Riemann Sums
Lecture Notes
page 1
Sample Problems
1
on the interval [1; 4].
x
a) Compute the left Riemann sum for f on this interval using a regular partition with n = 6
subintervals.
b) Compute the right Riemann sum for f on this interval using a regular partition with n = 6
subintervals.
1. Consider the function f (x) =
2. Consider the function f (x) = x2 on the interval [0; 6] :
a) Compute the left Riemann sum for f on this interval with n = 6 subintervals.
b) Compute the right Riemann sum for f on this interval with n = 6 subintervals.
c) Compute the left Riemann sum for f on this interval with n = 12 subintervals.
d) Compute the right Riemann sum for f on this interval with n = 12 subintervals.
e) Compute the left Riemann sum for f on this interval with n = 100 subintervals.
f) Compute the right Riemann sum for f on this interval with n = 100 subintervals.
g) Compute the left Riemann sum for f on this interval with n subintervals.
h) Compute the limit of the left Riemann sum for f on this interval with n intervals, as n approaches
in…nity.
i) Compute the right Riemann sum for f on this interval with n subintervals.
j) Compute the limit of the right Riemann sum for f on this interval with n intervals, as n approaches
in…nity.
Practice Problems
1. Consider the function f (x) =
a)
b)
c)
d)
p
x on the interval [0; 4] :
Compute the left Riemann sum for f on this interval with n = 4 subintervals.
Compute the right Riemann sum for f on this interval with n = 4 subintervals.
Compute the left Riemann sum for f on this interval with n = 10 subintervals.
Compute the right Riemann sum for f on this interval with n = 10 subintervals.
2. Consider the function f (x) = ln (x + 1) on the interval [0; 10] :
a) Compute the left Riemann sum for f on this interval with n = 10 subintervals.
b) Compute the right Riemann sum for f on this interval with n = 10 subintervals.
3. Consider the function f (x) = x3 on the interval [0; 1] :
a)
b)
c)
d)
e)
f)
Compute the left Riemann sum for f on this interval with n = 4 subintervals.
Compute the right Riemann sum for f on this interval with n = 4 subintervals.
Compute the left Riemann sum for f on this interval with n = 10 subintervals.
Compute the right Riemann sum for f on this interval with n = 10 subintervals.
Compute the left Riemann sum for f on this interval with n = 100 subintervals.
Compute the right Riemann sum for f on this interval with n = 100 subintervals.
c copyright Hidegkuti, Powell, 2012
Last revised: March 2, 2013
Riemann Sums
Lecture Notes
page 2
Answers - Sample Problems
1. a)
223
140
b)
1: 592 857
2. a) 55
b) 91
c)
341
280
1: 217 857
253
= 63: 25
4
f)
182 709
= 73: 084
2500
i)
72n2 + 108n + 36
36 (2n2 + 3n + 1)
=
n2
n2
g)
36 (n
d)
1) (2n
n2
325
= 81: 25
4
1)
72n2
=
e)
177 309
= 70: 924
2500
108n + 36
n2
h) 72
j) 72
Answers - Practice Problems
1. a) 1 +
p
2+
p
3
4: 146 264 37
b) 3 +
p
2+
p
p
p
p
p
p
p
2 10
c)
6+ 2+ 3+ 5+ 6+ 7+ 8
25
p
3
6: 146 264 37
4: 884 075
p
p
p
p
p
p
p
p
2 10
d)
6 + 2 + 3 + 5 + 6 + 7 + 8 + 10
25
2. a) ln (10!) = ln 3628 800
3. a)
9
= 0:140 625
64
4. e)
9801
= 0:245 025
40 000
15: 104 412 57
b)
f)
c copyright Hidegkuti, Powell, 2012
5: 684 075
b) ln (11!) = ln 39 916 800
25
= 0:390 625
64
c)
81
= 0:202 5
400
17: 502 307 85
d)
121
= 0:302 5
400
10 201
= 0:255 025
40 000
Last revised: March 2, 2013
Riemann Sums
Lecture Notes
page 3
Sample Problems - Solutions
1
on the interval [1; 4].
x
a) Compute the left Riemann sum for f on this interval using a regular partition with n = 6
subintervals.
1. Consider the function f (x) =
Solution: The interval [1; 4] is 3 units long. The regular partition will contain intervals of length
3
1
1
1
1
= . The partition consists of 1; 1 ; 2; 2 ; 3; 3 ; 4 Notice that these are seven numbers.
