Section 5.1 – Estimating with Finite Sums
Area
Find the area of the region S that lies under the curve y = f(x) from a to b. We will require that f (x) 0
on the interval.
a
b
We do know how to find the area of certain shapes. The easiest shape to find the area of is the
rectangle. Recall that A = l × w.
We will start by approximating the area using rectangles. Consider the following problem:
Find the area under the curve f ( x)  5  x 2 from x = 0 to x = 2.
5
4
3
2
1
0.5
1
1.5
2
Try using one rectangle to approximate the area. To determine the height of the rectangle, we evaluate
the function f at either the left endpoint or right endpoint of the interval.
1
5
5
4
4
3
3
2
2
1
1
0.5
1
1.5
2
0.5
1
1.5
2
Thus, the area is bounded.
We have two sums:
An upper sum (where we use the maximum point in each subinterval), and
A lower sum (where we use the minimum point in each subinterval).
Now, try using two rectangles to approximate the area. To determine the height of the rectangle, we
evaluate the function f at either the left endpoint or right endpoint of each interval.
ba 20

 1.
The length of each subinterval is x 
n
2
5
5
4
4
3
3
2
2
1
1
0.5
1
1.5
2
0.5
1
1.5
2
This is a better approximation. Again, the area is bounded.
2
Now try using four rectangles to approximate the area. To determine the height of the rectangle, we
evaluate the function f at either the left endpoint or right endpoint of each interval.
ba 20 1

 .
The length of each subinterval is x 
n
4
2
5
5
4
4
3
3
2
2
1
1
0.5
1
1.5
2
0.5
1
1.5
2
This approximation is even better. Once again, the area is bounded.
We can continue on with this process as long we want.
3
"Approximate area
"7.8125`
"Approximate area
"Approximate area
"7.578125`
"Approximate area
"6.8125`
"7.078125`
We could also use the midpoint of each subinterval as the height of the rectangle.
Example: Approximate the area under the curve f ( x)  5  x 2 from x = 0 to x = 2 using two
rectangles. Choose the midpoint of each subinterval as the sample points.
5
4
3
2
1
0.5
1
1.5
2
4
Displacement Versus Distance Traveled
If v = v(t) is the velocity function, then |v(t)| is the speed, and the distance traveled is approximated by
the sum v(t1 ) t  v(t2 ) t   v(tn ) t
If the position of the object is given by s = s(t), then the displacement is given by: s(b) – s(a)
Distance Traveled
If you are given the acceleration function over a time interval, the speed at time t is the area under the
curve from 0 to t.
If you are given the velocity function over a time interval, the total distance covered in time t is the
area under the curve from 0 to t.
Example: An object is dropped straight down from a helicopter. The object falls faster and faster but
its acceleration decreases over time because of air resistance. The acceleration is measured
in ft/sec3 and recorded every second after the drop for 5 seconds.
Time (sec)
Acceleration
0
32
1
2
19.41 11.77
3
7.14
4
4.33
5
2.63
a) Find an upper estimate for the speed when t = 5 seconds.
b) Find an upper estimate for the speed when t = 4 seconds.
c) Find an upper estimate for the total distance fallen when t = 3 seconds
Time (seconds)
Speed (upper est.)
0
0
1
32
2
51.41
3
63.18
4
70.32
5
74.65
5
Average Value of a Function: The average value of a non-negative function on an interval [a, b] is
the area under its graph divided by b – a.
Example: Find the average value of the function f ( x)  sin x on the interval [0, π].
Partition the interval into four subintervals and use midpoints for the height of each
rectangle.
6
Section 5.2 – Sigma Notation and Limits of Finite Sums
Finite Sums and Sigma Notation
Keep in mind that we can often rewrite a sum in a more compact notation using
n
notation. Recall that:
ak

k 1
 a1  a2  a3 

(summation)
 an where ak is the formula for the kth term, k is
the variable of summation, and in this case it starts at one, and goes to n (that is, we will replace k with
the integers from 1 to n in the formula).
Note: It is common to use i, j, or k as the variable of summation (though we could use any other letter
if we wanted to).
5
Example: Evaluate the sum:
k

k 1
2
 1
Example: Express the following sum using sigma notation:
3 + 5 + 7 + 9 + 11 + 13 + 15
Example: Express the following sum using sigma notation:
3 + 9 + 27 + 81
7
In order to evaluate the sums, we need some formulas.
Theorem – Algebra Rules for Finite Sums
n
1.
ak  b

k 1
k
n
2.
3.
c

k 1
4.
k

k 1
k

k 1
n
6.
2

 k3 
k 1
for any constant c.
k 1

n
5.
 c ak
 nc
n
n
n
cak

k 1
n
n
ak   bk

k 1
k 1

for any constant c.
n(n  1)
2
n(n  1)(2n  1)
6
 n(n  1) 


