lim
x 2 x  2
x 2
x 2  x  2

= -∞ +3
= -∞
Therefore, the line x = 2 is a vertical asymptote.
Note: for a function defined by a quotient of polynomials, a vertical asymptote will
occur at a point where the denominator is 0.
Step 6
1
 1

lim
 lim 3
lim 
 3 =
x  x  2
x 
x  x  2


We observe that y =
=
0 + lim 3
x 
=
Also
3.
1
 1

 lim 3
lim 
 3  = lim
x   x  2
x  
x  x  2


113
=
0 + lim 3
=
3.
x  
Hence the graph will settle down near the line y = 3 as x approaches ± ∞.
Therefore the line y = 3 is a horizontal asymptote.
Step 7
We can now sketch the graph.
y
y=
1
3
x2
y=3
(0, 2 ½)
x=2
0
1
3
( 1 ,0)
x
When a linear asymptote is not parallel to the x-axis or y-axis, it is called an oblique
asymptote (also called a slant asymptote).
If a rational function is a quotient of two polynomials that have no common factors, and
if the degree of the numerator is one larger than the degree of the denominator, the graph
will have an oblique asymptote.
Definition 8.11 (Oblique asymptote)
A function y = f(x) is asymptotic to the line y = mx +b if
lim[f(x)  (mx  b)]  0
x 
or
lim [f(x)  (mx  b)]  0 .
x  
Remark:
A line y = mx + b is never a vertical asymptote, but it can be a horizontal asymptote if
m = 0. In this case it is not an oblique asymptote.
114
Example
Sketch the graph of y =
x2  3
, indicating all the asymptotes.
2x  4
Step 1
If x = 0, y = 34 .
If y = 0, then x2 – 3 = 0, since 2x - 4 ≠ 0.
Therefore x =  3 = ± 1,73 (appr).
Hence the intercepts with the axes are ( 0, 34 ) , ( 3 , 0) and (- 3 , 0).
We observe that y is not defined at x = 2, and we expect the graph to have a vertical
asymptote at x = 2.
Step 2
Find the critical values:
dy d  x 2  3 



dx dx  2(x  2) 
2(x  2)(2x)  (x 2  3)(2)
=
4(x  2) 2
2x 2  8x  6
=
4(x  2) 2
(x  3)(x  1)
=
2(x  2) 2
dy (x  3)(x  1)
and
= 0 if (x - 3)(x - 1) = 0.

dx
2(x  2) 2
Hence critical points occur at the points (1, 1) and ( 3, 3).
Step 3
The critical points x = 1 and x = 3, as well as the point x = 2 where the function is not
defined, suggest the following intervals
(-∞, 1) ; ( 1,2); (2,3); (3, ∞)
x<1
1< x < 2
2<x<3
x>3
(e.g choose
(e.g. choose (e.g. choose (e.g. choose
x = 0)
x = 3/2)
x = 5/2)
x = 4)
dy
3
3
3
3
>0
<0
<0
>0
dx
16
4
4
16
We can draw the following conclusions:
1.
y = f(x) is increasing on the interval (- ∞, 1)
dy
2.
changes from (+) to (-) at x = 1, thus y = f(x) has a relative maximum at
dx
x = 1.
3.
y = f(x) is decreasing on the interval (1,2).
4.
y = f(x) is also decreasing on the interval (2,3).
115
5.
dy
changes from (-) to (+) at x = 3, thus y = f(x) has a relative minimum at
dx
x = 3.
Step 4
We investigate the regions of concavity.
d2y
d x 2  4x  3
[
]

dx 2 dx 2(x  2) 2
 2(x  2) 2 (2x  4)  (x 2  4x  3)4(x  2) 
= 

4(x  2) 4


3
2
 4(x  2)  4(x  2)(x  4x  3) 
= 

4(x  2) 4


1
=
(x  2) 3
1.
2.
3.
d2y
d2y
≠
0
for
all
values
of
x.
is not defined for x = 2, hence there is a
dx 2
dx 2
change in concavity at x = 2.
d2y
If x < 2,
< 0, hence y = f(x) is concave down on the interval ( - ∞,2).
dx 2
d2y
If x > 2,
> 0, hence y = f(x) is concave up on the interval (2, ∞).
dx 2
Step 5
Vertical asymptote:
x2  3
lim
x  2 2(x  2)
=
- ∞
x2  3
= ∞
lim
x  2  2(x  2)
Therefore x = 2 is a vertical asymptote for the graph of y = f(x).
Horizontal asymptote:
x(x  x3 )
x2  3
 -∞
lim
 lim
x    x(2  4 )
x   2x  4
x
x(x  x3 )
x2  3
 ∞.
 lim
x  x(2  4 )
x  2x  4
x
Therefore there is no horizontal asymptote for the graph of y = f(x).
and
lim
Since the degree of the numerator is one more than the degree of the denominator, there
will be an oblique asymptote.
Oblique asymptote:
Divide the numerator by the denominator using long division method:
116
x2  3
2x  4
We observe that
x2  3
2x  4
x
1
1
= (  1) 
= (½ x + 1) +
.
2
2x  4
2x  4
x
1
.
(  1) =
2
2x  4
 x 2  3  x 
1
= 0.
lim 
   1 = lim
x    2x  4
x



2
2x

4




2
 x  3  x 
1
lim 
   1 = lim
= 0.
x  2x  4
x  2x  4
 2 

Further
and
Thus the vertical distance between the curve y =
x2  3
x
and the line y =  1
2x  4
2
approaches zero as x → ± ∞
Therefore the line y =
1
2
x  1 is an oblique asymptote for the graph of y =
x2  3
.
2x  4
Step 6
We can now sketch the graph of the function.
y
x=2
y=
x2  3
2x  4
y=
1
2
(- 3 , 0)
( 0,
0
3
4
)
( 3 , 0)
1
2
x 1
x
2
EXERCISES
1.
Determine the
(i)
intercepts with axes
(ii)
critical points
(iii) intervals of increasing and decreasing
(iv)
relative maxima and minima
117
(v)
concavity
of the following functions. Then sketch the graphs.
(a)
f(x) = x3 – 3x
(b)
f(x) = -x4 + 6x3 - 9x2
(c)
f(x) = 3x5 – 5x3
(d)
(e)
2.
f(x) =
f(x) =
x

2
4 3
4x – 4x2
2
Obtain the information indicated in (1) of the following functions. Determine
asymptotes where appropriate. Then graph the functions:
x
(a)
f(x)
=
x2
x2
(b)
f(x)
=
x 1
2x 2  3
(c)
f(x)
=
x 2  4x  3
x2 1
(d)
f(x)
=
x
2
x 3
(e)
f(x)
=
x 1
118
CHAPTER 09
THE INDEFINITE INTEGRAL
We were led to the notion of a derivative by considering the problem of a tangent line to
a curve. By considering two new problems, we will be led to the notion of an integral.
The first problem of integral calculus is that of defining antiderivatives, i.e. the reversal
process of differentiation.
If G is a function such that
dG
= f(x)
dx
then G is said to be an antiderivative of f(x).
Thus, an antiderivative of f(x) is nothing more than a function G which, when
differentiated, gives f(x). The process of determining G from f(x) is called
antidifferentiation.
In calculus the symbols dG, dx, dy, etc. are referred to as differentials. We will not
discuss it except to mention that it is used to indicate a change in the variable.
dG
= f(x) can be considered as a quotient of differentials. Multiplying both
dx
members by dx yields
dG = f(x)dx.
Thus, on the differential level, a quite useful interpretation of an antiderivative of f(x) is
that it is a function whose differential is f(x)dx.
Since
Definition 9.1
An antiderivative of a function f(x) is a function G such that
dG
= f(x)
dx
or equivalently
dG = f(x)dx.
Example:
1.
2.
d x3
x3
2
( ) = x , then
If f(x) = x , then
is by definition an antiderivative of
3
dx 3
x2 .
1
d
1
1
If f(x) =
, then
(lnx)  , and ln x is an antiderivative of .
x
dx
x
x
2
119
The question now is: these antiderivatives , as obtained in the examples, are they unique?
Are they the only antiderivatives of these functions?
The answer is “no”!
Consider the following:
3
d x
(  4) = x2 .
dx 3
d x3
(  2) = x2
dx 3
d x3
(  4567,398) = x2
dx 3
d
1
(lnx  5) 
dx
x
d
1
(lnx  16 )  .
dx
x
3
3
3
x
x
x
Then
4,
 2 and
 4567,398 are also antiderivatives of x2 and
3
3
3
1
(lnx  5) as well as (lnx  16 ) are also antiderivatives of .
x
x3
Indeed, it can be shown that any antiderivative of x2 must be of the form (  c)
3
1
where c is a constant, and any antiderivative of
is of the form (ln x +k), where k
x
is a constant.
Thus any two antiderivatives of a function differ only by a constant.
The entire collection of antiderivatives of e.g. x2 is indicated by the symbol  x 2 dx
which is read “the indefinite integral of x2 “.
Hence
x3
2
=
 C,
x
dx

3
1
or
 x dx = ln x + C.
More generally the indefinite integral of an arbitrary function f(x) is the collection of
all antiderivatives of f(x). Furthermore, we write
 f(x)dx = G(x) + C
dG
= f(x).
dx
The symbol “ ∫ ” is called the integral sign, f(x) is called the integrand and C is called
the constant of integration. The process of finding the antiderivative is called
integration.
if and only if dG = f(x)dx or
120
Note that to integrate f, that is, to find ∫f(x)dx, it is sufficient to determine any
antiderivative g of f, and then by inserting the constant of integration all antiderivatives
are generated.
Using our repertoire of differentiation formulas that was established in the previous
chapters, we can compile a list of basic integration formulas:
1.
 kdx
=
kx + C
2.
n
 x dx
=
x n 1
+C
n 1
3.
e
dx
=
ex + C
4.
 x dx
=
ln x + C
5.
 kf(x)dx
=
k  f(x)dx + C
6.
 [f(x)  g(x)]dx =  f(x)dx ±  g(x)dx
x
1
In order to integrate trigonometric functions, we have to use the knowledge we have built
up about their derivatives.
d
7.
Since
sinx   cosx , then  cosxdx  sinx  C
dx
d
8.
Since
cosx   sin x , then  sinxdx   cos x  C
dx
d
9.
Since,
tanx  sec 2 x , then  sec 2 xdx  tanx  C
dx
d
10.
Since
secx   sec x tanx , then  secx tanx dx  secx  C
dx
d
cosecx   cosecx cotx , then  cosec x cotx dx dx  cosecx  C
11.
Since
dx
d
cotx  cosec 2 x , then  cosec 2 xdx  cotx  C
12.
Since
dx
Examples
Find the following indefinite integrals
3
(a)
 4 dx
3
By formula (1) with k =
follows that
4
121
3
 4 dx
=
3
x+C
4
(b)
 x dx
5
By formula (2) with n = 5. Hence
(c)
5
 x dx =
x 51
x6
C =
C
5 1
6
 8x dx
3
By formula (5) with k = 8 and f(x) = x3, hence  8x 3 dx = 8 x 3 dx
 x dx
3
But
by formula (2) with n = 3 is
 x dx
3
x 31
=
+ C1
3 1
x4
=
+ C1
4
where C1 is the constant of integration.
Therefore
3
=
8 x 3 dx
 8x dx
 x4

=
8
 C1 
 4

4
=
2x  8C1
=
2x 4  C , by setting 8C1 = C.
It is not necessary to write all the intermediate steps in the integration process. More
concisely we have
3
=
8 x 3 dx
 8x dx
=
(d)
4
  5e
x
dx
=
=
(e)

1
dy
=
4 x
e dx
5
4
 ex  C
5


2x 4  C
(by formula 5)
(by formula 3)
1
 y 2 dy
y
From formula (2) with n = -½ , follows
1

1
=
 y dy
 y 2 dy
1
 1
=
y 2
+C
 12  1
1
=
=
y2
1
2
+C
2 y C
122
(f)
 (x
3
 2x 2  4)dx
=
 x dx   2x
=
 x 21 
x 31
  C 2  4x  C 3
 C1  2
3 1
2

1


=
 x3 
x4
 C1  2   C 2  4x  C 3
4
 3 
3
2
dx   4dx
Letting C = C1 + C2 + C3, we can write
 x3 
x4
3
2
   4x  C

2
=
(x

2x

4)dx

4
 3 
x4 2 3
=
 x  4x  C
4 3
(g)
 (2
5
x  7x  10e  π)dx =
4
3
x
=
4
5
2 x dx  7  x 3 dx  10 e x dx   πdx
x
24
5
x
4
1
5
1
9
5
7
7
x 31
 10e x  πx  C
3 1
x4
 10e x  πx  C
4
=
2
=
10 5 7 4
x  x  10e x  πx  C
9
4
9
5
9
(h)

3
6x dx
=

=
3
3
=
1
3
6x 3 dx
1
3
6  x dx
6
x
4
3
4
3
C
4
33 6 3
x C
4
=
(i)

3x  x 
3
x 2
dx
6
By factoring out 1/6 and multiplying the binomials in the integrand we obtain
1


 1
 3x  x 2  x 3  2 
4
5
1




1  3


dx
6
=
3x  x  6x  2x 2 dx



6
6 

123
=
=
11
3
 7

1 x3 x 6
x2
x2 
 11  6
2 3 C
3
6  73
2
6
2 


7
11
3
1  9 3 6 6
4 2 
2
x  x  3x  x   C
6  7
11
3 
7
=
EXERCISES
Find the indefinite integrals.
1.
(x 3  4x 2 - 3)dx

2.
 3

  3  5x dx
x

3
 (3x 4  4x 3 )dx
4.
 (x
5.