6
2
2
2
2
We usually start labeling with zero. In this case, these seven numbers are fx0 ; x1 ; x2 ; x3 ; x4 ; x5 ; x6 g.
On each interval, we approximate the area under the graph by a rectangle as tall as the function value
of the left endpoint of the interval. For example, on the …rst interval, we approximate the area under
1
the graph using a rectangle with height = 1. On the second interval, the height of the rectangle is
1
1
2
= .
1
3
1
2
1
1
1
and height 1. The area is A1 =
1= .
2
2
2
1
2
1 2
1
1
and height
= . The area is A2 =
= .
The second rectangle has width
3
2
3
2 3
3
2
1
Let us notice that all rectangles have the same width of . This is an advantage of a regular partition.
2
1
1 1
1
The third rectangle has height . Its area is A3 =
= .
2
2 2
4
1
2
1 2
1
The fourth rectangle has height
= . Its area is A4 =
=
5
5
2 5
5
2
1
1 1
1
The …fth rectangle has height . Its area is A4 =
= .
3
2 3
6
1
2
1 2
1
The sixth rectangle has height
= . Its area is A4 =
= . In short, the left-hand
7
7
2 7
7
2
The …rst rectangle has width
c copyright Hidegkuti, Powell, 2012
Last revised: March 2, 2013
Riemann Sums
Lecture Notes
page 4
approximation is
1 1 1 1
1 1 1 1
1 1 1 1
1
+
+
+
+
+
=
2 1 2 1:5 2 2 2 2:5 2 3 2 3:5
2
2 1 2 1 2
223
1
=
1+ + + + +
=
1: 592 857
2
3 2 5 3 7
140
Lf;n=6 =
1
1
1
1
1
1
+
+ +
+ +
1 1:5 2 2:5 3 3:5
We can clearly see from the picture that this approximation is an overestimation of the area.
b) Compute the right Riemann sum for f on this interval using a regular partition with n = 6
subintervals.
1 1
1 1 1 1
1 1 1 1
1 1
1
+
+
+
+
+
=
2 1:5 2 2 2 2:5 2 3 2 3:5 2 4
2
341
1 2 1 2 1 2 1
=
+ + + + +
=
1: 217 857
2 3 2 5 3 7 4
280
R =
1
1
1
1
1
1
+ +
+ +
+
1:5 2 2:5 3 3:5 4
We can clearly see from the picture that this approximation is an underestimation of the area. Thus
we now know that the area under the graph is between those two values:
1: 217 857 < A < 1: 592 857
Note: We sometimes use summation notation when writing such expressions.
notation, these Riemann sums are
L =
5
X
1
k=0
and R =
2
6
X
1
k=1
2
1
2
1X
=
2 k=0
1
2
6
X
5
1
1+k
1
1+k
c copyright Hidegkuti, Powell, 2012
=
k=1
1+k
X 1
1X 1
1X 2
=
=
=
2 k=0 2 + k
2 k=0 2 + k
2+k
k=0
2
5
1
1
2
Using summation
5
5
1
2+k
Last revised: March 2, 2013
Lecture Notes
Riemann Sums
page 5
2. For this problem, we will need the following theorem: for all natural numbers n;
12 + 22 + 32 + ::: + n2 =
n (n + 1) (2n + 1)
6
Consider the function f (x) = x2 on the interval [0; 6] :
a) Compute the left Riemann sum for f on this interval with n = 6 subintervals.
Solution: Each subinterval is of length 1; and so the partition is f0; 1; 2; 3; 4; 5; 6g.
The left-hand sum is
Lf;n=6 = 1 02 + 1 12 + 1 22 + 1 32 + 1 42 + 1 52 = 1 + 4 + 9 + 16 + 25 = 55
Using summation notation,
Lf;n=6 =
5
X
2
1 k =
k=0
5
X
k 2 = 55
k=0
We can see on the picture that this Riemann sum underestimates the area.
b) Compute the right Riemann sum for f on this interval with n = 6 subintervals.
Rf;n=6 = 1 12 + 1 22 + 1 32 + 1 42 + 1 52 + 1 62 = 1 + 4 + 9 + 16 + 25 + 36 = 91
Using summation notation,
Rf;n=6 =
6
X
k=1
2
1 k =
6
X
k 2 = 91
k=1
We can see on the picture that this Riemann sum underestimates the area. Thus, we have that
55 < A < 91
c copyright Hidegkuti, Powell, 2012
Last revised: March 2, 2013
Riemann Sums
Lecture Notes
page 6
c) Compute the left Riemann sum for f on this interval with n = 12 subintervals.
