1
Section 4.2/4.3: Area, Riemann Sums and the Definite Integral
Practice HW from Larson Textbook (not to hand in)
p. 235 # 1, 3, 19-25, 43 (use Maple), 45 (use Maple) 53, 55
p. 245 # 31-38
Sigma Notation
The sum of n terms a1 , a2 ,, an is written as
n
 ai  a1  a2    an , where i = the index of summation
i 1
4
Example 1: Find the sum
 2i  1 .
i 1
Solution:
█
2
The Definite Integral
Suppose we have a function f ( x)  0 which is continuous, bounded, and increasing for
a  x  b.
Goal: Suppose we desire to find the area A under the graph of f from x = a to x = b.
3
To do this, we divide the interval for a  x  b into n equal subintervals and form n rectangles
(subintervals) under the graph of f . Let x0 , x1 , x2 , xn be the endpoints of each of the
subintervals.
Here,
Width of each
Width of each
ba

 x 
subinterva l
rectangle
n
Area of Rectangle 1  (Length)(W idth)  f ( x0 )x
Area of Rectangle 2  (Length)(W idth)  f ( x1 )x

Area of Rectangle n  (Length)(W idth)  f ( xn1 )x
Summing up the area of the n rectangles, we see
Total Area from
a xb
Left Hand

(Lower)
endpoint sum
n 1
  f ( xi )x  f ( x0 )x  f ( x1 )x    f ( xn1 )x
i 0
  f ( x0 )  f ( x1 )    f ( xn1 )x
4
We can also use the right endpoints of the intervals to find the length of the rectangles.
Area of Rectangle 1  (Length)(W idth)  f ( x1 )x
Area of Rectangle 2  (Length)(W idth)  f ( x2 )x

Area of Rectangle n  (Length)(W idth)  f ( xn )x
Summing up the area of the n rectangles, we see
Total Area from
a xb
Right Hand

Upper
endpoint sum
n
  f ( xi )x  f ( x1 )x  f ( x2 )x    f ( xn )x
i 1
  f ( x1 )  f ( x2 )    f ( xn )x
5
Note: When f is increasing,
Left Endpoint (Lower) Sum 
Total Area
 Right endpoint (Upper) sum
from a  x  b
If f is decreasing,
Right Endpoint (Upper) Sum 
Right Endpoint Sum 
Total Area
 Left endpoint (Lower) sum
from a  x  b
Total Area
 Left endpoint sum
from a  x  b
6
In summary, if we divide the interval for a  x  b into n equal subintervals and form n
rectangles (subintervals) under the graph of f . Let x0 , x1 , x2 , xn be the endpoints of each of
the subintervals.
Total Area from
a xb
Total Area from
a xb
Left Hand

(Lower)   f ( xi )x  f ( x0 )x  f ( x1 )x    f ( xn1 )x
endpoint sum i 0
  f ( x0 )  f ( x1 )    f ( xn1 )x
Right Hand

n 1
(Upper)
endpoint sum
n
  f ( xi )x  f ( x1 )x  f ( x2 )x    f ( xn )x
i 1
  f ( x1 )  f ( x2 )    f ( xn )x
The endpoints of the n subintervals contained within [a, b] are determined using the
formula
xi  x0  i x,
i  1, 2, n
wh ere
x0  a and x 
ba
n
7
Example 2: Use the left and right endpoint sums to approximate the area under
y  x 2  1 on the interval [0, 2] for n = 4 subintervals.
Solution:
8
9
█
10
Note: We can also approximate the area under a curve using the midpoint of the rectangles
to find the rectangle’s length.
n
Total Area from Midpoint

  f ( xi )x  f ( x1 )x  f ( x2 )x    f ( xn )x
a xb
sum
i 1
  f ( x1 )  f ( x2 )    f ( xn )x
where xi  x0  i x,
xi 
i  1, 2, n
where x 
ba
and
n
1
xi 1  xi   midpoint of [ xi 1 , xi ], i  1,2, n
2
11
Example 3: Use the midpoint rule to approximate the area under y  x 2  1 on the
interval [0, 2] for n = 4 subintervals.
Solution: Graphically, our goal is to find the area of the n = 4 rectangles for the interval
[a, b] = [0, 2] produced by the following graph.
In this problem,
Width of a Width of a
ba 20 1