 2 
2
100
Example: Evaluate the sum:
k

k 1
50
Example: Evaluate the sum:
2
 1

k 1 15
k
8
n
Example: Suppose that
 ak  4 and
k 1
n
bk  12 .

k 1
Find the values of:
n
a)
10ak

k 1
n
b)
a

k 1
c)
 3a

k 1
k
=
 bk  =
n
k
 2bk  =
Limits of Finite Sums
Recall that in Section 5.1, we could obtain better and better approximations of the area by taking more
and more rectangles. Thus, the length of the subintervals became smaller and smaller. It looks like we
should be using a limit as n goes to infinity.
Example: Find the area under the curve f ( x)  5  x 2 from x = 0 to x = 2.
9
Reimann Sums
Riemann sums were introduced by Bernhard Riemman, and these will be important to the theory of
definite integrals.
Let f be some function defined on [a, b]. Note that f can have negative as well as positive values on the
interval.
a
b
Divide [a, b] into n subintervals, not necessarily of equal length by choosing n – 1 points between a
and b such that a  x1  x2   xn1  b .
Let a  x0 and b  xn so that a  x0  x1  x2   xn1  xn  b
The set P  x0 , x1 , x2 ,
, xn  is called a partition of [a, b].
The partition P divides [a, b] into n closed subintervals:
 x0 , x1  ,  x1, x2 ,
,  xk 1 , xk  ,
,  xn1 , xn 
The interval  xk 1 , xk  , where k is an integer between 1 and n, is called the kth subinterval of P.
The widths of the subintervals are given by:
x1  x1  x0 , x2  x2  x1,
xk  xk  xk 1,
xn  xn  xn 1,
If the widths of all the subintervals are equal, then
ba
x1  x2 
 xk 
 xn  x 
n
Before, we selected either the left endpoint, right endpoint, or midpoint of each subinterval.
It turns out (thanks to Riemann) that we can select ANY point of the subinterval.
Let ck   xk 1 , xk  , that is, ck is some point in the kth subinterval
 xk 1, xk  .
If f (ck )  0 , then the area of the kth rectangle whose base is length xk and whose height is f (ck ) is
given by: f (ck )xk .
If f (ck )  0 , then the negative of the area of the kth rectangle whose base is length xk and whose
height is f (ck ) is given by: f (ck )xk .
10
n
Summing all the products for the n rectangles we obtain: S P   f (ck )xk
k 1
The sum S P is called a Riemann sum for f on the interval [a, b].
If the subintervals are of unequal width, then we define the norm of a partition P, denoted by ||P||, to be
the maximum length of the given subintervals, that is, P  max x1 , x2 , , xn 
Example: Graph the function f ( x)  x 2  1 over the interval [0, 2].
Partition the interval into four subintervals of equal length.
4
Sketch the rectangles associated with the Riemann sum S P   f (ck )xk if ck is the:
k 1
a) left-hand endpoint
b) right-hand endpoint














c) midpoint of the kth subinterval







11
Section 5.3 – The Definite Integral
Definition – The Definite Integral as a Limit of Reimann Sums
Let f (x) be a function defined on a closed interval [a, b]. We say that a number I is the definite
n
integral of f over [a, b] and that I is the limit of the Reimann sums
 f (c )x
k 1
k
k
if the following
conditions are satisfied:
Given any number  > 0 there is a corresponding number  > 0 such that for every given
partition P  x0 , x1 , x2 , , xn  of [a, b] with ||P|| <  and any choice of ck in  xk 1 , xk  we
n
 f (c )x
have:
k 1
k
k
 I  .
Thus, f is integrable on [a, b] if:
b
a
b
n
f ( x)dx  nlim
f (ck )xk or
 
a
k 1
f ( x)dx  lim
P 
n
f (ck )xk

k 1
n
Example: Express lim
P 
2ck xk

k 1
3
as a definite integral where P is a partition of [–1, 0].
The definite integral will depend on the function and the interval, not the variable.
All of the following have the same value:
b
a f ( x)dx
b