6
1
Solution: Each subinterval will have length
= . The partition is
12
2
f0; 0:5; 1; 1:5 ; 2 ; 2:5; 3; 3:5; 4; 4:5; 5; 5:5; 6g. The left Riemann sum is
1 2 1
1 2 1
1 2 1
1 2
1 2 1
0 +
0:52 +
1 +
1:52 +
2 +
2:52 +
3 +
3:52 +
4
2
2
2
2
2
2
2
2
2
1 2 1
1
+ 4:52 +
5 +
5:52
2
2
2
Although this looks like a lot of computation, it can be made quite simple using a bit of algebra and
1
the theorem stated above. We …rst factor out
and write the rest as fractions, with a common
2
denominator of 2.
Lf;n=12 =
Lf;n=12
1
1 2
=
0 + 0:52 + 12 + 1:52 + ::: + 5:52 =
2
2
1
2
112
1 12 22 32 42
+
+
+
+ ::: +
=
2 4
4
4
4
4
1 1 2
=
1 + 22 + 32 + ::: + 112
2 4
1 11 12 23
253
=
=
= 63: 25
8
6
4
The same computation, using summation notation:
Lf;n=12 =
11
X
1
k=0
2
1
k
2
2
2
+
2
2
2
+
factor out
3
2
2
+
4
2
2
+ ::: +
11
2
1
4
use theorem with n = 11
1 X k2
1 X 2 1 11 12 23
253
=
=
k =
=
= 63: 25
2 k=0 4
8 k=0
8
6
4
11
11
d) Compute the right Riemann sum for f on this interval with n = 12 subintervals.
Solution: The di¤erence between the left and right Riemann sums is just the …rst and the last
rectangle.
!
2
2
2
2
2
1
1
1
2
3
4
12
Rf;n=12 =
0:52 + 12 + ::: + 5:52 + 62 =
+
+
+
+ ::: +
2
2
2
2
2
2
2
1 12 22 32 42
122
+
+
+
+ ::: +
2 4
4
4
4
4
1 1 2
=
1 + 22 + 32 + ::: + 122
2 4
1 12 13 25
325
=
=
= 81: 25
8
6
4
factor out
=
1
4
use theorem with n = 12
Using summation notation,
Rf;n=12 =
12
X
1
k=1
2
1
k
2
2
1 X k2
1 X 2 1 12 13 25
325
=
=
k =
=
= 81: 25
2 k=1 4
8 k=1
8
6
4
12
12
Because this function is increasing, all left sums underestimate the area and all right sums overestimate
the area under the graph. Thus
63: 25 < A < 81: 25
c copyright Hidegkuti, Powell, 2012
Last revised: March 2, 2013
2
!
Riemann Sums
Lecture Notes
page 7
e) Compute the left Riemann sum for f on this interval with n = 100 subintervals.
6
Solution: Each subinterval is
units long. The partition is
100
6
12
6k
600
x0 = 0; x1 =
; x2 =
; ::::; xk =
; :::; x100 =
=6
100
100
100
100
The left Riemann sum is
6
6
=
02 +
100
100
Lf;n=100
6
=
100
0 +
6
100
6
100
=
2
6 1
100
2
6
+
100
2
12
100
2
6 2
100
+
+ :::: +
6
+ :::: +
100
!
2
6 99
100
2
(99 199) =
3
6
100
02 + 12 + 22 + :::: + 992 =
2
6
100
=
2
6
100
6 99
100
2
99 100 199
6
177 309
= 70: 924
2500
Using summation notation,
Lf;n=100
99
X
6
=
100
k=0
=
6 X
=
100 k=0
99
2
6
k
100
2
6
100
(99 199) =
2
6
100
6
100
2
k =
177 309
= 70: 924
2500
99
3X
2
k =
k=0
6
100
3
99 100 199
6
f) Compute the right Riemann sum for f on this interval with n = 100 subintervals.
The right Riemann sum is
6
100
Rf;n=100 =
6
=
100
6
100
=
6
100
=
6
100
2
+
2
6 1
100
6
100
6
100
+
2
12
100
6 2
100
+ :::: +
2
+ :::: +
6
100
6 100
100
2
02 + 12 + 22 + :::: + 1002 =
2
(101 201) =
2
6 100
100
!