 x 

  0.5 .
Subinterva l Rectangle
n
4
2
The endpoints of the n = 4 subintervals are calculated as follows using the formula
xi  x0  ix :
x0  a  0
x1  x0  (1)x  0  (1)(0.5)  0.5
x2  x0  (2)x  0  (2)(0.5)  1
x3  x0  (3)x  0  (3)(0.5)  1.5
x4  b  x0  (4)x  0  (4)(0.5)  2
12
In this problem, we must find the midpoints of the n = 4 subintervals using the formula
1
xi   xi 1  xi  . They are given as follows.
2
1
1
1
( x0  x1 )  (0  0.5)  (0.5)  0.25
2
2
2
1
1
1
x 2  ( x1  x 2 )  (0.5  1)  (1.5)  0.75
2
2
2
1
1
1
x3  ( x 2  x3 )  (1  1.5)  (2.5)  1.25
2
2
2
1
1
1
x 4  ( x3  x 4 )  (1.5  2)  (3.5)  1.75
2
2
2
x1 
Hence, the heights of the rectangles using the function f ( x)  x 2  1 are given as follows:
f ( x1 )  f (0.25)  (0.25) 2  1  0.0625  1  1.0625
f ( x2 )  f (0.75)  (0.75) 2  1  0.5625  1  1.5625
f ( x3 )  f (1.25)  (1.25) 2  1  1.5625  1  2.5625
f ( x4 )  f (1.75)  (1.75) 2  1  3.0625  1  4.0625
Midpoint Sum
Area of Area of Area of Area of




n  4 subinterva ls
Rect 1
Rect 2
Rect 3
Rect 4


n
 f ( xi )x
i 1
4
 f ( xi )x
i 1
  f ( x1 )x  f ( x 2 )x  f ( x3 )x  f ( x 4 )x 
 1.0625  1.5625  2.5625  4.0625(0.5)
 (9.25)(0.5 )
 4.625
13
Note: The left endpoint (lower), right endpoint (upper), and midpoint sum rules are all
special cases of what is known as Riemann sum.
Note: To increase accuracy, we need to increase the number of subintervals.
14
If we take n arbitrarily large, that is, take the limit of the left, midpoint, or right endpoint
sums as n   , the left, midpoint, and right hand sums will be equal. The common
value of the left, midpoint, and right endpoint sums is known as the definite integral.
Definition: The definite integral of f from a to b, written as
b
 f ( x) dx
a
is the limit of the left, midpoint, and right hand endpoint sums as n   . That is,
n1
n
n
b
lim  f ( xi )x  lim  f ( xi )x  lim  f ( xi )x   f ( x) dx
n i 0
n 1
n 1
 i
 i
 a
Left Hand Sum
Midpoint Sum
Right Hand Sum
Notes
1. Each sum (left, midpoint, and right) is called a Riemann sum.
2. The endpoints a and b are called the limits of integration.
3. If f ( x)  0 and continuous on [a, b], then
b

a
f ( x) dx 
Area under f
from a  x  b
4. The endpoints of the n subintervals contained within [a, b] are determined using the
ba
formula. Here, x 
.
n
Left and right endpoints: xi  x0  i x, i  1, 2, n
1
Midpoints: xi  xi 1  xi   midpoint of [ xi 1 , xi ], i  1,2, n
2
5. Evaluating a Riemann when the number of subintervals n   requires some tedious
algebra calculations. We will use Maple for this purpose. Taking a finite
number of subintervals only approximates the definite integral.
15
2
Example 4: Use the left, right, and midpoint sums to approximate
2
x
 e dx using
1
n = 5 subintervals.
Solution: On this one, we begin by finding the subintervals and corresponding functional
values for the endpoints of the n = 5 subintervals. First, note that the length of each
subinterval for the interval [a, b] = [-1, 2] is
x 
b  a 2  (1) 3

  0.6
n
5
5
Hence, the endpoints of the n = 5 subintervals using the formula xi  x0  i x and the
2
functional values using f ( x)  e  x at these endpoints are:
x0  a  1
2
f ( x0 )  f (1)  e ( 1)  e 1  0.3679 .

