Learning Objectives for Section 13.4
The Definite Integral
1. The student will be able to
approximate areas by using left and
right sums.
2. The student will be able to
compute the definite integral as a
limit of sums.
3. The student will be able to apply
the properties of the definite integral.
Barnett/Ziegler/Byleen Business Calculus 11e
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Introduction
We have been studying the indefinite integral or antiderivative
of a function.
We now introduce the
definite integral. This
integral will be the area
bounded by f (x), the x axis,
and the vertical lines x = a
and x = b, with notation

b
f ( x ) dx
a
Barnett/Ziegler/Byleen Business Calculus 11e
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5
Estimating  0 . 5 x
2
 2 dx
1
One way to approximate the area under
a curve is by filling the region with
rectangles and calculating the sum of the
areas of the rectangles.
Take the width of each rectangle to be
x = 1. If we use the left endpoints, the
heights of the four rectangles are
f (1), f (2), f (3) and f (4), respectively.
f (x) = 0.5 x2 + 2
1
2
3
4
5
L4 = f (1) Δx + f (2) Δx + f (3) Δx + f (4) Δx
= 2.5 + 4 + 6.5 + 10 = 23
Barnett/Ziegler/Byleen Business Calculus 11e
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Estimating Area
(continued)
We can repeat this using the right side of
each rectangle to determine the height.
The width of each rectangle is again
x = 1. The heights of each of the four
rectangles are now f (2), f (3), f (4) and
f (5), respectively.
f (x) = 0.5 x2 + 2
1
2
3
4
5
The sum of the rectangles is then
R4 = 4 + 6.5 + 10 + 14.5 = 35
The average of L4 and R4 would be an even better
approximation: Area  (23 + 35)/2 = 29.
Barnett/Ziegler/Byleen Business Calculus 11e
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Estimating Area
(continued)
The previous average of 29 is very close
to the actual area of 28.666….
f (x) = 0.5 x2 + 2
Our accuracy can be improved if we
increase the number or rectangles, and
let x get smaller.
1
2
3
4
5
The error in our process can be calculated if the function is
monotone. That is, if the function is only increasing or only
decreasing.
Let Ln and Rn be the approximate areas, using n rectangles of equal
width, and the left or right endpoints, respectively.
Barnett/Ziegler/Byleen Business Calculus 11e
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Estimating Area
(continued)
If the function is increasing, convince
yourself by looking at the picture that
f (x) = 0.5 x2 + 2
Ln  Area  Rn
If f is decreasing, the inequalities go
the other way.
1
If you use Ln to estimate the area, then
Error = |Area – Ln|  |Rn – Ln|.
If you use Rn to estimate the area, then
Error = |Area – Rn|  |Rn – Ln|.
Either way you get the same error bound.
Barnett/Ziegler/Byleen Business Calculus 11e
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3
4
5
6
Theorem 1
It is not hard to show that
|Rn – Ln| = | f (b) – f (a)| x,
and that for n equal subintervals,
x 
ba
.
n
Error  | f ( b )  f ( a ) | 
b  a
n
For our previous example:
Barnett/Ziegler/Byleen Business Calculus 11e
Theorem 1
Error  |14 . 5  2 . 5 |
5 1
 12
4
7
Theorem 2
If f (x) is either increasing or decreasing on [a, b], then its left
and right sums approach the same real number I as n  .
This number I is the area between
the graph of f and the x axis from
x = a to x = b.
Barnett/Ziegler/Byleen Business Calculus 11e
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Approximating Area
In the previous example, we had
Error  |14 . 5  2 . 5 |
5 1
f (x) = 0.5 x 2 + 2
 12
4
If we wanted a particular accuracy,
say 0.05, we could use the error
1
2
3
formula to calculate n, the number of
rectangles needed:
5 1
|14 . 5  2 . 5 |
 0 . 05
n
4
5
Solving for n yields n = 960. We would need at least 960
rectangles to guarantee an accuracy of 0.05.
Barnett/Ziegler/Byleen Business Calculus 11e
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Definite Integral as Limit of Sums
We now come to a general definition of the definite integral.
Let f be a function on interval [a, b]. Partition [a, b] into n
subintervals at points
a = x0 < x1 < x2 < … < xn–1 < xn = b.
The width of the kth subinterval is xk = (xk – xk-1).
In each subinterval, choose an arbitrary point ck
xk-1  ck  xk.
Barnett/Ziegler/Byleen Business Calculus 11e
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Definite Integral as Limit of Sums
(continued)
Then define
n
L n  f ( x 0 )  x  f ( x1 )  x  . . .  f ( x n  1 )  x 

