13.3 The Definite Integral
Question 1: What is a definite integral?
Question 2: How is the definite integral related to the approximate area?
In this section we continue to compute areas using rectangles. Doing this allows us to
undo the process of taking the derivative of a function. From the derivative of a function,
we can estimate changes in the function and make the estimates arbitrarily close.
By using more and more rectangles over an interval, the area of the rectangles
approach the area between the derivative function and the x axis over the interval. This
area is the exact change in the original function and is represented by a definite integral.
1
Question 1: What is a definite integral?
A definite integral is a notation used to describe the area between a function and the x
axis over some interval. A definite integral uses the integral sign that was introduced
earlier, but adds the endpoints of the interval. Suppose we have a non-negative function
f  x  with the area between x  a to x  b shaded.
Figure 1 – The shaded area between f (x) and the x axis over the interval [a, b].
The definite integral for the area of this shaded region is
b

f ( x) dx
a
The constants a and b are called the limits of integration. They define the interval  a, b 
for the shaded region. The function in the integrand, f  x  , forms the top boundary of
the shaded region. Since this function is positive over the entire interval, the area of this
region is positive.
A function that is negative has area that is negative.
2
Figure 2 – The area enclosed by y = -x – 5 and the x axis, over the interval [1, 3].
The definite integral for the shaded region in Figure 2 is
3
   x  5 dx .
1
Since the function is negative over 1, 3 , the shaded area is negative.
Example 1
Area Between a Function and the x Axis
For each definite integral, graph the region corresponding to the area.
Determine whether the definite integral is positive or negative.
2
a.
x
2
 4  dx
0
Solution The definite integral corresponds the region between the x axis
and y  x 2  4 from x  0 to x  2 .
3
Figure 3 – The shaded region for the definite integral in part a.
Since the region is below the x axis, this area is negative.
3
b.
x
2
 4  dx
2
Solution The definite integral corresponds the region between the x axis
and y  x 2  4 from x  2 to x  3 .
4
Figure 4 – The shaded region for the definite integral in part b.
Since the region is above the x axis, the area is positive.
3
c.
x
2
 4  dx
0
Solution This region consists of the two regions from parts a and b.
5
The definite integral is the sum of the area over  0, 2 and  2, 3 . Since
the negative area on
0, 2
is greater than the positive area on  2, 3 ,
the definite integral is negative.
For some simple functions, the shaded region is a shape like a circle, triangle, or
rectangle. For these regions, we can find the area exactly using area formulas from
geometry.
Example 2
Evaluate the Definite Integral
5
Use area formulas to evaluate the definite integral
  x  1 dx .
0
Solution The definite integral corresponds to the area between y  x  1
and the x axis from x  0 and x  5 .
Figure 5 – The shaded region can be thought of as a rectangle capped by a triangle.
From the graph, we can see that this region consists of a rectangle and
a triangle. Since we know how to find the area of each of these
geometric figures, we can find the area of the region exactly.
6
The rectangle has a width of 5 units and length of 1 unit. Its area is
Area of Rectangle  length  width

 5

1
5
The triangle has a base that is 5 units long. Its height is also 5 units
long. The area is
1
 base  height
2
1
=
 5  5
2
25