a f ( y)dy

b
a f ( z)dz

b
a f (t )dt
(The variable is called a dummy variable)
Notation and the Existence of the Definite Integral
b
The notation for the definite integral is due to Gottfried Leibniz.
a f ( x)dx
The function f(x) is called the integrand, and the numbers a and b are called the lower and upper limits,
respectively.
When the definition is satisfied, we say that the Riemann sums of f on [a, b] converge to the definite
b
integral I 
a f ( x)dx
and that f is integrable over [a, b].
12
Theorem 1 – The Existence of Definite Integrals
A continuous function is integrable. That is, if a function f is continuous on an interval [a, b],
then its definite integral over [a, b] exists.
Recall that in Section 5.1, we choose ck such that f (ck ) was either a maximum or a minimum on the
subinterval  xk 1 , xk  . Using the maximum we obtained an upper sum and using the minimum we
obtained a lower sum. The value of all other Riemann sums should fall in between the value of the
lower sum and the value of the upper sum.
Integrable and Nonintegrable Functions
Theorem 1 states that if f is continuous, then f is integrable. If f is not continuous on the interval, it may
or may not be integrable.
If f is piecewise continuous with a finite number of discontinuities, then it could be integrable.
If f is continuous except at a point where it has a vertical asymptote, it could be integrable.
Properties of Definite Integrals
Theorem 2 –
When f and g are integrable, the definite integral satisfies the following:
b
1.
a f ( x)dx
a


b f ( x)dx
a
2.
a f ( x)dx
 0
b
3.
b
a k f ( x)dx
a

k f ( x)dx
b
4.
b
a  f ( x)dx


a f ( x)dx
b
5.
a  f ( x)  g ( x) dx
b
6.
a f ( x)dx

c
b

a f ( x)dx
b f ( x)dx


b
a g ( x)dx
c
a f ( x)dx
13
7. If f has a maximum value max f and a minimum value min f on [a, b], then
b
min f  (b  a ) 
 f ( x)dx
 max f  (b  a )
a
b
8. If f(x) g(x) on [a, b] then
b
a f ( x)dx

a g ( x)dx
b
If f(x) 0
on [a, b] then
a f ( x)dx
 0
(See Page 375 for a geometric interpretation of Rules 2–7)
Example: Suppose that f and h are integrable and that
9
1
9
f ( x)dx   1,
7
9
f ( x)dx  5,
7 h( x)dx
 4
9
a) Find
1 2 f ( x)dx
9
b) Find
7  f ( x)  h( x) dx
9
c) Find
7 2 f ( x)  3h( x) dx
1
d) Find
9 f ( x)dx
7
e) Find
1 f ( x)dx
14
Area Under the Graph of a Nonnegative Function
Definition – Area Under a Curve as a Definite Integral
If y = f(x) is nonnegative and integrable over a closed interval [a, b], then the area under the
b
curve y = f(x) over [a, b] is the integral of f from a to b, A   f ( x )dx
a
b
Example: Compute
0 x dx and find the area A under y = x over the interval [0, b], b > 0.
b
Thus,
0 x dx
 ______ .
b
a
Example: Compute x dx and find the area A under y = x over the interval [a, b], 0 < a < b.
15
Therefore we have the following rules:
b
1.
a c dx
 c(b  a )
b
2.
 x dx 
a
b
3.
for any constant c
b2 a 2

2 2
when a  b
b3 a 3

3 3
when a  b
2
 x dx 
a
Example: Use the above rules to evaluate the following integrals.
2.5
a)
 x dx