6.
 
3
3 2

1
 x)(x
1
3
 2)dx
(4  x ) 2 dx
4
x3  3
1 
dx
x
2
7.
8.
9.
 1

  5 x  1 dx
u
43
 6e  u u  5 du



x (x 5  3 x  x 1 )
dx
3
124
11
3
3x 3 x 6 x 2 2x 2



C
14
11
2
9
CHAPTER 10
METHODS OF INTEGRATION
As we have seen in the previous chapter, the problem of finding an indefinite integral is
solved if we can find one antiderivative for the given function f(x). There are many other
functions for which antiderivatives can be found, but learning to recognize them takes a
lot of experience. For those functions which we cannot immediately recognize the
antiderivatives, we discuss some methods to find the antiderivatives.
1.
INTEGRATION BY U-SUBSTITUTION.
In an integration process, replacing a function of x by another variable may reduce the
integral to one of the previous forms and as a result simplify the work.
Examples:
20
(a)
 (x  1) dx
Let us set u = x+1; then
du
 1 or du = dx. If we substitute this into the integral
dx
we obtain
 (x  1)
20
u
=
dx
=
 2xe
x2
du
u 21
C
21
1
(x  1) 21  C .
21
=
(b)
20
dx
We remember that
d 2
du
(x )  2x , thus let us put u = x2, then
 2x and
dx
dx
du = 2xdx.
Hence
 2xe
x2
dx
=
=
=
=
(c)
x
e
ue
x2
u
(2xdx)
du
e +C
2
ex  C .
x 2  5 dx
d 2
du
(x  5)  2x , we can let u = x2 + 5, then
 2x or du = 2xdx, which
dx
dx
du
gives
 dx.
2x
Therefore,
Since
125
1
2
 x x  5 dx
=
2
 x(x  5) 2 dx
1
2
1
du
2x
=
 x.u
=
1 2
u du
2
=
1 u2
C
2 32
=
1 2
(x  5) 2  C .
3
1
3
3
(d)

3
6x dx
Let u = 6x, then
du
du
 6 or
 dx
dx
6
Thus,

3
6x dx
1
3
=
 (6x)
=
u3
=
1 3
u du
6
=
1 u3
C
6 43
=
1
(6x) 3  C .
8
1
dx
du
6
1
4
4
(e)
 (x
2
 1) e x
3 3x
dx
Let u = x3 + 3x, then
du
du
 dx
 3x 2  3 or
dx
3(x 2  1)
Then
 (x
2
 1) e x
3 3x
dx
=
=
=
=
126
 (x
2
 1) e u
du
3(x 2  1)
1 u
e du
3
1 u
e C
3
1 x 3 3x
e
 C.
3
(f)
2x 3  3x
 x 4  3x 2  7dx
Let u = x4 +3x2 +7, then
or
du
 4x 3  6x = 2(2x3+3x)
dx
du
 dx
2(2x 3  3x)
Thus,
2x 3  3x
 x 4  3x 2  7dx
2x 3  3x
du
=
 u 2(2x 3  3x)
1 du
=
2 u
1
=
ln u  C
2
1
=
ln x 4  3x 2  7  C
2
4
2
Since x +3x +7 > 0 for all x R, we can omit the vertical bars and write
1
2x 3  3x
ln(x 4  3x 2  7)  C .
 x 4  3x 2  7dx =
2
(g)
 sin(4x  3)dx
Let u = 4x+3, then
du
d
du

(4x  3) = 4 or
 dx
dx dx
4
Thus,
 sin(4x  3)dx
=
=
1
sin u du
4
 14 cosu  C
=
 14 cos(4x  3)  C .
=
(h)

du
 sinu. 4
sinx cosx dx
Let u = sin x, then
du
d
du

sin x = cos x, or
 dx
dx dx
cosx
Thus,

sinx cosx dx
=
=
127

u cos x
 (u)
1
2
du
du
cosx
=
u
3
2
3
2
C
3
=
2
(sinx) 2  C .
3
EXERCISES
Evaluate the following integrals using a suitable substitution.
1.

x(x 2  1) 99 dx
x2
2.

3.

x

(1  x ) 9
4.
5.

dx
2  2x 3
1
2
3 (x 3
 1) 2 dx
dx
x
x n 1
dx
a  bx n
6.

7.
 (3x
8.
 4s
9.

10.
 tan x sec
3
(7  x) 2 dx
2
 14x)(x 3  7x 2  1)dx
4s  1
ds
 2s  1
2
cosx
1  sinx
2
dx
x dx
2.
INTEGRATION BY PARTS
This integration technique is particularly useful for integrating the product of two
functions. The formula for integration by parts comes from the product rule in
differentiation.
Let r(x) = g(x)f(x), then r′(x) = g′(x)f(x) + g(x)f ′(x).
Thus,
128
 r (x)dx
=
=
Hence
=
 g(x)f(x)dx =
 [g(x)f(x)  g(x) f  (x)]dx
 [g(x)f(x)]dx   [g(x) f  (x)]dx
g(x)f(x) + C.
g(x)f(x) -  g(x) f  (x)dx .
 g(x)dx  g(x) we can write
f(x)  g (x)dx
 g(x)f(x)dx =
But since
 f  (x)  g(x)dx dx
-
Let g′(x) = h(x), then we can also state the formula for integration by parts as follows:
 f(x)h(x)dx
=
 f (x) h(x)dxdx
f(x)  h(x)dx -
(A)
Remarks:
1. The method of integration by parts is suitable for integrals of which the integrand
is a product of two unrelated functions, i.e. functions of which one is not the
derivative or a multiple of the derivative of the other function.
2. The first function in the product does not necessarily have to be taken as the
function f(x). Usually h(x) is chosen as the factor which can be integrated the
easiest.
Examples:
x
(a)
 xe dx
Let f(x) = x and h(x) = ex in (A).
Then
 xe
(b)
x
dx
=
x  e x dx -
=
xex -  e x dx
d
  dx x  e
=
xex – ex +C.
=
ex  sinx dx -
x

dx  dx

 e sin x dx
x
 e sin x dx
x
 e  sinx dx dx
x
=
ex(-cosx)-  e x (cosx)dx
=
-excosx +
e
x
cosx dx .
We observe that the second integral is again a candidate for integration by parts.
We therefore repeat the procedure on this integral with f(x) = ex and h(x) = cosx
x
-excosx + [ex  cosxdx -  e x  cosxdx dx
 e sin x dx =

=
-excosx + [exsinx -  e x sin x dx ]
=
-excosx + exsinx -  e x sin x dx .
129

We now observe that the integral on the right hand side is exactly the same as
the one on the left hand side. Hence we can add  e x sin x dx on both sides.
Thus,
2  e x sin x dx =
-excosx + exsinx
and thus
 e sin x dx
x
=
½ [-excosx + exsinx ] + C.
 lnx dx
(c)
The integrand (lnx) is not a product of two functions, but it is not the derivative of
any of our known functions either. We can, however, transform it to a product of
two functions, by taking the second function as 1. Hence
=
 lnx dx
 lnx .1.dx
Let f(x) = lnx and h(x) = 1
Then
 lnx dx
lnx  1dx -
=
lnx.x -
=
lnx.x -
=
lnx.x - x + C
=
x[lnx -1] + C.
3.
4.
5.
6.
7.
8.







x2lnxdx
ln(-x)dx
e 2xsin3xdx
x x  1dx
sin-1xdx
t2 ln t dt
sin(lnx)dx
130

  x 1dx  dx
1
 x xdx
 1dx
EXERCISES
Evaluate the following integrals
1.  xe- 3xdx
2.
1
=
3. INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL
FRACTIONS.
We now consider the integral of a quotient of two polynomials, that is, of a rational
function.
Example:
The following are rational functions:
3x 4  5x 3  2x  1
x 3  2x  1
1
,
,
,
(x  1)(x 2  4) 2
x2  4
x2
while
x2
sinx
1
,
,
,
3
log e x
x
x
are not rational functions.
Without loss of generality, we may assume the numerator N(x) and the denominator D(x)
have no common polynomial factor and that the degree of N(x) is less than the degree of
N(x)
D(x), that is,
is a proper rational function. If the numerator is of a higher degree
D(x)
than the denominator, we may, using long division, divide N(x) by D(x).
To integrate a rational function it is usually necessary to first rewrite it as a polynomial
( which may be identical to 0) plus fractions of the form
A
Bx  C
or
2
k
(x  bx  c) k
(x  a)
with the quadratic x2 + bx + c irreducible, that is, not factorizable into linear terms with
real coefficients. Such fractions are called partial fractions.
3.1
The denominator splits into linear factors.
Each distinct linear factor (x-a) in the denominator yields a term of the form
A
.
xa
Example:
4x 2  14x  6
 x(x  1)(x  3)dx
Observe that the denominator of the integrand consists only of distinct linear factors, each
factor occurring only once. If there are n such distinct linear factors, there will be n
partial fractions, e.g.
4x 2  14x  6
A
B
C
=
(1)


x x 1 x  3
x(x  1)(x  3)
To determine the constants A, B and C, we first combine the terms on the right hand side.
4x 2  14x  6
A(x  1)(x - 3)  Bx(x  3)  Cx(x  1)
=
(2)
x(x  1)(x  3)
x(x  1)(x  3)
1. Evaluate
131
Since the denominators of both sides are equal, we can equate their numerators:
4x2 – 14x – 6 =
A(x+1) (x-3) + Bx(x-3) + Cx(x+1)
(3)
Although equation (1) is not defined for x = 0, x = -1 and x = 3, we want to choose values
for A, B and C which will make equation (2) true for all values of x. By successively
setting x in equation (3) equal to any three distinct numbers, we can obtain a system of
equations which can be solved for A, B and C. In particular, the work can be simplified
by letting x be the roots of D(x) = x(x+1)(x-3) = 0, i.e. x = 0, -1, 3.
Using equation (3)
If x = 0:
-6
=
A(1)(-3) + B(0) + C(0) = -3A
A
=
2

If x = -1,
12
=
A(0) + B(-1)(-4) + C(0) = 4B
B
=
3

If x = 3,
-12
=
A(0) + B(0) + C(3)(4) = 12C
C
=
-1

Thus equation (1) becomes
4x 2  14x  6
2
3
-1
=


x x 1 x  3
x(x  1)(x  3)
Hence,
4x 2  14x  6
 x(x  1)(x  3)dx =
=
=
=
2
3
1
 xdx   x  1dx   x - 3dx
1
1
1
2 dx  3
dx  
dx
x
x 1
x -3
2lnx + 3lnx+1 - lnx-3 + C
x 2 (x  1) 3
ln
+ C.
x 3
x5  2
 x 2  1dx
Since the degree of the numerator (x5 +2) is greater than the denominator (x2-1) we first
divide (x5+2) by (x2-1), using long division, to obtain
x5  2
x2
 x3  x  2
2
x 1
x 1
5
x 2
x2
3
and
(1)
 x 2  1dx =  (x  x)dx   x 2  1 dx .
x2
x2
For the integral  2
dx
dx =