6
100
3
2
100 101 201
6
182 709
= 73: 083 6
2500
Using summation notation,
Rf;n=100
100
X
6
=
100
k=1
=
6
100
6
k
100
2
2
6 X
=
100 k=1
(100 201) =
100
6
100
2
2
k =
6
100
182 709
= 73: 084
2500
100
3X
k=1
k2 =
6
100
3
100 101 201
6
Thus
70: 924 < A < 73: 084
c copyright Hidegkuti, Powell, 2012
Last revised: March 2, 2013
Riemann Sums
Lecture Notes
page 8
g) Compute the left Riemann sum for f on this interval with n subintervals.
6
Solution: Each subinterval is units long. The numbers in the partition are
n
6
6
6
6
6
x0 = 0; x1 = ; x2 = 2
; x3 = 3
; x4 = 4
; ::::; xn = n
=6
n
n
n
n
n
The left Riemann sum is
Lf;n
6 2 6
0 +
=
n
n
6
=
n
0 +
12
6
n
6
n
=
2
6
n
2
6
=
n
2
6
n
6
n
6
+
n
6
+ 2
n
2
6
n
2
2
+ 22
2
6
2
n
6
+
n
6
+ 3
n
6
n
+ 32
2
6
3
n
2
3
(n
2
3
6
n
1)
(n
2
1) ((n
2
6
(n 1) n (2n 1)
6
=
=
(n 1) (2n
n
6
n
72n2 108n + 36
36 (2n2 3n + 1)
=
=
n2
n2
1) =
2
6
1)
n
!
6
n
2
+ :::: + (n
1)2 =
2
6
1)
n
+ :::: + (n
2
12 + 22 + 32 + :::: + (n
6
+ :::: +
n
!
1) + 1) (2 (n
6
36 (n
1) (2n
n2
1) + 1)
1)
Using summation notation,
Lf;n
n 1
X
6
=
n
k=0
=
6
n
3
6
k
n
(n
2
6X
=
n k=0
n 1
1) n (2n
6
2
6
n
1)
2
k =
(n
k =
1) (2n
3
6
n
2
k=0
2
6
n
=
n 1
3X
6
n
1) =
(n
36 (n
1) ((n
1) (2n
n2
1) + 1) (2 (n
6
1)
=
72n2
1) + 1)
108n + 36
n2
h) Compute the limit of the left Riemann sum for f on this interval with n intervals, as n approaches
in…nity.
Solution:
lim
72n2
108n + 36
= lim 72
n!1
n2
n!1
108 36
+ 2
n
n
= 72
i) Compute the right Riemann sum for f on this interval with n subintervals.
Solution: The right Riemann sum is
Rf;n =
6
n
6
=
n
=
6
n
c copyright Hidegkuti, Powell, 2012
2
6
n
6
n
2
6
+ 2
n
2
6
n
+
6
n
12
2
6
n
2
2
+ 22
+
6
n
3
6
+ 3
n
6
n
2
+ 32
6
n
2
2
+ :::: +
6
n
n
6
+ :::: + n
n
6
n
2
+ :::: + n2
2
2
6
n
!
6
n
2
!
Last revised: March 2, 2013
Riemann Sums
Lecture Notes
Rf;n =
=
6
n
6
n
6
n
2
6
n
12 + 22 + 32 + :::: + n2 =
2
(n + 1) (2n + 1) =
3
page 9
n (n + 1) (2n + 1)
6
36 (n + 1) (2n + 1)
36 (2n2 + 3n + 1)
72n2 + 108n + 36
=
=
n2
n2
n2
Using summation notation,
Rf;n
n
X
6
=
n
k=1
=
6
n
6
k
n
2
2
6X
=
n k=1
n
(n + 1) (2n + 1) =
6
n
2
2
k =
6
n
3
n
X
2
k =
k=1
6
n
3
n (n + 1) (2n + 1)
6
36 (n + 1) (2n + 1)
72n2 + 108n + 36
=
n2
n2
j) Compute the limit of the right Riemann sum for f on this interval with n intervals, as n approaches
in…nity.
Solution:
108 36
72n2 + 108n + 36
= lim 72 +
+ 2
2
n!1
n!1
n
n
n
lim
= 72
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c copyright Hidegkuti, Powell, 2012
Last revised: March 2, 2013