0.5

b)
2
0 
2
d 
5
c)
0 2x dx

Definition – The Average or Mean Value of a Continuous Function
If f is integrable on [a, b], then its average value on [a, b] (also called its mean value)
b
1
f ( x)dx
is: av( f ) 
b  a a
Example: Find the average value of f(x) = 2 x 2  1 on [0, 5].
16
Section 5.4 – The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects the two processes: integration and differentiation.
The Mean Value Theorem for Definite Integrals
Recall that the Mean Value Theorem presented in Section 4.2 stated that if f was continuous on
[a, b] and differentiable on (a, b) then there was at least one number c in the interval (a, b) for which
f (b)  f (a )
f (c) 
. We now state an analogous theorem for definite integrals.
ba
Theorem 3 – The Mean Value Theorem for Definite Integrals
b
If f is continuous on [a, b], then at some point c in [a, b], f (c) 
1
f ( x)dx
b  a a
Notice the right hand side is just the average value of f on [a, b].
b
Written another way, the equation is f (c)(b  a )   f ( x )dx .
a
Example: Verify that f(x) = x on [0,5] meets the conditions of Mean Value Theorem for Definite
Integrals and then find c.
5
The integral
 x dx gives us the area under f(x) = x above the x-axis on [0, 5].
0
 2  (5  0) is the area under the horizontal line y  f  5 2  above the x-axis on [0, 5].
The expression f 5
5
f   (5  0) , then the two areas are equal.
2
0
Thus, f(c) is the height of the rectangle (on the same interval) whose area is equal to the area under the
curve y = f(x).
5
Since
 x dx =
5
5
4
4
3
3
2
2
1
1
1
2
3
4
5
1
2
3
4
5
17
Theorem 4 – The Fundamental Theorem of Calculus, Part 1
x
If f is continuous on [a, b], then F ( x)   f (t )dt is continuous on [a, b] and differentiable on
a
x
d
f (t )dt  f ( x)
(a, b) and its derivative is f(x): F ( x) 
dx a
Example: Find F' (x) by using the Fundamental Theorem of Calculus.
x
F ( x)   sin t dt
0
Example: Find F' (x) by using the Fundamental Theorem of Calculus.
x2
F ( x)   3t dt
1
Theorem 4 – The Fundamental Theorem of Calculus, Part 2
If f is continuous at every point of [a, b] and F is any antiderivative of f on [a, b], then
b
 f ( x)dx  F (b)  F (a) .
a
In other words, to evaluate the definite integral, find difference of the values of the
antiderivative at the upper limit and at the lower limit.
b
Note: The evaluation of the definite integral is also denoted by
 f ( x)dx
 F ( x) a  F (b)  F (a)
b
a
Example: Evaluate the following definite integrals.
2
a)
0 3x dx
2

b)
2
0 cos x dx
18
1
c)
0 e
2x
dx
1
 3 x2 
d)  2   dx
x
5 
1 

Example: Find the area under the curve f ( x)  5  x 2 from x = 0 to x = 2.
19
d
Example: Find
dx
sin x
1 3t dt
2
a) by evaluating the integral and differentiating the result.
b) by differentiating the integral directly (using the Fundamental Theorem of Calculus)
Total Area
If a function y = f(x) is both positive and negative on some interval [a, b], then the definite integral will
not give us the total area of the region bounded by f on [a, b], but rather the net signed area (area of the
portion above the x-axis is positive while the area of the portion below the x-axis is negative).
b
To find the total area, we must consider the integral A   f ( x ) dx
a
Steps: 1. Subdivide [a, b] at the zeros of f.
2. Integrate f over each subinterval.
3. Add the absolute values of the integrals
Example: Find the area under the curve f ( x)  5  x 2 from x = 0 to x = 3.
20
Section 5.5 – Indefinite Integrals and the Substitution Rule
The Substitution Rule
d
 f ( g ( x))  f ( g ( x))  g ( x)
dx
We used the chain rule to come up with general differentiation rules, like the general power rule,
d n
du
(u )  nu n 1 
dx
dx
Recall the Chain Rule:
This allowed us to differentiate an expression like x3  1 .
1/ 2
d  3 
1
3x 2
x  1   x3  1  3x 2 

dx 
2
2 x3  1
Because of this, we expect
2
3x 2
x 1
3
dx 
x3  1  C
n
 u du 
So we should have a general power rule for integration that states
u n 1
 C , n  1
n 1
Theorem 5 – The Substitution Rule
If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then
 f ( g ( x))  g ( x)dx   f (u )du
The Substitution Rule is just the Chain Rule for integration. We can state all of our integration rules in
terms of the Chain Rule now.
u n 1
 C , n  1
1.  u du 
n 1
n
1
8.
 csc
 sec u tan u du
2
u du   cot u  C
2.
 u du
 ln u  C
9.
3.
 e du
 eu  C
10.
 csc u cot u du
11.