f ( xk  1 )x
k 1
n
R n  f ( x1 )  x  f ( x 2 )  x  . . .  f ( x n )  x 

f ( xk )x
k 1
n
S n  f ( c1 )  x  f ( c 2 )  x  . . .  f ( c n )  x 

f (ck ) x
k 1
Sn is called a Riemann sum. Notice that Ln and Rn are both
special cases of a Riemann sum.
Barnett/Ziegler/Byleen Business Calculus 11e
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A Visual Presentation
of a Riemann Sum
Δx
f (c 2)
f (c 1)
a = x0
x1
c1
x2
...
c2
The area under the curve is
approximated by the Riemann sum
Barnett/Ziegler/Byleen Business Calculus 11e
x n -1
xn = b
cn
n

f (ck )  xk
k 1
12
Area (Revisited)
Let’s revisit our original problem and
calculate the Riemann sum using the
midpoints for ck.
The width of each rectangle is again
x = 1. The heights of the four rectangles are now f (1.5), f (2.5), f (3.5)
and f (4.5), respectively.
f (x) = 0.5 x2 + 2
1
2
3
4
5
The sum of the rectangles is then
S4 = 3.125 + 5.125 + 8.125 + 12.125 = 28.5
This is quite close to the actual area of 28.666. . .
Barnett/Ziegler/Byleen Business Calculus 11e
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The Definite Integral
Theorem 3. Let f be a continuous function on [a, b], then
the Riemann sums for f on [a, b] approach a real number
limit I as max xk  0.
This limit I of the Riemann sums for f on [a, b] is
called the definite integral of f from a to b, denoted

b
f ( x ) dx
a
The integrand is f (x), the lower limit of integration
is a, and the upper limit of integration is b.
Barnett/Ziegler/Byleen Business Calculus 11e
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Negative Values
If f (x) is positive for some values of x
on [a, b] and negative for others, then
the definite integral symbol

y = f (x)
b
f ( x ) dx
represents the cumulative sum of the
signed areas between the graph of
f (x) and the x axis, where areas
above are positive and areas below
negative.
Barnett/Ziegler/Byleen Business Calculus 11e
B
a
a
A

b
b
f ( x ) dx   A  B
a
15
Examples
Calculate the definite integrals by
referring to the figure with the
indicated areas.

b

Area B = 12
f ( x ) dx   3 . 5
a

Area A = 3.5
B
a
c
f ( x ) dx  12
b
c
A
b
c
y = f (x)
f ( x ) dx   3 . 5  12  8 . 5
a
Barnett/Ziegler/Byleen Business Calculus 11e
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Definite Integral Properties




a
f ( x ) dx  0
a
b
f ( x ) dx  
a
b
a
b

a
f ( x ) dx
b
k  f ( x ) dx  k  
b
f ( x ) dx
a
[ f ( x )  g ( x ) ] dx 
a
b

a
f ( x ) dx 

c

b
f ( x ) dx 
a
f ( x ) dx 
a
Barnett/Ziegler/Byleen Business Calculus 11e


b
g ( x ) dx
a
b
f ( x ) dx
c
17
Examples
Assume we know that

3
x dx 
9
2
0
,

3
x
dx  9 , and
2
0

4
x
2
dx 
37
.
3
3
Then
A)
B)

3
4x
3
dx  4  x
2
0

2
dx  4 ( 9 )  36
0
3
0
3
( 3 x  2 x ) dx  3  x
2
0
2
3
dx  2  x dx 
0
9
Barnett/Ziegler/Byleen Business Calculus 11e
3 ( 9 )  2 ( )  18
2
18
Examples
(continued)

3
9
x dx 
,
2
0

C)
3
x
2

3
x
2
0
dx  
4

D)
E)

4
0

dx  9 , and

4
x
2
dx  
x
2
x
2
dx 
3
37
.
3
37
3
3
4
4
dx  0
4
3
4
3 x dx  3  x dx  3  x dx  3  9  3 
2
2
0
Barnett/Ziegler/Byleen Business Calculus 11e
3
2
37
 64
3
19
Summary
■ We summed rectangles under a curve using both the left
and right ends and the centers and found that as the number
of rectangles increased, accuracy of the area under the
curve increased.
■ We found error bounds for these sums.
■ We defined the definite integral as the limit of these sums
and found that it represented the area between the function
and the x axis.
■ We learned how to compute areas under the x axis.
Barnett/Ziegler/Byleen Business Calculus 11e
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