(x  1)(x  1)
x 1
we can proceed as in example (1).
x2
A
B
=
(2)

x 1 x 1
(x  1)(x  1)
2. Evaluate
132
A(x  1)  B(x  1)
.
(x  1)(x  1)
A(x+1) + B(x-1)
=
Then
x+2
If x = 1 in (3) we obtain 3
or
=
=
A =
(3)
2A
3
2
If x = -1 in (3)
1 = -2 B
or
B = -½
Substituting these values in (2) we obtain
3
-1
x2
2
=
 2
x 1 x 1
(x  1)(x  1)
The integral (3) can now be evaluated.
x5  2
x2
3
 x 2  1dx =  (x  x)dx   x 2  1 dx
dx
dx
=  (x 3  x)dx  32 
 12 
x 1
x 1
=
¼x4 + ½ x2 +
3
lnx-1 - ½ ln x+1 + C
2
3
2
=
¼x + ½ x + ln (x  1)  ln (x  1)
=
¼x4 + ½ x2 + ln
4
2
1
2
+C
(x  1) 3
+ C.
(x  1)
3.2
The denominator has a repeated linear factor.
N(x)
If the denominator of
contains a linear factor which is repeated, then for each
D(x)
factor (x-a)k , where k is the maximum number of times (x-a) occurs as a factor, there
will correspond the sum of k partial fractions:
A3
A1
A2
Ak


 ................ 
2
3
(x  a) (x  a)
(x  a)
(x  a) k
Example
6x 2  13x  6
 (x  1) 2 (x  2) dx .
Since the degree of N(x) is less than the degree of D(x), no long division is necessary. In
D(x) the factor (x+2) occurs once and the factor (x+1) twice. There will be three partial
fractions and three constants to determine.
6x 2  13x  6
A
B
C
=
(1)


2
x  2 x  1 (x  1) 2
(x  2)(x  1)
Evaluate
133
which gives
6x2 + 13x + 6 =
If x = -2 in (2)
A =
If x = -1 in (2)
C =
Choose x = 0 for the third value:
A + 2B + 2C = 6

4 + 2B - 2 = 6

B = 2
Substituting these into the integral:
6x 2  13x  6
 (x  1) 2 (x  2) dx
A(x+1)2 + B(x+1)(x+2) + C(x+2)
4
-1
=
=
=
4
2
(2)
1
 x  2 dx   x  1dx   (x  1)
2
dx
1
+C
x 1
1
ln[(x+2)4(x+1)2] +
+ C.
x 1
4lnx+2 + 2 lnx+1 +
3.3
The denominator has an irreducible quadratic factor
Suppose a quadratic factor (ax2 + bx +c) occurs in D(x) such that ax2 + bx + c cannot be
expressed as a product of two linear factors with real coefficients. Such a factor is called
irreducible over the real numbers. To each irreducible quadratic factor that occurs only
once in D(x), there will correspond a partial fraction of the form
Ax  B
ax 2  bx  c
Example
 2x  4
Find  3
dx .
x  x2  x
Since x3 + x2 + x = x(x2+x+1), we have the linear factor x and the quadratic factor
x2 + x + 1 which does not seem factorizable by inspection, that is, x2 + x + 1 is an
irreducible quadratic factor. Thus there will be two partial fractions and three constants
to determine
A
Bx  C
 2x  4
=
(1)
 2
2
x x  x 1
x(x  x  1)
-2x – 4
=
A(x2 + x + 1) + (Bx+C)x

-2x - 4
=
Ax2 + Ax + A + Bx2 + Cx
=
(A + B)x2 + (A + C)x + A
From (2) follows, equating coefficients of like powers of x
A+B
=
0
A
+C
=
-2
A
=
-4
134
(2)
Solving this gives us, A = - 4, B = 4 and C = 2.
Substituting this in the partial fractions of (1) we obtain
 2x  4
4
4x  2
=
 x 3  x 2  xdx
 x dx   x 2  x  1 dx
dx
2x  1
=
 4
 2 2
dx
x
x  x 1
=
-4lnx + 2lnx2+x+1 + C
=
ln(x-4)(x2+x+1)2 + C
 (x 2  x  1) 2 
=
ln 
 + C.
x4


3.4
The denominator has a repeated irreducible quadratic factor
If the denominator D(x) contains irreducible quadratic factors, some of which are
repeated, then for each (ax2+bx+c)k , where k is the maximum number of times
(ax2+bx+c) occurs, there will correspond a sum of k partial fractions
A1 x  B1
A 2 x  B2
A k x  Bk
.

 ................. 
2
2
2
ax  bx  c (ax  bx  c)
(ax 2  bx  c) k
Example
3x 4  x 3  20x 2  3x  31
dx

(x  1)(x 2  4) 2
B x  C1 B 2 x  C 2
A
3x 4  x 3  20x 2  3x  31
=
.
 12
 2
2
2
x 1
(x  1)(x  4)
x 4
(x  4) 2
This gives
3x4+x3+20x2+3x+31 = A(x2+4)2 + (B1x+C1)(x+1)(x2+4)+(B2x+C2)(x+1)
We can rewrite the equation above as
(A+B1)x4+(B1+C1)x3+(8A+4B1+C1+B2)x2+(4B1+4C1+B2+C2)x+(16A+4C1+C2)
= 3x4+x3+20x2+3x+31
This gives us the quations
A + B1
=
3
B1 + C1
=
1
8A + 4B1 + C1 + B2
=
20
4B1 +4C1 + B2 + C2
=
3
16A+
4C1
+ C2
=
31
Solving this system (with a lot of patience!) we obtain
A = 2; B = 1; C = 0; D = 0 E = -1
This gives the following decomposition
3x 4  x 3  20x 2  3x  31
2
x
1
dx = 
dx   2
dx   2
dx
2
2

x 1
(x  1)(x  4)
x 4
(x  4) 2
2x
1
dx -  2
= 2lnx+1 + ½  2
dx
x 4
(x  4) 2
135
= 2lnx+1 + ½lnx2+4 -
 (x
2
1
dx .
 4) 2
Remark:
1
dx , we need a trigonometric substitution
 4) 2
which will only be discussed in the next chapter.
To calculate an integral of the form
 (x
2
EXERCISES
Calculate the following integrals:
1.

x 3  x 2  12x  1
2.

dx
(x  1)(x  2)(x  3)
3.
4.


x 2  x  12
4y 2  7y  12
y 3  y 2  6y
x2
(x  3)(x  2) 2
x 2  2x  7
5.

x3  x2  2
6.

x(x  1) 2
1
2
dy
dx
dx
dx
136
CHAPTER 11
INTEGRATION OF THE TRIGONOMETRIC
FUNCTIONS.
11.1
(a)
INTEGRALS OF TANX, COTX, SECX AND COSECX.
sinx
=
 cosx dx
 tanxdx
du
 du
 sinx or
 dx.
Let u = cosx, then
dx
sinx
Thus,
sinx
=
 cosx dx
 tanxdx
sinx du
=
 u  sinx
du

=
u
=
-ln cosx +C.
Or
 tanxdx
ln1 - ln cos x +C
=
1
C
cosx
ln secx + C.
=
ln
=
(b)
 cotxdx
=
Let u = sinx, then
cosx
 sinx dx
du
du
 cosx or
 dx
dx
cosx
Then
 cotxdx
=
=
=
=
(c)
cosx
 sinx dx
cosx du
u cosx
du
u
ln sinx + C.

 sec x dx
The method used in the previous two examples, will not be appropriate here. In
this case we use one of the mathematical techniques we have used earlier :
multiply by a special form of the number “1”.
137
 sec x dx
secx  tanx
=
 secx. secx  tanxdx
=
sec 2 x  sec x tan x
 secx  tanx dx.
Let u = secx + tanx, then
du
du
 sec x tan x  sec 2 x , or
 dx
dx
secxtanx  sec 2 x
Hence,
 sec x dx
(d)
 cosecxdx
=
sec 2 x  sec x tan x
 secx  tanx dx
sec 2 x  secx tan x
du
=

u
secx tanx  sec 2 x
du
=
u
=
ln secx + tanx + C.
cosecx  cotx
 cosecx cosecx  cotx dx
=
cosec 2 x  cosecx cotx
 cosecx  cotx dx.
du
 cotxcosecx  (cosec 2 x) = cosec2x - cotx
Let u = cosecx-cotx, then
dx
cosecx.
du
Then
 dx
2
cosec x  cotx cosecx
Thus,
cosec 2 x  cosecx cotx
=
cosecxdx
 cosecx  cotx dx

cosec 2 x  cotxcosecx
du
=
2

u
cosec x  cotxcosecx
du
=
u
=
ln cosecx  cotx + C.
=
11.2
PRODUCTS AND POWERS OF TRIGONOMETRIC FUNCTIONS
(a)
Integrals of the form  sinm xcos n xdx
(i)
If the power of Sine is odd, m = 2n+1, save one sine factor and
use
sin2x = 1-cos2x to express the remaining factors in terms of cosine.
138
 sin
2n1
xdx
 sin xsinxdx
 (sin x) sinxdx
 (1  cos x) sinxdx.
2n
=
2
=
n
2
=
n
du
du
 sinx and
 dx
dx
 sinx
2
n
Then  sin 2n1 xdx
=
 (1  cos x) sinxdx
Let u = cosx, then
du
 (1  u ) sinx - sinx
-  (1  u ) du
2 n
=
2 n
=
The expression (1-u2)n can be expanded by the binomial theorem (See chapter 5)
and the result integrated as a polynomial in u.
Example:
1.
 sin
5
=
xdx
=
Let u = cosx, then
 sin x sin xdx
 (1  cos x) sinx dx
4
2
2
du
du
 sinx and
 dx
dx
 sinx
Thus,
 sin
5
xdx
=
 (1  cos
=
 (1  u ) sinx  sinx
  (1  2u  u )du
x) 2 sinx dx
du
2 2
2
=
4

u3 u5 
 u  2    C
3
5

2
1
 cosx  cos 3 x  cos 5 x  C.
3
5
=
=
(ii)
2
If the power of Cosine is odd, (n = 2k+1), save one cosine factor and use
cos2x = 1 –sin2x to express the remaining factors in terms of sine.
Examples:
1.
 cos
3
3xdx
=
=
Let u = sin3x, then
 (cos 3x)cos3xdx
 (1  sin 3x)cos3xdx
2
2
du
du
 3cos3x and
 dx
dx
3cos3x
Hence,
139
 cos
3
3xdx
=
 (1  sin
=
 (1  u
2
2
3x)cos3xdx
)cos3x
du
3cos3x
1
(1  u 2 )du

3
1
u3 
 u    C
=
3
3 
1
1
sin3x  sin 3 3x  C.
=
3
9
2
4
 sin x cos x cosx dx
=
2.
 sin
2
xcos 5 xdx
=
=
=
=
=
 sin x(1  sin x) cosxdx
 sin x(1  2sin x  sin x)cosxdx
 (sin xcosx  2sin xcosx  sin xcosx )dx
2
2
2
2
2
1
3
2
4
4
6
sin 3 x  52 sin 5 x  17 sin 7 x  C
Summary:
How to integrate odd positive powers of sines and cosines
To evaluate
Write
2n1
(sin2x)nsinxdx
sin
xdx

=(1-cos2x)nsinxdx
2n1
(cos2x)n cosxdx
cos
xdx

=(1-sin2x)ncosxdx
(b)
Substitution
u = cosx
du = -sinxdx
u = sinx
du = cosxdx
Even powers of Sine and Cosine
We are familiar with the following identities:
(i)
sin2θ + cos2θ = 1
(iv)
sinxcosx = ½ sin2x
1

cos
2

1  cos 2
(ii)
sin2θ =
(v)
cos2θ =
2
2
1

cos
2
m

1

cos
2n
(iii)
sin2mθ =
(vi)
cos2nθ =
2
2
Examples:
1.
 cos
2
=
1  cos4x
dx
2
1


2  (1  cos4x)dx
=
½ x + ½ (¼ sin4x) +C
2xdx =

140
=
2.
 sin
4
3xdx =
½ x + ⅛sin4x + C .
 (sin
2
3x) 2 dx
=
 1  cos6x 
  2  dx
=
¼  1  2cos6x  cos 2 6x dx
=
¼  1  2cos6x  12 (1  cos12x)dx
=
¼  (1  2cos6x  12  12 cos12x)dx
=
3
¼  (  2cos6x  12 cos12x)dx
2
2
=
=
3.
 sin
4
xcos 2 xdx