2
x1  x0  (1)x  1  (1)(0.6)  0.4

f ( x1 )  f (0.4)  e ( 0.4)  e .16  0.8521
x2  x0  (2)x  1  (2)(0.6)  0.2

f ( x2 )  f (0.2)  e (0.2)  e 0.04  0.9608
x3  x0  (3)x  1  (3)(0.6)  0.8

f ( x3 )  f (0.8)  e (0.8)  e 0.64  0.5273
x4  x0  (4)x  1  (4)(0.6)  1.4

f ( x4 )  f (1.4)  e (1.4)  e 1.96  0.1409
x5  b  x0  (5)x  1  (5)(0.6)  2

2
2
2
2
f ( x5 )  f (2)  e ( 2)  e 4  0.0183
Then
Left Sum for

n  5 subinterva ls

n 1
 f ( xi )x
i 0
4
 f ( xi )x
i 0
  f ( x0 )x  f ( x1 )x  f ( x 2 )x  f ( x3 )x  f ( x 4 )x 
  f ( x0 )  f ( x1 )  f ( x 2 )  f ( x3 )  f ( x 4 )x
 0.3679  0.8521  0.9608  0.5273  0.1409(0.6)
 (2.849)(0. 6)
 1.7094
16
n
Right Sum for

n  5 subinterva ls

 f ( xi )x
i 1
5
 f ( xi )x
i 1
  f ( x1 )x  f ( x 2 )x  f ( x3 )x  f ( x 4 )x  f ( x5 )x 
  f ( x1 )  f ( x 2 )  f ( x3 )  f ( x 4 )  f ( x5 )x
 0.8521  0.9608  0.5273  0.1409  0.0183(0.6)
 (2.4994)(0 .6)
 1.49964
To get the midpoint sum, we need to find the midpoints of the subintervals using the
1
formula xi  xi 1  xi  . Recall from above that the endpoints of the subintervals are
2
x0  1 , x1  0.4 , x2  0.2 , x3  0.8 , x4  1.4 , and x5  2 . The following calculation
finds the midpoints and evaluates the functional values at these midpoints.
x1 
1
1
( x0  x1 )  (1  0.4)  0.7
2
2
x2 
1
1
1
( x1  x 2 )  (0.4  0.2)  (0.2)  0.1 
2
2
2
x3 
1
1
1
( x 2  x3 )  (0.2  0.8)  (1)  0.5
2
2
2
x4 
1
1
1
( x3  x 4 )  (0.8  1.4)  (2.2)  1.1
2
2
2
x5 
1
1
1
( x 4  x5 )  (1.4  2)  (3.4)  1.7
2
2
2
Midpoint Sum for

n  5 subinterva ls


2
f ( x1 )  f (0.7)  e ( 0.7)  e 0.49  0.6126



2
f ( x 2 )  f (0.1)  e ( 0.1)  e 0.01  0.99
2
f ( x3 )  f (0.5)  e (0.5)  e 0.25  0.7788
2
f ( x 4 )  f (1.1)  e (1.1)  e 1.21  0.2982
2
f ( x5 )  f (1.7)  e (1.7)  e 2.89  0.0556
n
 f ( xi )x
i 1
5
 f ( xi )x
i 1
  f ( x1 )x  f ( x 2 )x  f ( x3 )x  f ( x 4 )x  f ( x5 )x 
  f ( x1 )  f ( x 2 )  f ( x3 )  f ( x 4 )  f ( x5 )x
 0.6126  0.99  0.7788  0.2982  0.0556(0.6)
 (2.73516)(0.6)
 1.641096
█
17
To increase accuracy, we need to make the number of subintervals n (the number of
rectangles) larger. Maple can be used to do this. If we let n   , then the resulting limit of
the left endpoint, midpoint, or right endpoint sum will give the exact value of the definite
integral. Recall that
n1
n
n
b
lim  f ( xi )x  lim  f ( xi )x  lim  f ( xi )x   f ( x) dx
n i 0
n 1
n 1
 i
 i
 a
Left Hand Sum
Midpoint Sum
Right Hand Sum
The following example will illustrate how this infinite limit can be set up using the right
hand sum.
Example 5: Set up the right hand (upper) sum limit for finding the exact value of
2
 (x
2
 1) dx for n total subintervals. Then use Maple to evaluate the limit.
0
Solution: For n subintervals, the formula for the right hand sum is given by
n
 f ( xi )x where
n
lim
x 
i 1
ba
and xi  x0  i x,
n
i  1, 2, n
2
For the definite integral  ( x 2  1) dx , x0  a  0 and b  2 . Thus for n subintervals,
0
x 
ba 20 2
2 2

