CalculusVolume2-SASG-01-02
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OpenStax Calculus Volume 2
Student Answer and Solution Guide
Chapter 1
Integration
1.2. The Definite Integral
Section Exercises
In the following exercises, express the limits as integrals.
61.
over Answer: 63.
over Answer: In the following exercises, given L
n
or R
n
as indicated, express their limits as as definite integrals, identifying the correct intervals
.
65.
Answer: 67.
Answer: 69.
Answer:
OpenStax Calculus Volume 2
Student Answer and Solution Guide
In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x
-axis.
71.
Answer: 73.
Answer: 75.
Answer:
OpenStax Calculus Volume 2
Student Answer and Solution Guide
In the following exercises, evaluate the integral using area formulas.
77.
Answer: The integral is the area of the triangle, 79.
Answer: The integral is the area of the triangle, 9.
81.
Answer: The integral is the area .
83.
Answer: The integral is the area of the “big” triangle less the “missing” triangle, In the following exercises, use averages of values at the left (
L
) and right (
R
) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.
85.
over Answer: , , 87.
over Answer: , , Suppose that and and and . In the following exercises, compute the integrals.
89.
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OpenStax Calculus Volume 2
Student Answer and Solution Guide
Answer: 91.
Answer: 93.
Answer: In the following exercise, use the identity to compute the integrals.
95.
Answer: The integrand is odd; the integral is zero.
In the following exercise, find the net signed area between f
(
x
) and the x
-axis.
97.
(
Hint:
Look at the graph of f
.)
Answer: The integrand is antisymmetric with respect to . The integral is zero.
In the following exercises, given that , and , compute the integrals
.
99.
Answer: 101.
Answer:
OpenStax Calculus Volume 2
Student Answer and Solution Guide
103.
Answer:
OpenStax Calculus Volume 2
Student Answer and Solution Guide
In the following exercises, use the comparison theorem
105.
Show that .
Answer: The integrand is negative over .
107.
Show that .
Answer: over , so over .
109.
Show that .
Answer: . Multiply by the length of the interval to get the inequality.
In the following exercises, find the average value f
ave
of f
between a
and b
, and find a point c
, where .
111.
, , Answer: ; 113.
, , Answer: when 115.
, , Answer: ; In the following exercises, approximate the average value using Riemann sums L
100
and R
100
. How does your answer compare with the exact given answer?
117.
[T]
over the interval ; the exact solution is .
Answer: , ; the exact average is between these values.
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OpenStax Calculus Volume 2
Student Answer and Solution Guide
119.
[T]
over the interval ; the exact solution is .
Answer: , In the following exercises, compute the average value using the left Riemann sums L
N
for
. How does the accuracy compare with the given exact value?
121.
[T]
over the interval ; the exact solution is .
Answer: , , . The exact answer , so L
100
is not accurate.
123.
[T]
over the interval ; the exact solution is .
Answer: , , . The exact answer
, so L
100
is not accurate to first decimal.
125.
Suppose that and . Show that .
Answer: Use . Then, .
127.
Show that the average value of over is equal to . Without further calculation, determine whether the average value of over is also equal to Answer: so divide by the length 2
π
of the interval. has period π
, so yes, it
is true.
129.
Suppose that parabola opens downward (
) and has a vertex of
. For which interval is as large as possible?
OpenStax Calculus Volume 2
Student Answer and Solution Guide
Answer: The integral is maximized when one uses the largest interval on which p
is nonnegative.
Thus, and .
131.
Suppose f
and g
are continuous functions such that for every subinterval of . Explain why for all values of x
.
Answer: If for some , then since is continuous, there is an interval containing t
0
such that over the interval , and then
over this interval.
133.
Suppose that can be partitioned. taking such that the average value of f
over each subinterval is equal to 1 for each Explain why the average value of f
over is also equal to 1.
Answer: The integral of f
over an interval is the same as the integral of the average of f
over that interval. Thus,
Dividing through by gives the desired identity.
135.
Suppose that for each i
such that one has . Show that
.
Answer: 137.
[T] Compute the left and right Riemann sums, L
10
and R
10
, and their average for
over . Given that , to how many decimal places is
accurate?
OpenStax Calculus Volume 2
Student Answer and Solution Guide
Answer: , , so the estimate is accurate to two decimal places.
139.
Estimate using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value ?
Answer: The average is , which is equal to the integral in this case.
141.
From the graph of shown:
a.
Explain why .
b.
Explain why, in general, for any value of a
.
Answer: a. The graph is antisymmetric with respect to over so the average value is zero. b. For any value of a
, the graph between is a shift of the graph over , so the
net areas above and below the axis do not change and the average remains zero.
143.
If f
is 1-periodic and , is it necessarily true that for all A
?
Answer: Yes, the integral over any interval of length 1 is the same.
This file is copyright 2016, Rice University. All Rights Reserved.
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