1 3
4 2
3
8

1 sin12x +C
x  62 sin6x  2.12
x  121 sin6x  961 sin12x + C.
=
 (sinxcosx)
=
(
1
2
sin2x) 2 ( 12  12 cos2x)dx
=
[
1
8
sin 2 2x  18sin 2 2xcos2x]dx
=
1
8
=
1
16
x  161 14 sin4x  18 . 13 . 12 sin 3 2x  C
=
1
16
1
1
x  64
sin4x  48
sin 3 2x  C
[
1
2
sin 2 xdx
 12 cos4x]dx  18  sin 2 2xcos2xdx
Integrals of the form  sinmxcosnxdx;
(c)
2
 sinmxsinnxdx;  cosmxcosnxdx.
We use the following identities:
1. sin(mx)cos(nx)
=
½ sin[(m+n)x] + ½ sin[(m-n)x]
2. sin(mx)sin(nx)
=
½ cos[(m-n)x] - ½ cos[(m+n)x]
3. cos(mx)cos(nx)
=
½ cos[(m+n)x] + ½ cos[(m-n)x]
=

=
½  [sin9x  sin(x)]dx
Examples
1.
 sin4xcos5x dx
1
2
[sin(4x  5x)  sin(4x  5x)]dx
141
2.
(d)
 cos3xcos4x
=
½  [sin9x  sinx]dx
=
½ [- 19 cos9x  cosx] + C
=
 181 cos9x  12 cosx  C.
=

=
1
2
=
1
2
=
1
14
1
2
[cos(3x  4x)  cos(3x  4x)]dx
[cos7x  cos(x)]dx
[cos7x  cosx]dx
sin7x  12 sinx  C .
Integrals of the form  tan n xdx and  cot n xdx
To integrate tannx, set
tannx = tann-2xtan2x = (tann-2x)(sec2x-1) = tann-2xsec2x-tann-2x
To integrate cotnx, set
cotnx = cotn-2xcot2x = (cotn-2x)(cosec2x-1) = cotn-2xcosec2x – cotn-2x
Example:
 tan xdx
5
Let tanx = u, then
=
 tan xtan
=
 tan x(sec
=
 tan xsec
=
 tan xsec
=
 tan xsec
3
2
3
3
3
3
2
2
2
xdx
2
x 1)dx
xdx   tan 3 xdx
xdx   tanxtan 2 xdx
x   tanx(sec 2 x  1)dx .
du
du
.
 sec 2 x and dx 
dx
sec 2 x
Substitute this into the first two integrals and using a result in the first section of
this chapter , we obtain
 tan xdx
5
=
 u du   udu  ln secx
=
¼u4 - ½ u2 + lnsecx + C
=
¼tan4x – ½ tan2 x + lnsecx + C.
3
142
+C
Integrals of the form  secn xdx and  cosecn xdx
(e)
If the power is an even integer, we write
secnx = secn-2xsec2x = (tan2x+1)1/2(n-2)sec2x
cosecnx = cosecn-2xcosec2x = (cot2x+1)1/2(n-2)cosec2x
Example
 cosec xdx
4
=
 cosec xcosec
=
 (cot
=
 cot
=
-⅓cot3x – cotx + C.
2
2
2
2
xdx
x  1)cosec 2 xdx
xcosec 2 xdx   cosec 2 xdx
If the power is odd, we use integration by parts.
Example
Find
 sec xdx .
Let
f(x) = secx and h(x) = sec2xdx.
Then
f (x)  secxtanx and
 h(x)dx
Thus
 sec xdx
=
secxtanx -  tan 2 xsecxdx
=
secxtanx -
 (sec x - 1)secxdx
=
secxtanx -
 (sec x - secx)dx
3
3
=
Hence 2  sec 3 xdx
or
 sec xdx
3
= tanx.
2
3
 sec xdx  
3
secxtanx -
secxdx .
=
secxtanx + lnsecx + tanx
=
½ [secxtanx + lnsecx + tanx ] + C.
(f) Integrals of the form  tan m xsecn xdx
(i)
If the power of tangent is odd (m = 2k+1), save a factor of secxtanx and
use tan2x = sec2x -1 to express the remaining factor in terms of secx.
Example:
143
 tanxsec
5
xdx =
Let u = secx, then
 tanxsec
(ii)
5
xdx =
 tanx secx sec xdx
4
du
du
= secxtanx, hence dx 
dx
secxtanx
 tanx secx sec xdx
4
=
 tanx secx u
=
u
=
1
5
u5  C
=
1
5
sec 5 x  C.
4
4
du
secx tanx
du
If the power of secant is even (n = 2k), save a factor of sec2x and use
Sec2x = 1 + tan2x to express the remaining factors in terms of tanx.
Example
 tan
6
xsec 4 xdx
Let u = tanx, then
=
 tan xsec
=
 tan x(1  tan
=
 (tan x  tan x)sec
=
 (tan xsec
6
2
xsec 2 xdx
6
6
6
2
x)sec 2 xdx
8
2
2
xdx
x  tan 8 xsec 2 x)dx .
du
du
 sec 2 x , which gives dx =
dx
sec 2 x
Then
 tan
6
xsec 4 xdx
(iii)
=
 (tan xsec
=
 (u
=
1
7
u 7  19 u 9  C
=
1
7
tan 7 x  19 tan 9 x  C.
6
6
2
x  tan 8 xsec 2 x)dx
 u 8 )du
If the power of secant is odd and the power of tangent is even, we
express the integrand totally in terms of secx.
144
Example
 tan
2
xsecxdx
=
 (sec x - 1)secxdx
=
 (sec x - secx)dx
=
 sec xdx - 
2
3
3
secxdx
=
½ [secxtanx + lnsecx + tanx ] - lnsecx + tanx +C
=
½ secxtanx
- ½ lnsecx + tanx + C .
EXERCISES
Evaluate the following integrals using an appropriate substitution
1.

2.
 cot
3.

4.
 sec
5.

6.
 sin
cos4xsin3xdx
5
3xdx
sin3xcos3xdx
7
5xdx
cos5xsin2xdx
3
x cosxdx
7.

cos2xcos3xdx
8.

sin 72 x sin 12 xdx
9.

tan5xsec4xdx
10.
 sec xdx
11.

tan5xsec3xdx
12.

cot2xcosec2xdx
6
11.3 TRIGONOMETRIC SUBSTITUTION
Integrals involving
a 2  x 2 and
x 2  a 2 can usually be calculated by using
trigonometric substitutions:
(i)
For
a2  x2
we set x = asinu
(ii)
For
a2  x2
we set x = atanu
(iii)
For
x2  a2
we set x = asecu
where a > 0 in each case.
Example:
145
1. Find
dx
 (5  x
2 3/2
)
Solution:
This is of the form (i) with a =
5.
dx
 (5  x
Then
2 3/2
)
Set x =
5 sinu , then dx =
5cosu
du
2
u) 3/2
=
 (5  5sin
=
 [5(1  sin
=
5
=
1
cosu
du

5 (cos 2 u) 3/2
=
1
sec 2 udu

5
=
1
tanu  C
5
5cosu
du
2
u)]3/2
5cosu
5 (1  sin 2 u) 3/2
Using the diagram
x
5
u
5  x2
we obtain
dx
 (5  x
2. Find
2 3/2
)
=
1
tanu  C
5

=
x
5 5  x2
4  x 2 dx
Solution:
146
C
du
5 cosudu
This is of the form (ii) with a = 2. Set x = 2tanu, then

Then
4  x 2 dx
=

dx
 2sec 2 u or dx = 2sec2xdu
du
4  4tan 2 u 2sec 2 u du
=

=
4  (1  tan 2 u) sec 2 udu
=
4
=
4(1  tan 2 u) 2sec 2 udu
 secu.sec u du
2
4  sec 3 u du
From the example in the previous section we have
 sec u du
=
½ [secutanu + lnsecu + tanu ]

=
4[½ [secutanu + lnsecu + tanu] + C
=
2[secutanu + lnsecu + tanu ] + C.
3
Hence
4  x 2 dx
Considering the following triangle we can substitute values for secu and tanu.
x
4  x2
u
2
Hence

4  x 2 dx
= 2[secutanu + lnsecu + tanu ] + C
=
 4  x2  x 
4  x2 x 
 C
2
   ln
2
2
2
 2

4  x2  x
+C
2
=
½ x 4  x 2 + 2 ln
=
½ x 4  x 2 + 2 ln( x+ 4  x 2 ) – 2ln2 +C
We can absorb the constant (– 2ln2) into the constant C, to obtain
147

(3)
Find

4  x 2 dx
x
x 2  2x  3
= ½ x 4  x 2 + 2 ln( x+ 4  x 2 ) + C
dx .
Solution
First complete the square under the square root

x
x 2  2x  3
dx
=

=

x
(x 2  2x  1) - 1  3
x
(x 2  1) 2 - 4
dx
dx
This square root is of the form (iii) with a = 2.
Set (x+1) = 2secu, then x = 2secu-1 , dx = 2secu tanu du and secu 
x 1
2
Then

x
(x  1) - 4
2
2
dx
=

(2secu  1)
=

(2secu  1)
=

=
 (2sec u  secu)du
=
2tanu - lnsecu + tanu + C
4sec 2 u  4
2secu tanu du
4(sec 2 u  1)
2secu tanu du
2secu tanu(2secu - 1)
du
2tanu
2
Using the right angled triangle
x+1
(x  1) 2  4
2
follows

x
(x 2  1) 2 - 4
dx
=
u
2tanu - lnsecu + tanu + C
148
=
(x  1) 2  4
(x  1) 2  4
x 1
2
 ln

C
2
2
2
x  1  (x  1) 2  4
+C
( x  1)  4  ln
2
2
=
=
(x  1) 2  4  ln x  1  (x  1) 2  4  ln2  C
(x  1) 2  4  ln x  1  (x  1) 2  4  C .
=
11.4
INTEGRATION OF INVERSE TRIGONOMETRIC FUNCTIONS
Since
d
[sin 1 x]
dx
1
=
1 x
2
1

, it follows easily that
1 x
2
dx = sin-1 + C
The following follow in the same way:
d
[cos 1 x]
dx
=
d
[tan 1 x]
dx
=
d
[cot 1 x]
dx
=
d
[sec 1 x]
dx
=
-
-
1
1 x
2
dx
 4  9x
2
1 x
1 x
1
1 x2
implies
1 x
x x 1
1
x x 1
2
implies
implies
Examples
(1) Find
-1
implies
2
-

1
1 x2
1
d
[cosec 1 x] =
dx
implies
.
149
2
1
2
dx = tan-1x + C
2
dx = cot-1x + C
-1
x
x
dx = cos-1x + C
1
x 1
2
dx = sec-1x+ C
-1
x 1
2
dx = cosec-1x+ C
dx
 4  9x
2
1
=

=
1
1
dx

4 1  3x 2
=
1 2
.
4 3
 9x 2 
41 

4 

dx
2 
3
=
1
6
(2) Find


dx
 5  6x  x 2
dx
 5  6x  x 2
dx

2
 3x 
1  
2
tan 1
3x
2
2
dx
C.
dx .
=

dx
 5  (6x  x 2 )
Complete the square on (6x+x2).

dx
 5  6x  x 2
Let u =
dx
=

=

=

=

=
1
2
dx
 5  [(9  6x  x 2 ) - 9]
dx
 5  (9  6x  x 2 )  9
dx
4  (x  3) 2
dx
 (x  3) 2 

41 

4


dx
 x  3
1 

 2 
x3
du
, then
 ½ or 2du = dx.
2
dx
150
.
2
Thus,

dx
 5  6x  x
2
dx
1

=
½
=

=
sin-1u + C
=
sin-1
2du
1 u2
1
1 u2
du
x3
+ C.
2
EXERCISES
Verify the following:
1
1.

2.
 13  4x  x
3.

4.
x
5.