 and xi  x0  i x  0  i  i
n
n
n
n n
Thus, for f ( x)  x 2  1 ,
2
2
4
2 
f ( xi )  f ( i )   i   1  2 i 2  1
n
n 
n
Hence, the right hand sum is given by
(continued on next page)
18
2
 (x
2
 1) dx  Right Hand Sum
0
n
 f ( xi )x
n
 lim
i 1
n
 4
2
 8
2 
  n 2 i 2  1 n i
n
 lim
i 1
n
 3 i3  i 

n
n 
n
 lim
i 1
This result and the resulting limit value can be generated using the following Maple
commands:
> f := x -> x^2 + 1;
> deltax := 2/n;
> x[i] := 0 + i*2/n;
> s := Sum(f(x[i])*deltax, i = 1..n);
> rsum := Limit(s, n = infinity);
> value(rsum);
█
19
Summarizing, using Maple, we can find the following information for approximating
2
2
 ( x  1) dx and
0
2
e
 x2
dx .
1
2
 (x
2
 1) dx
0
n=4
subintervals
n=10
subintervals
n=30
subintervals
n=100
subintervals
n=1000
subintervals
Exact Value
n 
2
e
 x2
Left Endpoint
3.75
4.28
Midpoint
4.625
Right Endpoint
5.75
4.66
5.08
4.534814815
4.665925926
4.801481482
4.626800000
4.666600000
4.706800000
4.662668000
4.666666000
4.670668000
14
 4. 6
3
14
 4. 6
3
14
 4. 6
3
dx
1
Left Endpoint
Midpoint
1.709378108
1.641150303
Right Endpoint
1.499639827
1.675264205
1.631946545
1.570395065
n=30
subintervals
1.645709427
1.629242706
1.610753045
n=100
subintervals
1.634088303
1.628935862
1.623601388
n=1000
subintervals
1.629429262
1.628905837
1.628380572
Value
n 
1.628905524
1.628905524
1.628905524
n=5
subintervals
n=10
subintervals
20
Properties of the Definite Integral
If f and g are integrable functions on [a, b] , then
a
1.
 f ( x) dx  0
a
a
2.
 f ( x) dx  0
a
3.
b
a
a
b
 f ( x) dx   f ( x) dx
b
b
a
a
4.  k f ( x) dx  k  f ( x) dx , where k is a number.
b
b
b
a
a
a
5.  ( f ( x)  g(x))dx   f ( x) dx   g ( x) dx
5
Example 6: Given

1
1
a.
 f ( x) dx
5
5
f ( x) dx  6 and
 g ( x) dx  2 , find
1
5
b.  (3 f ( x)  g ( x)) dx
1
Solution:
█
21
Additive Interval Property
Fact: If f and g are integrable functions on [a, b] and a  c  b , then
b

a
Example 7: Given
c
b
a
c
f ( x) dx   f ( x) dx   f ( x) dx
5
4
5
1
1
4
 f ( x) dx  6 and  f ( x) dx  4.5 , find  f ( x) dx .
Solution: By the additive interval property of integrals, we can say that
4
5
5
1
4
1
5
4
1
1
 f ( x) dx   f ( x) dx   f ( x) dx
Hence,
5

4
f ( x) dx   f ( x) dx   f ( x) dx
 6  4 .5
 1 .5
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