13.
u
u
au
C
4.  a du 
ln a
u
5.
 sin u du
6.
 cos u du  sin u  C
7.
 sec
2
  cos u  C
1
 sec u  C
  csc u  C
du  sin 1 u  C
1 u
1
du  tan 1 u  C
12. 
1 u2
2
1
u 1
2
du  sec 1 u  C
u du  tan u  C
21
Example: Use the Substitution Rule to evaluate the integral.
  2x  1
5
dx
Example: Evaluate the integral.
 3x 2 
  2 x3  1  dx
22
Example: Evaluate the integral.
 xe
x2
dx
Example: Evaluate the integral.
 1 
  3x  1  dx
23
Example: Evaluate the integral.
 1 
  4 x  3  dx
The Integrals of sin2x and cos2x
Recall from trigonometry, cos 2 x  1  2sin 2 x  2cos 2 x  1 .
If we solve for sin2x and cos2x in terms of cos2x then we obtain the identities:
cos 2 x 
1  cos 2 x
2
and
sin 2 x 
1  cos 2 x
2
We use these identities to evaluate integrals that of sin2u and cos2u .
Example: Evaluate the integral.
2
sin
 x dx
24
Example: Solve the initial value problem:
dy
1
 4 x( x 2  8) 3 , y (0)  0
dx
25
Section 5.6 – Substitution and Area Between Curves
Recall that we found introduced the idea of the definite integral by considering the problem of finding
the area under the curve y = f(x) on the interval a x b. We now consider the more general problem
of finding the area of a region bounded by two or more curves. This could lead to more complicated
integrals, so we also introduce a Substitution Rule for Definite Integrals.
Substitution Formula
Theorem 6 – Substitution in Definite Integrals
If g' is continuous on the interval [a, b] and f is continuous on the range of g, then
b

f ( g ( x))  g ( x)dx 
a
g (b )

f (u )du
g (a)
Example: Evaluate the definite integral.
1
0 ( x  7) dx
9
Example: Evaluate the definite integral.
14
1
3
2 x  1 dx
26
Example: Evaluate the definite integral.
1
0 e
5x
dx
Definite Integrals of Symmetric Functions
Theorem 7 –
Let f be continuous on the symmetric interval [–a, a].
a

a) If f is an even function, then
a
a
f ( x)dx  2  f ( x)dx
0
a
b) If f is an odd function, then

f ( x)dx  0
a
Example: Evaluate the definite integral.

 sin x dx

Example: Evaluate the definite integral.

 cos x dx

27
Areas Between Curves
Now consider the problem of finding the area between two curves y = f(x) and y = g(x) on the interval
a x b.
a
b
Lets assume that f(x) g(x) on the interval a x b. We can approximate the area between two curves
using rectangles, just as we did when we approximated the area under one curve.
a
b
As before, divide the interval [a, b] into n subintervals of equal length x where x 
ba
.
n
Label the endpoints of the subintervals as a  x0 , x1 , x2 , , xn1 , xn  b .
The width of each rectangle is just the length of each subinterval, x.
Let ci be the sample point chosen from the ith interval  xi 1 , xi  .
The height of each rectangle will be determined by f (ci )  g (ci ) .
Thus, the area of each rectangle is Ai   f (ci )  g (ci ) x
n
And the area of the whole region is approximated by A    f (ci )  g (ci ) x .
i 1
By taking larger and larger values of n, we will get better and better approximations.
n
Thus, the area is given by A  lim   f (ci )  g (ci ) x .
n 
i 1
28
Definition – Area Between Curves
If f and g are continuous with f(x)  g(x) throughout [a, b], then the area of the region between
the curves y = f(x) and y = g(x) from a to b is the integral of (f – g) from a to b:
b
A    f ( x)  g ( x)  dx
a
Note that we can use this formula to solve the original area problem of finding the area under the curve
y = f(x) on the interval [a, b]. The x-axis is the horizontal line y = 0, so take g(x) = 0, then we obtain
b
  f ( x)  0 dx
a
a
A    f ( x)  g ( x )  dx 
b
b

 f ( x)dx
which is just what we used earlier in this
a
chapter.
Example:
Find the area of the region bounded by y = x2 and y = x + 6.
10
8
6
4
2
-3
-2
-1
1
2
3
4
29
Integration with Respect to y
In some cases in is more advantageous to integrate with respect to y. Consider the region bounded by
x = f(y) and x = g(y) from y = c to y = d.
d
c
We must view this problem horizontally instead of vertically.
right
Before, we could generalize our rule as
A
 upper curve  lower curve dx
left
upper
So viewing the problem horizontally, we can say A 
 right curve  left curve dy
lower
Example: Find the area of the region bounded by x = y2 and y = x – 2.
4
3
2
1
-1
-1
1
2
3
4
5
-2
-3
30