3  2x  x
2
dx
2
1
1  2x - 3x
1
x2  4
2
dx
dx
1
21  18 x  2 x 2
sin 1

dx
dx
x 1
C
2
=
1
 x  2
tan 1 
 +C
3
 3 

1
=
1
x
sec 1
2
2
3

sin 1
1
2
3x  1
C
2
sin 1
151
 C
2x  9
123
C
CHAPTER 12
THE DEFINITE INTEGRAL
12.1 AN AREA PROBLEM
We are already acquainted with the computation of the areas of rectangles,
parallelograms, triangles and in general of figures bounded by straight lines.
However, when a figure is bounded by one or more curves, it is not immediately obvious
what we mean by its area and how we would determine such an area.
Consider the figure below.
f(x) =
2x
2
1
0
x
1
2
It shows a right angled triangle formed by the lines y = 2x, y = 0 and x = 1.
Area of the triangle
=
½ x base x height
=
½ (1)(2)
=
1 square unit.
We shall now determine the area of this region by another method which, as you will see,
is applicable to regions of a more complex nature. This method involves summation of
areas of rectangles.
Let us divide the interval [0,1] on the x-axis into four sub-intervals of equal length ∆x.
This
is achieved by means of the equally spaced points x0 = 0, x1 = ¼, x2 = 2/4 = ½ , x3 = ¾
and
x4 = 4/4 = 1. Each sub-interval has length ∆x = ¼ . These sub-intervals determine four
subregions R1, R2, R3 and R4. With each subregion we can associate a circumscribed
rectangle, i.e. a rectangle whose base is the corresponding sub-interval and whose height
is the maximum value of f(x) on that interval.
f(1)
Area of R1 = ¼ f(¼)
Area of R2 = ¼ f(½)
Area of R3 = ¼ f(¾)
Area of R4 = ¼ f(1)
y
f(¾
) ()
f(½)
R4
R3
f(¼
f (0)
R
R)1
2
x0
x¼1
x14 x
x½2
x3
¾
0
Since the area of each rectangle is an approximation to the area of its corresponding
subregion, the sum of the areas of these rectangles, denoted by S 4 , is an approximation to
the area A of the triangle.
S4
=
¼ f(¼) + ¼ f(½) + ¼ f(¾) + ¼ f(1)
152
=
=
¼ [2(¼) + 2(½) + 2(¾) + 2(1)]
5/4 .
4
You can verify that we write S 4 as S 4 =
 f(x )Δ x , and we can call this an upper sum.
i
i 1
The fact that S 4 is greater than the actual area of the triangle might have been expected
since S 4 includes the area of regions which are not in the triangle.
On the other hand, with each subregion we can also associate an inscribed rectangle, i.e. a
rectangle whose base is the corresponding subinterval and whose height is the minimum
value of f(x) on that interval. Thus the areas of the four inscribed rectangles associated
with R1, R2, R3 and R4 are ¼ f(0), ¼ f(¼), ¼ f(½) and ¼ f( ¾) respectively.
Their sum, denoted by S 4 , is also an approximation to the area A of the triangle.
=
¼ f(0) + ¼ f(¼) + ¼ f (½) + ¼ f(¾)
S4
=
¼ [2(0) + 2(¼) + 2(½) + 2(¾)]
=
¾
y
f(¾
)
f(½
0
Using the sigma notation, )f(we¼)can write
R4
R2
R3
4
S 4 as S 4 =
 f(x
i 1
i 1
)Δ x , which can be called a
lower sum.
x
Clearly S 4 is less than the area of the triangle
because
½ the rectangles
1do not account for
¼
¾
that portion of the triangle which is not included by the rectangles.
Note that
S 4 = ¾ ≤ A ≤ S 4 = 5/4 .
We say S 4 is an approximation to the area A from below and S 4 is an approximation to
A from above.
Of course, if [0,1] is divided into more subintervals, better approximations to A will be
obtained. For example, suppose we consider the case of six subintervals of equal length
∆x = 1/6. Then S 6 , the total areas of six circumscribed rectangles and S 6 , the total area
of six inscribed rectangles are respectively given by
1 1 1 2 1 3 1 4 1 5 1 6
S6
=
f( )  f( )  f( )  f( )  f( )  f( )
6 6 6 6 6 6 6 6 6 6 6 6
1 1
1
1
2
5

=
2( )  2( )  2( )  2( )  2( )  2(1)

6 6
3
2
3
6

=
7/6
1
1 1 1 2 1 3 1 4 1 5
S6
and
=
f(0)  f( )  f( )  f( )  f( )  f( )
6
6 6 6 6 6 6 6 6 6 6
1
1
1
1
2
5 
=
2(0)  2( )  2( )  2( )  2( )  2( )

6
6
3
2
3
6 
=
5/6.
153
Note that
S 6 = 5/6 ≤ A ≤ S 6 = 7/6
and, with appropriate labelling, both S 6 and S 6 will be of the form
2
2
y
 f(x) Δ x .
y
S6
S6
More generally, if we divide the interval [0,1] into n subintervals of equal length ∆x, then
∆x = n1 and the endpoints of the subintervals are x = 0, n1 , n2 , n3 ,........., nn1 , nn  1 . The total
0
0
area of the n circumscribed rectangles determined by these n subintervals is
x
1 4 1 ) 5 1 f( 2 )  ..........  1 f( n )
S n 1 =2
3 n f(
3
5
1
2
4
n 6 n n1 x
n n
1
6
6
6
6
6
6
6
6
6
1
1
2
n
=
n 2 n  2 n  ............  2 n 
   
=

2 [1 + 2 + ......+ n]
n2
2 n(n  1)
2
n2
n 1
=
.
n
For n inscribed rectangles, the total area determined by the subintervals is
1
1 1
1 n-1
=
Sn
n f(0)  n f( n )  ..........  n f( n )
=
=
=

 

1
1
n-1
n 20  2 n  ............  2 n
2 [1 + 2 + .....+ (n-1)]
n2
2 n(n  1)
2
n2
n 1
=
.
n
From the nature of S n and S n , it seems reasonable ( and it is indeed true!) that
=
Sn ≤
A
≤ Sn .
As n becomes larger, S n and S n become better approximations to the area A from above
and below respectively. In fact, let us take the limit of S n and S n as n approached
infinity () through positive integer values
n 1
 1
lim S n = lim
 lim 1    1 .
n 
n  n
n 
 n
n 1
 1
lim S n = lim
 lim 1    1 .
n 
n  n
n 
 n
Since S n and S n have the same common limit, as n approached infinity, i.e.
154
=
lim S n
n 
lim S n
n 
= 1
and
Sn ≤ A ≤ Sn ,
we shall take this limit to be the area of the triangle. Thus A = 1 square unit, which
agrees with our prior findings.
Mathematically, the upper sum S n and lower sum S n , as well as their common limit,
have a meaning which is independent of area. For the function f(x) = 2x over the interval
[0,1], we shall define the common limit of S n and S n , namely 1, to be the definite
integral of
f(x) = 2x from x = 0 to x =1. We shall abbreviate this symbolically by writing
1
 2xdx  1 .
0
The numbers 0 and 1 appearing with the integral sign

are called the limits of
integration,
with 0 as the lower limit and 1 as the upper limit. The expression 2x is called the
integrand.
In general, the definite integral of a function f(x) over the interval from x = a to x = b,
a ≤ b, is the common limit of S n and S n , if it exists, and is written
b
 f(x)dx .
a
12.2 THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS
Until now, the limiting processes of both the derivative and definite integral have been
considered as distinct concepts. We shall now bring these fundamental ideas together
and establish the relationship that exists between them. As a result, definite integrals may
be evaluated more efficiently without the tedious procedures of the last section.
Assume f to be a function which is continuous on the interval [a,b] and, for the sake of
simplicity, also assume that f(x) ≥ 0 on [a,b]. Let A(x) be the area function for the
region bounded by the graph of f and the x-axis between a and b where a ≤ x ≤ b.
Since the area of this region can also be obtained by the limit of a sum, we can also refer
x
to it by
 f(x)dx .
a
x
Hence
A(x)
=
 f(x)dx .
a
We know that A′(x) = f(x).
On the other hand, if F(x) is any anti-derivative of f(x), then F′(x) = f(x).
Since both A(x) and F(x) are anti-derivatives of the same function, we can conclude that
they must differ by a constant C:
A(x) = F(x) + C.
Since A(a) = 0, then
0
= F(a) + C
155
or
Thus,
C
A(x)
= - F(a)
=
F(x) - F(a).
x
 f(x)dx
Hence
= F(x) - F(a).
a
b
 f(x)dx
If x = b,
=
F(b) - F(a).
a
Thus a relationship between a definite integral and antidifferentiation has become clear.
b
To find  f(x)dx it is sufficient to find an antiderivative of f(x), say G(x) and subtract its
a
value at the lower limit a from its value at the upper limit b. Our result can be stated
more generally as follows:
Theorem 12.1 (Fundamental Theorem of Integral Calculus)
Let f be a continuous function on the closed interval [a,b]. If F is an antiderivative of f on
[a,b], then
b
 f(x)dx
=
F(b) - F(a).
a
Some Properties of the Definite Integral
b
1.
 f(x)dx
a
b

b
 f(t)dt 
 f(z)dz
a
a
2. If f is an integrable function on [a,b] and if a < c < b, then f is integrable on [a,c]
and [c,b] and
b
c
b
a
a
c
 f(x)dx =  f(x)dx +  f(x)dx
3. If f is an integrable function on [a,b] and k is any constant, then kf is also an
integrable function on [a,b] and
b
 kf(x)dx
b
= k  f(x)dx
a
b
4.
If a < b, then  f(x)dx = a
a
a
 f(x)dx
b
a
5. If a = b, then  f(x)dx = 0
a
156
6. If f and g are both integrable functions on [a,b], then f ± g is also an integrable
function on [a,b] and
b
b
b
a
a
a
 (f  g)(x)dx =  f(x)dx ±  g(x)dx
EXAMPLES
x
2
 (2x  6x
(a)
4
 5)dx =
2

2
 65 x 5  5x 1
1
[4 + 65 (32) - 10] - [1 + 65 - 5]
32 52 - (- 2 54 )
35 15 .
=
=
=

(b)
π
=
x2
 2
  3sinx 
2
0
=
12
 3(1)  (0  3.0)
2 4
2
 (x  3cosx)dx
0
2
=
2
(c)

1
8
 3.
2
x3
1 x4
dx

1
3
4
 x (1  x ) 2 dx
=
1
To find the antiderivative of the integrand x3(1+x4)-½ , we can first calculate the
indefinite integral

Let u = 1 + x4, then
du
du
, which gives
 4x 3 and dx =
dx
4x 3
x
2
Hence

1
x3
1 x4
3

1
2

1
2
1
2
(1  x ) dx   u du  u  (1  x
4
1
4
1
2
1
2
2
=
1
1

4 2
 2 (1  x ) 

 1
1
1
2
2 17  2
=
1
2
dx =
 13

2
6
1 2x  x(x  1) dx


( 17  2 ) .
2
2
(d)
1
3
4
 x (1  x ) 2 dx using substitution.
=
1
2
2 x 3 dx  12  2x(x 2  1) 6 dx
1
1
2
=
 4
2 34 x 3  +

1
157
 (x
1 1
2 7
2
 1) 7

2
1
1
4) 2
.
=
=
=
1
(e)
 2e
 43  1 7
7
2 1 + 14 5  2


5 7  128
3 3
3
(
16
)


2
2
14
7
3
3( 16 ) 5  149
.