2
Area of Triangle 
The definite integral is the sum of these areas,
5
  x  1 dx  5 
0
25
 17.5
2
For regions with more complicated shapes, we can use the strategy from 13.3 to
approximate the area using rectangles.
7
Question 2: How is the definite integral related to the approximate area?
In Section 13.2, we approximated the area under the derivative
R  Q   60.0200 Q 0.0432 thousand dollars per electric car
to find the change in revenue when production is changed from 90 cars to 110 cars.
Using left and right sums, we were able to find an upper and lower bound for this
change. When the number of rectangles is increased, these bounds move closer and
closer together. For a large enough number of rectangles, we can make the estimates
match to any number of decimal places.
As the number of rectangles increases, it is hard to show every term in a left of right
sum. Symbolically, we can use summation notation to indicate a sum. In summation
notation, the Greek letter sigma is used to indicate a sum. After the sigma a template for
the terms is written to establish the pattern in the sum. For instance, the expression
f ( x1 ) x  f ( x2 )x    f ( xn ) x
can be written in summation notation as
n
 f  x  x
i 1
i
Each term in the sum is a product of a function value f  xi  and  x . The index i
indicates what changes in each term of the sum. In this case, i changes from 1 to n so
the terms contain different values of x from x1 through xn .
Example 3
Write with Summation Notation
In Example 4, we formed a sum
R  Q1  Q  R  Q2  Q    R  Q8  Q Write this sum with summation notation.
8
Solution This sum has eight terms. Each term is a product of the rate
R and Q . This forms the basic pattern to the terms. Each term has a
different Q value so we can write the sum as
8
 R  Q  Q i 1
i
As the number of rectangles increases, the left and right hand estimates get closer
together. In Example 2 and Example 4 of Section 13.2, we found the left and right hand
sums for 2, 4 and 8 rectangles. The table below shows these estimates and several
others for larger numbers of rectangles.
Number of Rectangles
Left Hand Sum
Right Hand Sum
2
986.090
981.820
4
984.993
982.861
8
984.453
983.387
16
984.185
983.652
32
984.051
983.784
64
983.984
983.851
As the number of rectangles is increased, the left and right hand sums get closer and
closer together. Rounded to the nearest integer, the sums are exactly the same, 984.
For larger and larger number of rectangles, the sums will match to more and more
decimal places. If we were to fit 4500 rectangle from 90 to 110, both estimates would be
the same to three decimal places, 983.918.
Let’s look at some of these estimates graphically.
9
Number of
Rectangles
Left Hand Sum
Right Hand Sum
4
8
16
As the number of rectangles increases, the top of each rectangle looks more and more
like the rate function. The area of the rectangles gets closer and closer to the area
under the function and above the horizontal axis between 90 and 110.
If we were able to use an infinite number of rectangles, each with an infinitely small
width, the area of these rectangles would be equal to the area under the function.
10
Figure 6 – As the number of rectangles increases, the area of the rectangles
gets closer and closer to the area under the function, above the horizontal
axis, from 90 to 110.
Although we can’t sum up the infinite number of rectangles directly, we can symbolize
this sum using a limit,
Exact Area  lim
n 
n
 R  Q  Q i 1
i
In the expression on the right, we are summing up the area of n rectangles whose
heights are R  Qi  and widths are Q . By taking the limit as n approaches infinity, we
are describing the values the sum of the areas approach as we use more and more
rectangles. In practice, we don’t need to actually compute this sum for an infinite
number of rectangles. We simply use enough rectangles so that the sums match to a
specified number of decimal places.
A definite integral is used to symbolize the exact area under a function over some
interval.
11
Let f ( x) be a function defined on an interval  a, b  . The
definite integral of f from a to b,
b
 f ( x) dx
a
denotes the exact area under the function f ( x) , above the x
axis, and between a and b. This area is defined by
n
lim
n 
where  x 
 f  x  x
i 1
i
ba
and xi is an x value in the ith rectangle.
n
This limit must exist for the area to be defined.
The function f ( x) is called the integrand and the symbol

is called an integral.
We can use the definite integral notation to locate the area on a graph. For the change
in revenue from Q  90 to Q  110 , we write
110

90
R  Q  dQ  lim
n 
n
 R  Q  Q
i 1
i
The definite integral on the left represents the exact area under R  Q  and above the
horizontal axis over the interval 90, 110 . The right hand sum represents the area of
rectangles whose heights are determined by R  Qi  and widths are Q . For now, we’ll
use left, right, or midpoint sums to estimate the area of this region. However, in later
sections we’ll find the exact value of the area using antiderivatives.
12
Example 4
Estimate the Definite Integral
5
  x  1 dx
Estimate the definite integral
using left and right hand sums
0
with 5 rectangles.
Solution To estimate the area under f  x   x  1 , we will calculate the
Riemann sum
5
 f  x   x i 1
i
The width of each rectangle is  x 
50
1
5
Figure 7 – The left hand sum for 5 rectangles on the interval [0,5].
From the graph in Figure 7, we see that x1  0 , x2  1 , x3  2 , x4  3 ,
and x5  4 . This means that
13
5
 f  x   x  f  0   x  f 1  x  f  2   x  f  3  x  f  4   x
i 1
i

1 1

15

2 1 
3 1

4 1

5 1
For the right hand sum, x1  1 , x2  2 , x3  3 , x4  4 , and x5  5 .
Figure 8 - The right hand sum for 5 rectangles on the interval [0,5].
In this case, the sum is
5
 f  x   x  f 1  x  f  2   x  f  3  x  f  4   x  f  5  x
i 1
i

2 1 

20
3 1 
4 1

5 1

6 1
These estimates place the exact area between 15 and 20. In Example
2, we found the exact area geometrically as 17.5. In general, we can’t
find the area geometrically so the left and right hand sums give us a
convenient way of estimating the area. In section 13.4, we’ll learn how
to find the area exactly for integrands whose antiderivative we can
compute.
14
Example 5
Estimate the Definite Integral
The definite integral
3
 x dx
2
1
denotes the area of some region.
a. Graph the region corresponding to the definite integral.
Solution The integrand of this definite integral is f  x   x 2 . The region
corresponding to the definite integral is the area between the function
and the x-axis from x  1 to x  3 .
b. Estimate the definite integral using a midpoint sum with four
rectangles.
Solution Each rectangle must have a width of  x 
3 1
 0.5 . The
4
rectangles heights come from the rate at the midpoint of each rectangle.
For instance, the first rectangle extends from x  1 to x  1.5 so the
15
midpoint is x  1.25 . The area of the corresponding rectangle is
1.252  0.5 .
The midpoint sum is
4
 f  x   x  1.25
i 1
2
i
 0.5  1.752  0.5  2.252  0.5  2.752  0.5
 8.625
Figure 9 – The rectangles for the midpoint sum where the width of each
rectangle is 0.5.
16