2
14

3
2

1
3t
=
dt
1
2  e 3t dt
1
1
=
=
=
2
3e 3t dt

3 1
2 3t 1
e 1
3
2 3
(e  e  3 ) .
3
 
EXERCISES
Evaluate the following definite integrals
1
3
1.
 (3x  5x
3
6.
)dx
1
2
4
5
2.
x dx
7.
1
 34

2 

x

2
x
0 
dx


8.
1
3
4. 
1
t 1
t 2  2t  3
3 tan 1x
dx

2
0
1 x
b
1
3.
dt
 4  (t  1) 2
(
b  x )dx
0
(x  1)(x  2) 3
2 x 2  x  2 dx
3
9.
dt
e2 ln( 2 x )
5. 
dx
x
2
10.
1
2
 x 2x  1 dx
1
158
CHAPTER 13
APPLICATIONS OF THE DEFINITE INTEGRAL
13.1 AREA BETWEEN CURVES
In this section we consider the problem of finding the area of a region enclosed by several
curves.
1. If the function f is continuous such that f(x) ≥ 0 for all x in the interval [a,b], then
b
the definite integral
 f(x)dx gives the area enclosed by
a
(i)
the curve of f
(ii)
the x-axis (the line y = 0)
(iii)
the line x = a
(iv)
the line x = b
2. If f and g are continuous functions on [a,b] and f(x) ≥ g(x) for all x in [a,b], then
the area A of the region bounded by their graphs on [a,b] is given by
b
 [f(x)  g(x)]dx
A=
a
Our procedure will be the following:
(a) Draw a sketch of the area. We must know which is the upper curve and which is
the lower curve. The lower curve must be subtracted from the upper curve.
(b) Find the limits of integration by equating the two equations.
Example
(a) Find the area of the region bounded above by the curve y = 4 – x2 and below by
the
x-axis.
Solution:
To sketch the area we must know the intersections with the axes.
y
If x = 0, y = 4
The x-axis is the line y = 0.
Hence , if y = 0
4 – x2 = 0
4
(2-x)(2+x)
= 0
0
Which gives x = ± 2.
-2
 (4  x

2
Area
=
2
)  0 dx
2
2
=
 (4  x
2
) dx
2
2
=
1 3

4x  3 x 

 2
159
2
x
=
8 
8

8      8  
3 
3

=
16 -
16
32 2
u
3
=
3
Note:
Since we calculate an area, the answer should be in square units. The units of
measurement are not mentioned in the problem, hence we use the general u2.
(b) Find the area of the region bounded by the curves y = x2 – x – 2 and y = 0, from x
= -2 to x = 2.
Solution:
We first find the intersections with the axes:
If x = 0 then y = - 2.
If y = 0, then x2 – x – 2 = (x-2)(x+1) = 0 which gives
x = 2 or x = -1.
y
A
2
-1
0
x
2
B
-2
We observe that the area to determine consists of 2 parts: the part A enclosed by the
lines
x = -2, y = 0 and the curve y = x2 – x -2; and the part B between y = 0 and y = x2 – x -2.
Area = area A + area B
1
=
 (x
2

2
1
=
 (x

 x  2) dx
1
2
 x  2)dx 
 (x
2
 x  2)dx
1
1
=
2
2
2
=
 0  (x
2
 x  2)  0 dx 
2
1 3 1 2

1 3 1 2

 3 x  2 x  2x    3 x  2 x  2x 

 2 
 1
 1 1
  8 4
  8 4
  1 1

  3  2  2     3  2  4    3  2  4     3  2  2 
 
 
 


160
11  9 
  
6  2
19 2
=
u
3
(c) Find the area bounded by the graphs y = x2 + 2x and y = -x + 4 on the closed
interval [ -4, 2]
=
Solution:
Points of intersection:
x2 + 2x = -x + 4

x2 +3x - 4 = 0

(x+4)(x-1) = 0.
This gives x = - 4 or x = 1.
Points of intersection are therefore (-4, 8) and (1,3)
Intersections with the axes
Intersection with x – axis : x2 +2x = 0
 x(x+2) = 0
 x = 0 or x = -2.
-x + 4 = 0
 x=4
y
4
A
x
0
-2
-4
B
2
1
The area to be determined consists of the area A plus the area B.
 (x  4)  (x
1
Area
=
4

2


 2x) dx   (x 2  2x)  ( x  4) dx
1
  x
1
=
2
4
2

2


 3x  4 dx   x 2  3x  4 dx
1
1
2
 1 3 3 2

1 3 3 2

=
 3 x  2 x  4x    3 x  2 x  4x 

 4 
1
 1 3
  64
  8
 1 3

=     4     24  16     6  8      4 
  3
  3
 3 2

 3 2
161
 65 3
 7 3

 3  2  44 +  3  2  2

 

58
71 2
+ 43 =
u

3
3
=
=
EXERCISES
First sketch the region enclosed by the given curves!
1. Find the area of the region enclosed by y = x+1, y = 9 – x2, x = -1 , x = 2
2. Calculate the area between y = x and y = x2.
3. Calculate the area bounded by y = x + 1, y = (x-1)2, x = -1 and x = 2.
4. Find the area enclosed by x = 1 – y2 and x = y2 – 1.
5. Find the area of the region enclosed by y = 2x – x2 , y = x - 2 and x = 3
13.2
ARC LENGTH
In this section we show how to calculate the length of a curve in the xy-plane (or
Cartesian plane). The procedure for defining arc length is very similar to the procedure
we used for
defining area.
Figure 1
11
∆yi
y
∆si
y= f(x)
∆xi
0
a
b
x
To find the length of the curve in Figure 1, we go through the same process as for
integration. We begin by dividing the curve into small pieces. We do so by partitioning
the interval [a,b] into n subintervals of equal length. We then find the approximate
length ∆si of the arc in each subinterval and add up these lengths. The appearance of the
curve near a typical subinterval is sketched in Figure 2.
162
(xi,f(xi))
Figure 2
∆si
y = f(x)
∆yi
(xi-1,f(x i-1))
∆xi
If ∆x is small, and if f(x) is continuous in the interval [xi-1,xi], then the length of the curve
between xi-1 and xi, which we will denote by ∆si, can be approximated by the straight-line
segment joining the points (xi-1, f(xi-1)) and (xi,f(xi)). The length of this segment is, by
Δx 2  Δyi , where ∆yi = f(xi) – f(xi-1).
2
the Pythagoras Theorem, given by
Thus,
∆si ≈ Δx 2  Δyi .
(A)
If f is a continuous function on [a,b] and a differentiable function on (a,b), then f is also
continuous on the smaller interval [xi-1,xi] and differentiable on (xi-1,xi). By the Mean
Value Theorem, there is a number ci in (xi-1,xi) such that
∆yi = f(xi) – f(xi-1) = f ′(ci)(xi - xi-1) = f ′(ci) ∆x .
(B)
From (A) and (B) we have
2
Δx 2  (f (c i )) 2 Δx 2
∆si ≈
=
Δx 2 (1  [f (c i )]2 )
=
1  [f (c i )]2 ∆x.
So that
n
s
=
 Δsi 
i 1
n

1  [f (c i )]2 Δx .
i 1
If f ′ is continuous in [a,b], then (f ′)2 is continuous in [a,b], and 1  [f (c i )]2 is
continuous in [a,b]. Thus,
b

1  [f (x)] 2 dx
a
exists.
But,
b

1  [f (x)] 2 dx = lim
Δx0
a
n

i 1
1  [f (c i )]2 Δx .
Definition 13.1:
Suppose that f and f ′ are continuous functions in [a,b]. Then the length of the curve (arc
length) f(x) between x = a and x = b is defined by
b
S =

1  [f (x)] 2 dx
a
163
If we use the Leibniz notation for derivatives, we can write the arc length formula as
follows:
b

S =
a
2
 dy 
1    dx
 dx 
Examples
3
2
1. Find the length of the arc of the graph of f(x) = 4x between x = 0 and x =
Solution:
b
Length of arc

=
1  [f (x)] 2 dx
a
3
2
1
2
Since f(x) = 4x , it follows that f ′x) = 6x .
2
3
Hence
S
=
1

1  (6x 2 ) 2 dx
0
2
3

=
1  36x dx
0
2
3 3
 1 2

2
(
)(
)(1

36x)
 36 3


0
=
1 
3
2 2
1  36( )   1  36(0)  2
54 
3 
54
=
1
=
54
(125  1)
62
.
27
=
1
2. Find the length of the curve cos x = ey for x between


and
.
6
3
Solution:
b
Length of curve is given by

1  [f (x)] 2 dx .
a
Since cos x = ey it follows that f(x) = y = ln(cosx)
dy
1
and f ′(x) =

(sinx ) = - tanx
dx cosx
π
3
Hence
S
=

1  ( tanx) 2 dx
π
6
164
3
2
3
.
π
3
=

sec 2 x dx
π
6
π
3
=
 sec x dx
π
6
π
3
π
6
=
ln secx  tanx 
=
ln 2  3  ln
=
ln 2  3  ln 3
=
2 3

2 
 = ln 1 
 .
ln 

3
3 


2
3
(See Chapter 11.1)

1
3
EXERCISES
Find the length of the given curve:
x3
1
1. f(x) =
for x from 1 to 3.

6 2x
2. y =
x4
1
 2 for 1 ≤ x ≤ 3
4 8x
3. f(x) =

1
2  x2
3
4. y = ln(sinx) ,

3
2
between x = 0 and x = 1
π
π
x
6
3
5. 1 = ey + x2 between x = 0 and x = ½
13.3 SURFACE AREA OF A SOLID OF REVOLUTION
A surface of revolution is formed when a curve is rotated about a line. Such a surface is
the lateral boundary of a solid of revolution. We can think of it as peeling away the outer
layer of the solid of evolution and laying it out flat so that we can measure its area.
Figure 3
Some surfaces of revolution.
165
If the graph of y = f(x) between x = a and x = b is rotated about the x-axis, and if f ′(x) is
continuous on [a,b] and f(x) ≥ 0 on [a,b], then the lateral surface area (excluding the
ends) of the resulting solid is given by
b
S
2π  f(x) 1  f (x)  dx .
2
=
a
y
y=
f(x)
x
a
b
S
y
a
b
Examples
1. Find the lateral surface area of the solid obtained by rotating the graph of y =
2 x between x = 3 and x = 8 about the x-axis.
Solution:
1

dy
If y = 2 x then y′ =
 x 2.
dx
b
Hence
S
=
2π  y 1  y dx
2
a
2
=
 -1 
2π  2 x 1   x 2  dx
3
 
=
4π  x 1 
8
8
3
8
=
1
x
4π  x  1 dx
3
166
dx
8
=
=
3
 2

2
4 ( x  1) 
 3
3
8
3
=
=
8
3
2
(9) 
8
3
(4)
3
2
(27  8)
3
152
.
3
2. The region bounded by the graph of y = x3, the y - axis and x = ½ is rotated about
the x-axis. Find the lateral surface area of the resulting solid.
Solution:
If y = x3, then y′ = 3x2. The y-axis is the line x = 0.
b
Then
S
=
2π  y 1  y dx
2
a
1
2
=
 
2π  x 3 1  3x 2
2
dx
0
1
2
=
2π  x 3 1  9x 4 dx
0
=
=
=
=
1
2
 1 2

2π   (1  9x 4 ) 
 36  3 
0
3


1  25  2
1

2π
   (1)
 54  16 
54 


3

  5 
   1
27  4 

3
2
  61 
61
=
.
1728
27  64 
3. Find a formula for the lateral surface area of a cone.
Solution:
A cone of base radius r and height h can be generated by revolving the graph of
r
f(x) =
from x = 0 to x = h
x
h
about the x-axis.
167
y
y
f (x)
=
r
h
x
r
0
h
0
x
h
For such a cone we have
b
Lateral area =
S
=
2π  f(x) 1  f (x)  dx
2
a
2
r
r
2π  x 1    dx
h
h 
0
h
=
r
h2  r2
x
dx
h
h2
0
h
=
2π 
=
2π
=
2 r h  r  x 2 
 
h2
 2 0
=
 r h2  r2 .
r
h2
h
h 2  r 2  x dx
0
2
2
h
EXERCISES
1. The curve y = 4  x 2 between x = - 1 and x = 1 is rotated about the x-axis.
Find the surface area of the resulting solid.
Ans = 8π
1
between x = 1 and x = 2, is rotated about the x12x
axis. Find the lateral surface area of the resulting solid.
Ans = 12289π/192
2. The graph of
y = x3 
3. The graph of y = sinx , for x between 0 and π, is rotated about the x-axis. Find the
lateral surface area of the resulting solid.
Ans = 59/24
4. Find the area of the surface by revolving the curve x2 + y2 = r2 for - ½ r ≤ x ≤ ½
r
around the x-axis.
Ans = 2πr2
5. Find the surface area of the resulting solid if the curve 2y = x + 4 between x= 0
and x = 2 is rotated about the x-axis.
Ans = 5 5
168
x
6. Prove that the surface area of a sphere of radius r is 4πr2.
169
CHAPTER 14
POLAR COORDINATES
14.1 POLAR COORDINATES AND EQUATIONS
A polar coordinate system consists of a fixed point 0, called the pole, and a ray,
emanating from 0; called the polar axis.
(a) Polar coordinates
To locate a point P in the plane, we draw the segment OP. Let r be the length of OP and
let θ be the angle from the polar axis to OP, measured in a counter clockwise direction.
(See Figure 1)
Figure 1
r units
P(r,θ)
θ
0
Polar axis
Pol
e
Then P may be described
by any of the following ordered pairs:
(r,θ), (r, θ ± 2kπ) or ( -r , θ ± (2k-1)π) where k is any integer.
If we want to graph a given ordered pair (r, θ), we first draw θ, with its initial side along
the polar axis and its vertex at the pole. Then, if r > 0, mark off r units along the terminal
side of θ, starting from the pole. If r < 0, we first extend the terminal side of θ back
through the pole (this is equivalent to drawing θ + π) and then mark off r units. (See
Figure 2)
Figure 3
Figure 2
P(x,y) or P(r,θ)
P(r,θ) or P(-r,
θ+π)
θ
0
y
y
Polar axis
θ
0
θ+π
x
x
Q(r,θ+π) or Q(r,θ)
If we superimpose a polar system on a rectangular system, with the pole coinciding with
the origin and the polar axis along the positive x-axis, we can find the equations that
relate (r,θ) of the polar system to (x,y) of the rectangular system, (see Figure 3 above):
170
x  rcosθ 


 y  rsinθ 
r 2  x 2  y 2 

y 
 tanθ  x 
and
Example 1.
Plot the following points
(a) (2,
8
3
)
(b)
(-1,

4
)
(c)
Solution:
The four points are plotted in Figure 4
(a)
B
P=(2,
(3, -22½ π)
(d)
(-2, -½ π)
(b)
8
3
8
3
)
2
3
0

0
A
4
A
P=(1, 4 )
(c)
0
 2
(d)
A
P=(-2,- 2 )
P=(3, -22½π )
0
A
2

Figure 4
Remark:
1. In (a), since 8π/3 = 2π + (2π/3), the line OB makes an angle of 2π/3 with OA and
the point P lies 2 units along this line.
2. We often use the triangles for the special angles 6 , 4 , 3
171


6
4
2
3
2

1

3
4
1
1
Example 2
A point P has polar coordinates (3, 6 ). Find its rectangular coordinates, and graph the
point.
Note: To convert (x,y) to (r,θ) we must be careful in our choice of θ. Be sure that θ lies
in the proper quadrant.
Solution:
We have r = 3 and θ = 6 , so using the triangle for the special angle 6 we obtain
x
y
=
r cosθ = 3 cos 6 = 3.
=
r sinθ = 3 sin 6
Therefore P has rectangular coordinates (
3
2
=
3 3
2
= 3. ½ = 32
3 3 3
,2)
2
P(3, 6 )
y

0
6
x
Polar axis
Example 3
A point P has polar coordinates (4,
2
). Find its rectangular coordinates and graph the
3
point.
Solution:
We have r = 4 and θ =
2
, so
3
x = rcosθ
=
4cos
2
3
172
= - 4cos

3
= -4( ½ ) = - 2
y = rsinθ
=
4sin
2
= 4sin
3
 3
= 4  = 2 3
 2 

3
Therefore P has rectangular coordinates ( -2, 2 3 ).
y
P(4,
2
2
)
3
3
0
x
Example 4
A point P has polar coordinates (- 6,

).
Convert to rectangular coordinates.
4
Solution:

Since r = - 6 < 0, we extend the terminal side of θ =
through the pole and obtain
4
(- 6,

) = (6,
4

+ π) = ( 6,
4
5
).
4
So we have
x
=
6cos
5
=
6( - cos
=
6sin
5
)
= - 6(
4
4
y

=
6(-sin

)
=
- 6(
4
4
1
2
1
2
) = -3 2
) = -3 2
Thus (x,y) = (-3 2 , -3 2 ).
Example 5
Convert the point (2,0), which is in polar coordinates, to rectangular coordinates.
Solution:
x
=
2cos0 =
2(1) =
2
and
y
=
2sin0 =
2(0) =
0, so that
(x,y) = (2,0)
Example 6
Convert ( 1, -π) to rectangular coordinates.
Solution:
Since (1, -π) = (1 , -π + 2π) = (1, π), we have
x
=
1.cos π
=
1(-1)
173
=
-1, and
=
1.sin π
(x,y) = (-1,0)
y
=
1(0)
=
0, so taht
Example 7
Convert (1, 3 ) to polar coordinates with r > 0 and 0 ≤ θ ≤ 2π.
Solution:
r
=
tanθ
=
x 2  y2
3
=
=
12  ( 3 ) 2
=
1 3 =
4 = 2 and
3
1
Since the point (1, 3 ) lies in the first quadrant, we have θ =
π
, (see the triangles in the
3
remark) so that
π
).
3
(r, θ) = (2,
Example 8
Convert (2,-2) to polar coordinates with r > 0 and 0 ≤ θ ≤ 2π.
Solution:
r =
22  22 =
x 2  y2 =
tanθ =
2
44 =
8 = 2 2 and
= -1
2
Since (2,-2) lies in the fourth quadrant, and 0 ≤ θ ≤ 2π , we have θ =
7
, and
4
(r, θ) = (2 2 ,
7
)
4
Example 9
Convert (-4 3 , -4) to polar coordinates with r > 0 and 0 ≤ θ ≤ 2π
Solution:
r
=
tanθ
=
(-4 3 ) 2  (-4) 2 =
4
=
4 3
48  16 = 8, and
1
3
Since the point (-4 3 , -4) lies in the third quadrant and 0 ≤ θ ≤ 2π, we have
θ = tan-1 (
1
)+π=
6
3
(r, θ)
= (8,

7
+π =
7
, and
6
)
6
(b) Polar equations
We may transform an equation in x and y into an equation in r and θ, (or vice versa).
174
Examples
1. A curve has equation 2x2 + 3y2 = 4 in rectangular coordinates. Find its equation
in polar coordinates.
Solution:
Since x = rcosθ and y = rsinθ, we have
2x2 + 3y2 = 2(rcosθ)2 + 3(rsinθ)2 = 2r2cos2θ + 3r2sin2θ = 4
Solving for r in terms of θ, gives
4
r2 =
2
2 cos   3 sin 2 
2. A curve has equation r = 3sinθ in polar coordinates. Find its equation in
rectangular coordinates.
Solution
Multiplying the given equation by r gives r2 = 3rsinθ.
Since x2 + y2 = r2 and y = rsinθ, it follows easily that
x2 + y2 = 3y.
EXERCISES
1. Plot the following points:
(a) (4,
5
)
(b) (-1, -

)
(c) (-1, -
5
3
4
) (d) (⅓ , -
2
)
3
4
2. Find the rectangular coordinates of the given points:
(a) (4,

)
(b)
6
3.
(6,
7
)
(c)
(-3, -
6

) (d)
(-2, -
3
3
)
4
(e) (2,3π) (f)
(-1, - π).
Write the given points in polar coordinates with r > 0 and 0 ≤ θ ≤ 2π.
(a)
(-3,0)
(b)
(2,- 2)
(c)
(3, - 3 3 )
(d)
(4 3 ,
4)
4.
(e)
(-1, 3 )
(f)
(-8 3 , -8)
Find the equation in rectangular coordinates of each of the following:
(a)
r = acosθ + bsinθ
(b)
r = cos2θ
(c)
r2 = 2cos2θ
(d)
5.
rsinθ = 4
(e)
r=
1
2  3 sin 
(f)
Find an equation in polar coordinates for each of the following:
(a)
x2 + y2 = 2x
(b)
2xy = 1
(c)
(d)
x=1
(e)
x2 – y2 = 4
175
r2sin2θ = 1
y2 = 2x
14.2 GRAPHS OF POLAR EQUATIONS
In rectangular coordinates we define the graph of the equation y = f(x) as the set of points
whose coordinates satisfy the equation. In polar coordinates, however, we must be
careful. Each point in the plane has an infinite number of representations.
The graph of an equation r = f(θ) is the set of ordered pairs (r, θ), found by
assigning values to θ and finding corresponding values of r.
Examples:
1. Graph r = 2sinθ
Solution:
Take θ at intervals of

, starting with θ = 0, and make a table of corresponding
6
values of r:
Θ
0


6
3
r = 2sinθ 0 1

2
2
3
2
3
π
5
6
1
3
7
4
6
3
0 -1
- 3
3
5
2
3
-2
11
2π
6
-1
- 3
We plot these points carefully. Notice that, since r is negative for π < θ < 2π, we get the
same points for 0 < θ < π, the curve traces itself twice.
( 3,
2
3
(0,1
)
)
( 3, 3 )
( 2, 4 )
( 2 , 34 )
(1, 56 )
(1, 6 )
(-1,0)
0
Graph of r =
2sinθ
Polar axis
(1,0
)
2. Graph r = 1 – 2cosθ
Θ
Solution:
0 
6
r = 1-2cosθ
1
1- 3


3
2
0
1
2
5
3
6
2
π
7
4
6
1+ 3
176
3
1+ 3
3
2
3
5
11
3
6
2
1
0
1- 3
2π
0
(2, 23 )
(1+
5
6
3,
(1, 2 )
)
(1- 3 , 116 )
(-1,0)
(13 , 6 )
(1,
(3, π)
(1+
3 , 76 )
Polar
axis
3
2
)
(2, 43 )
Graph of r= 1 – 2cosθ
14.3 AREA IN POLAR COORDINATES
We can use an integral to find an area bounded by curves whose equations are given in
polar form. The formula can be found from an approximation using sums of circular
sectors. As shown below, the area bounded by the graph of r = f(θ) and by the lines θ =
α and θ = β
( α < β) is given by

A=
1
2
 [f ( )] d
2
 
r=
f(θ)
θ =α
Polar axis
0
Examples:
1. Find the area of the region enclosed by r = 1 + sinθ
Solution:
Let us first draw the graph.
177
Θ
0


6
3
r=
1+sinθ
1
1,5
1+
3
2

2
2
3
2
1+
5
π
7
1
0,5
6
6
3
2
4
1,5
3
5
2
3
3
1-
0
1-
3
2
11
6
0,5
3
2
(2, 2 )
Polar axis
(1,π)
(1,0
)
Note the symmetry of the region about the line θ =
-

2
≤θ≤

is equal to the area for which

2
≤θ≤
2
 2

1

2
A = 2  2  f ( ) d  =
 

 2


, that is , the area for which
2
3
.
We can write this area as
2

2
1  sin   d
2

2

 1  2 sin   sin   d
2
=
2

2


2
=
 1  2 sin  d


 sin   d
2
+
2

2
2


2
=
 1  2 sin  d


  2 cos  2
2
=
=
 (1  cos 2 ) d

2

=
2
+ ½
2

 ½   12 sin 2  2

2

 
 
- 0 -    + 0 + ½ [ - 0 -    +0
2
2
 2
 2

3
2
178
.
2π
1
2. Find the area of the inside loop of r = 1 – 2cosθ
Solution:
Check the graph of r = 1 – 2cosθ in the previous section.
(2, 23 )
(1+
5
6
3,
(1, 2 )
)
(1- 3 , 116 )
(-1,0)
(13 , 6 )
(1,
(3, π)
(1+
3 , 76 )
Polar
axis
3
2
)
(2, 43 )
r= 1 – 2cosθ
Notice that the loop is described for 

3

 
.
Thus
3

3
Area
=
1
2
 (1  2 cos  ) d
2

3

3
=
1
2
 (1  4 cos   4 cos

=
1
2

3
3
 (1  4 cos  )d +  (1  2 cos 2 )d

3

1
2
 ) d
3


=
2
  4 sin  3 
3

+   12 sin 2  3 
3

3
=
1
2

 
 3   
  3 
3
3  
  4   12    4     12      12 

3

 2

 2 
 2 
2
2
3
3



 
  


179
=
EXERCISES
1. Graph the following:
(a) r + 2 = sinθ
(b) r = 2 cosθ
(c) r = 3 – 2cosθ
(d) r = ½ + cosθ
2. (a)
(b)
(c)
(d)

3 3
2 .
(e)
(f)
(g)
r2 = 4sin2θ
r + ½ cosθ = 1
r = 2θ
Find the area enclosed by the graph of r + 2 = sinθ.
Find the area of the region that consists of all points that lie within the
graph of r = 2cosθ and outside r = 1.
Find the area inside r = 1 + sinθ, outside r = 1 – cos θ and above the line
θ = 0.
Find the area enclosed by one loop of the four leaved rose r = cos2θ.
180
CHAPTER 15
DIFFERENTIAL EQUATIONS
15.1 INTRODUCTION
15.1.1 DEFINITIONS
Many problems in the physical and life sciences, engineering and the social sciences,
involve change: change in size, shape, motion, and so forth. Since a derivative is “a rate
of change”, the mathematical formulations of such problems are usually described by
differential equations. We intend dealing with an elementary introduction to a study of
ordinary differential equations.
Definition 15.1
An ordinary differential equation (ODE) is an equation that contains an unknown
function together with one or more of its derivatives.
Definition 15.2
The order of a differential equation is the order of the highest derivative of the
unknown function that appears in the equation.
Examples:
dy
1. x
 y  ex
dx
2. y′ = x2y2
(first order)
(first order)
2
d y
dy
2 y0
2
dx
dx
4. x2y″+xy′ + x2y = sin2x
5. y″′ + xy″-yy′ = ysinx
3.
(second order)
(second order)
(third order)
We notice in the above equations, x is the independent variable and y = y(x) is the
unknown function. Differential equations are different from the kinds of equations we
encountered so far in that the unknown is a function and not a number as in an equation
such as
x2 + 7x - 6 = 0.
15.1.2 SOLUTIONS OF DIFFERENTIAL EQUATIONS
The question that we want to consider is that of “solving” a given differential equation.
Definition 15.3
A solution of a differential equation is a function f defined on some interval, with the
property that f and its derivatives satisfy the equation.
To determine whether a given function satisfies a certain differential equation, we simply
substitute the function and its derivatives into the equation. If the result is an identity,
then the function is a solution; otherwise it is not.
181
Examples:
In each of the following, determine whether the given function satisfies the
corresponding differential equation.
dy
(a) y(x) = x2 – 5x
; x
 y  x2
dx
x
2
(b) y(x) = e – x
; (x-1)y″ - xy′ +y =(x-1)2
dy
(c) y(x) = e2x
;
 y  e 2x
dx
Solution:
(a) Substituting y(x) = x2 -5x and y′ (x) = 2x – 5 into the left hand side of the
dy
differential equation x
 y  x 2 , we get
dx
LHS
=
=
=
=
x(2x-5) – (x2 – 5x)
2x2 – 5x – x2 + 5x
x2
RHS
Thus, y(x) = x2 – 5x is a solution of the differential equation.
It can be verified in the same manner that y = x2 – Cx is a solution of the
differential equation for any constant C.
(b) Substituting y(x) = ex – x2 and its derivatives y′(x) = ex – 2x and y″(x) = ex - 2
into the differential equation (x-1)y″ - xy′ +y =(x-1)2, we get
LHS =
(x-1)( ex - 2) - x(ex – 2x) + ex – x2
=
xex -2x – ex + 2 – xex + 2x2 + ex – x2
=
x2 – 2x + 2
≠
(x-1)2
=
RHS.
Therefore, y(x) = ex – x2 is not a solution of the differential equation.
(c) Substituting y(x) = e2x ant its derivative y′(x) = 2e2x into the left hand side of the
dy
differential equation
 y  e 2x , we get
dx
LHS
=
2e2x - e2x
=
e2x
=
RHS.
Therefore, y(x) = e2x is a solution of the differential equation.
Remark:
Note that y(x) = e2x in (c) above is not the only solution of the differential equation
182
dy
 y  e 2x on the interval ( - ∞, ∞); for example, the function y(x) = Cex + e2x is also a
dx
solution for every real value of the constant C, since
dy
 y  Ce x  2e 2x  Ce x  e 2x = e2x
dx
dy
One can prove that all solutions of
 y  e 2x on (- ∞, ∞) can be obtained by
dx
substituting values for the constant C in y = Cex + e2x . On a given interval, a solution of
a differential equation from which all solutions on that interval can be derived by
substituting values of arbitrary constants, is called the general solution of the equation
on the interval.
dy
Therefore, y(x) = Cex + e2x is the general solution of
 y  e 2x on the interval (- ∞,
dx
∞).
15.1.3 INITIAL CONDITION
When an applied problem leads to a differential equal, there are usually conditions in the
problem that determine specific values for the arbitrary constants. As a rule of thumb, it
requires n conditions for all n arbitrary constants in the general solution of an nth-order
differential equation (one condition for each constant). For a first-order equation, the
single arbitrary constant can be determined by specifying the value of the unknown
function y(x) at an arbitrary point x0, say y(x0) = y0. This is called an initial condition,
and the problem of solving a first-order equation subject to an initial condition is called a
first-order initial-value problem.
Example
The solution of the initial-value problem
dy
 y  e 2x , y(0) = 3
dx
can be obtained by substituting the initial condition x = 0 and y = 3 in the general
solution
y = Cex + e2x to find C. We obtain
3 = Ce0 + e0 = C + 1
which gives C = 2.
Thus, the solution of the initial value problem, which is obtained by substituting this
value of C in y = Cex + e2x, is
y = 2e + e2x .
15.2
SOLUTIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS
15.2.1 FIRST-ORDER SEPARABLE EQUATIONS
The simplest form of first-order differential equations are those that can be written in the
dy
form f(x) 
.
dx
Such equations can often be solved by integration.
183
Example
dy
Solve
= x3
dx
Solution:
We can write the equation in the form
dy = x3dx
  dy   x 3 dx
 y = ¼ x4 + C
which is the general solution of
dy
= x3 on the interval ( - ∞, ∞)
dx
dy
 g(x), then we say that the equation
dx
is separable, and we can often find the general solution by first rewriting the eqauation in
the differential form
h(y)dy = g(x)dx,
i.e. all y’s are placed on the one side and all x’s on the other side.
Then both sides can be integrated to obtain
 h(y)dy   g(x)dx .
If the equation can be expressed in the form h(y)
This method of solving a differential equation is called separation of variables.
Examples:
1. Solve the differential equation
dy
= -4xy2,
dx
and then solve the initial value problem
dy
= -4xy2, y(0) = 1.
dx
Solution:
Separating variables yields
1
dy  4xdx .
y2
Integrating both sides
1
 y 2 dy  4 xdx
1
gives
-  2x 2  C .
y
Solving for y as a function of x, we obtain
1
y =
2
2x  C
184
The initial condition y(0) = 1 requires that y = 1 when x = 0. Substituting these values in
1
y =
2
2x  C
1
yields
1 =
, or C = -1.
C
Thus, the solution of the initial- value problem is
1
y =
2
2x  1
2. Solve the initial value problem
(4y-cosy)
dy
- 3x2 = 0, y(0) = 0.
dx
Solution:
Separating variables and integrating yields:
(4y-cosy)dy = 3x2dx
2

 (4y  cosy)dy  3 x dx

2y2 –siny
= x3 + C
(A)
The initial condition y(0) = 0 requires that y = 0 if x = 0. Substituting these values in
(A)
to determine the constant of integration
2(0) –sin(0) = 0 + C
or
C
= 0.
Therefore the solution of the initial-value problem is
2y2 – sin y = x3.
EXERCISES
1. Show that the members of the family are solutions of the differential equation and
find a member of the family that satisfies the initial condition(s).
(a) y = Ce5x;
y′ = 5y,
y(0) = 2
x2 C
 ;
3 x
1
(c) y = 1 3
;
 3x xC
(b) y =
(d) y = xln
C
;
x
(e) x = -ln(e-C –sint) ;
(f) x = e
t
sin2t
C
2
;
(g) y = C1x + C2x ½ ;
xy′ +y = x2,
y(3) = 2
y′ = y2(1+x2), y(0) = 1
y′ =
yx
,
x
dx
 e x cost ,
dt
y(2) = 4
x(0) = 3
dx
 e x sin t , x(0) = 1
dt
2x2y″-xy′ + y = 0, y(4) = 1, y′(4) = -2
185

y″ + 9y = 0,
(i) y = C1x2 + C2x2lnx;
x2y″-3xy′ + 4y = 0, y(1) = 0, y′(1) = 1.
y(
2
) = y′(

(h) y = C1sin3x + C2cos3x;
)=1
2
2. Solve the following differential equations
(a)
dy
=
y
dx
x
(b) (1+x4)
dy
=
dx
x3
y
(c) (1+y2)y′ = exy
(d) e-ysinx - y′cos2x = 0
(e) 3tany - y′secx = 0
4x 2
, y(1) = π
y  cosy
(f) y′ =
(g) y′ - xey = 2ey, y(0) = 0
(h)
dy

dt
2t  1
2y  2
, y(0) = -1
15.3 FIRST- ORDER LINEAR EQUATIONS
Not every first-order differential equation is separable, e.g the variables in the equation
dy
dx
 2xy  xe  x
2
cannot be separated.
However, this equation can be solved by a different method that we will now consider.
Definition 15.3.1
A first-order differential equation is linear if it is expressible in the form
dy
dx
 p(x)y  q(x)
where the functions p(x) and q(x) are continuous and may or may not be constant.
Examples:
(a)
(b)
dy
dx
dy
dx
 x 2 y  ex
 (sinx)y  x 3  0
186
(c)
dy
dx
 5y  2
One procedure for solving
dy
 p(x)y  q(x) is based on the following observation:
dx
If we define μ = μ(x) by
p(x)dx
μ = e
p(x)dx d
dμ
 e
.  p(x)dx  μ.p(x) .
dx
dx
d
dy dμ
dy
(  y)  μ

yμ
 μ.p(x).y
dx
dx dx
dx
then
Thus,
If
dy
dx
 p(x)y  q(x) is multiplied throughout by μ, and then simplified using
d
dy
(  y)  μ
 μ.p(x).y
dx
dx
d
dy
we get
(  y)  μ
 μ.p(x).y  .q(x) .
dx
dx
We can now solve equation (A) by integrating the left and right sides to obtain
d
 dx ( y)dx   .q(x)dx

 y   .q(x)dx  C

y
1

(A)
  q(x)dx  C.
Conclusion:
A first-order linear differential equation of the form
dy
dx
 p(x)y  q(x) can be solved
using the method of an integrating factor, given by the following three steps:
 p(x)dx


e
(a)
Calculate
.
(b)
Multiply both sides of
dy
dx
(c )
 p(x)y  q(x) by μ and express the results as
d
(  y)  .q(x) .
dx
Integrate both sides of the equation obtained in (2) and then solve for y.
Examples
1. Solve the differential equation
dy
dx
 2y
0
187
Solution:
The method of separation by variables can also be used for this equation, but
since it is of the form
dy
dx
 p(x)y  q(x)
with p(x) = -2 and q(x) = 0; it can also be solved by the method of integrating
factors.
(a) The integrating factor is
  e
p(x)dx
 2dx
= e 
= e-2x
dy
 2 y  0 throughout
dx
dy
μ (  2y  0 )
dx
(b)
Multiply


by μ:
dy
 2 y  0
dx
d
(from step (2) of the method)
 y  0
dx
d 2x

(e y)  0 (from step (1) above).
dx
(c )
Integrating both sides of this equation
d 2x
 dx (e y)dx   0dx

e-2xy = C, which can be written as
y
= Ce2x.
gives
2. Find all solutions to the equation
dy
dx

4
x
y  3x 2
and the unique solution that satisfies y(1) = 2.
Solution:
4
4
p(x)dx
 dx
4ln x
The integrating factor is   e 
e x e
 e lnx  x 4 .
Multiplying both sides of the equation by the integrating factor gives
x4
dy
dx
 4x 3 y  3x 6
which according to (2) of the method gives us
d 4

x y   3x 6 .
dx
188
Integrating both sides gives us
x4y
=
3
7
3
x7  C
C
.
7
x4
For the unique solution substitute x = 1 and y = 2 into the general solution:
3
C
2 = .1 
7
1
or
y

C
Therefore
=
=
y=
=
x3 
11
7
3
7
x3 
1 11
x
4 7
1
11 
3
 3x  4 
7
x 
ds
 2s  6t
dt
Solution:
In this problem t is the independent variable and s = s(t) is the function of t. Thus
t plays the role of x and s the role of y in our formula.
To use the method of integrating factors, the equation must be in the correct form.
Thus we first divide left and right by t:
ds 2
 s  6.
dt t
The integrating factor is now
3. Solve t
 e
2
 t dt
=e
= e lnt  t 2
Multiplying both sides by the integrating factor   t 2 gives
ds
t2
 2ts  6t 2
dt
d 2
or
(t s)  6t 2 .
dt
Integrating both sides of the last equation gives us
t 2 s   6t 2 dt  2t 3  C
or
2
2ln t
s = 2t + Ct-2.
EXERCISES
189
Find all solutions to the given differential equations. If an initial condition is specified,
then find the particular function which solves that initial-value problem.
dy
1.
 4y .
dx
dy
2. (1  x 3 )
 3x 2 y
dx
dy
3. (1  e x )
 ex y  0
dx
dx
4.
 x  1 ; x(0) = 1.
dt
dx
5.
 x  sint ; x(0) = 1.
dt
dy
 
6.
 ( tanx)y  2xsecx ; y   1 .
dx
4
dx
4t
7.
 2
x  3t ; x(0) = 4
dt t  1
dr
  1
8.
 r tant  cost ; r  
dt
3 2
dy
9. ( x 2  4)
 2x  8xy
dx
190
191