Math 131 Fall 2012
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CHAPTER 5 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
Problem 1. A skydiver jumps from an airplane and her velocity is then v(t ) ft/s after t seconds
of freefall in a tucked position, so that she continually speeds up.
a. Use an integral to express the distance the skydiver has fallen after 30 seconds of freefall.
(Assume that distance is measured from the jump altitude so that the velocity v(t ) is
positive.)
b. Given the following data, find both an underestimate and an overestimate for the distance
the skydiver has fallen after 30 seconds of freefall.
time
t
(seconds)
velocity v(t ) (ft/s)
0
0
10
160
20
192
30
198
c. If speed data were available every 2 seconds, what would be the difference between the
underestimate and overestimate based on 15 subdivisions of the time interval?
d. How many subintervals would be needed to have the underestimate and overestimate
accurate to within 10 ft?
Solution.
a. We do not have an analytic expression for v(t ) , and so it is not possible to find an
expression for the antiderivative, but we can still represent the distance fallen the
skydiver has fallen using a definite integral:
distance =
∫
30
0
v(t ) dt
b. Using the left and the right Riemann sums we can find the under- and over-estimate for
the distance traveled, and subtract these from the initial height.
distance > v(0) ⋅10 + v(10) ⋅10 + v (20) ⋅10
= 0 + 1600 + 1920 = 3520 feet
and
distance < v (10) ⋅10 + v(20) ⋅10 + v(30) ⋅10
= 1600 + 1920 + 1980 = 5500 feet
c. All of the same terms would appear in both the left and right Riemann sums except for
the v(0) ⋅ 2 = 0 term in underestimate and the v(30) ⋅ 2 = 396 term in the overestimate.
Thus the difference between the overestimate and underestimate would be 396 feet.
d. All of the same terms would appear in both the left and right Riemann sums except for
the v (0) ⋅ ∆t = 0 term in underestimate and the v(30) ⋅ ∆t = 198∆t term in the
overestimate. Thus the error is smaller than 198∆t . Setting this to the desired bound of
30
10 feet, we get ∆t ≤ 10 /198 = 0.05 . This would require n ≥
= 594 subintervals.
0.05
Math 131 Fall 2012
CHAPTER 5 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
2
Problem 2. True or False?
a. If f ( x) > 0 , then
∫
b
a
f ( x)dx measures the area under the graph of f between x = a and
x=b.
b. If F ′( x) = f ( x) , then ∫ f ( x) dx measures the total change in F between x = a and
b
a
x=b.
Solution.
a. True.
b. True.
Problem 3. Suppose C (t ) is the power, in kJ/h, produced by a solar panel t hours after sunrise
on a typical summer day. Give practical interpretations of
∫
2
0
Solution.
∫
2
C (t ) dt = 270 and
1 12
C (t ) dt = 288 .
12 ∫0
C (t ) dt = 270 is the energy produced by the solar panel in the first two hours after
1 12
C (t ) dt = 288 is the average power produced by the solar panel
12 ∫0
during the first 12 hours after sunrise measured in kJ/h.
0
sunrise measured in kJ.
Problem 4. The table below gives the expected growth rate, g (t ) , in ounces per week, of the
weight of a baby in its first 54 weeks of life (which is slightly more than a year).1 Assume for
this problem that g (t ) is a decreasing function.
week t
growth rate g (t )
0
6
9
6
18
4.5
27
3
36
3
45
3
54
2
a. Using six subdivisions, find an overestimate and underestimate for the total weight
gained by a baby over its first 54 weeks of life.
b. How frequently over the 54 week period would you need the data for g (t ) to be
measured to find overestimates and underestimates for the total weight gain over this time
period that differ by 0.5 lb (8 oz)?
1
Riordan J. Breastfeeding and Human Lactation, 3rd ed. Boston: Jones and Bartlett, 2005, p.103, 512-513.
Math 131 Fall 2012
CHAPTER 5 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
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Solution.
a. The growth rate is a decreasing function, so a left-hand sum will be an overestimate and a
right-hand sum an underestimate. The left-hand sum is
∫
54
∫
54
0
g (t ) dt ≈ (6 + 6 + 4.5 + 3 + 3 + 3)(9) = 229.5 oz.
and the right-hand sum is
0
g (t ) dt ≈ (6 + 4.5 + 3 + 3 + 3 + 2)(9) = 193.5 oz.
That is, we expect the weight gain to be between 12 and 14 lb!
b. We know that the difference between the over- and underestimates is
over − under = g (54) − g (0) ∆t .
Thus we need ∆t ≤ 8 / (6 − 2) = 2 weeks. So we would need data for g (t ) every two
weeks.
Problem 5.
a. Using 4 equal subdivisions, find a Riemann sum which is an underestimate for
∫
4
2
ln( x) dx .
Sketch a graphical representation of your Riemann sum on the axes below, and write
“LHS” or “RHS” next to your figure to indicate whether you are using a left-hand sum or
a right-hand sum. Write out the terms of the Riemann sum using exact values (no
calculator approximations). There is no need to simplify the sum.
b. Show that F ( x) = x ln( x) − x + C is an antiderivative for ln( x) , where C is any constant.
In other words, show F ′( x ) = ln( x) .
c. Using Part b, find the exact value of the integral
∫
4
2
ln( x) dx .
Solution.
a.
∫
4
2
ln( x) dx ≈ 0.5ln 2 + 0.5ln 2.5 + 0.5ln 3 + 0.5ln 3.5
Math 131 Fall 2012
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CHAPTER 5 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
b. We need to check that f ′( x) = ln x . Using the product rule, we have
1
f ′( x) = x ⋅ + 1 ⋅ ln x − 1 + 0 = 1 + ln x − 1 = ln x .
x
∫
c. By the Fundamental Theorem and Part b,
4
2
ln( x) dx = [ 4 ln 4 − 4 + C ] − [ 2ln 2 − 2 + C ]
= 4 ln 4 − 2 ln 2 − 2
= ln 43 − 2
Problem 6. The rate at which the world’s oil is being consumed is continuously increasing.
Suppose the rate of oil consumption (in billions of barrels per year) is given by the function r (t ) ,
where t is measured in years and t = 0 is the start of 2004.
a. Write a definite integral which represents the total quantity of oil used between the start
of 2004 and the start of 2014.
b. Suppose r (t ) = 32e0.05t . Using a left-hand sum and a right hand sum with five
subdivisions find an approximate value for the total quantity of oil used between the start
of 2004 and the start of 2014.
c. Find an error bound for your estimates in part (b).
d. Give an algebraic expression for the error bound for any number of rectangles.
e. How many subintervals of the 10-year time span would you need in order to ensure an
error less than 0.05 billion barrels.
f. Interpret each of the five terms in the left-hand sum from part (b) in terms of oil
consumption.
Solution.
a.
∫
10
0
r (t ) dt
b. Consider the following table of values:
time: t (years)
rate: r (billion
barrels per year)
∫
10
0
∫
10
0
0
32
2
35.365
4
39.085
6
43.195
8
47.738
10
52.759
r (t ) dt ≈ 32 ( 2 ) + 35.365 ( 2 ) + 39.085 ( 2 ) + 43.195 ( 2 ) + 47.738 ( 2 ) = 394.766 = LHS
r (t ) dt ≈ 35.365 ( 2 ) + 39.085 ( 2 ) + 43.195 ( 2 ) + 47.738 ( 2 ) + 52.759 ( 2 ) = 436.356 = RHS
c. Error bound = 436.356 − 394.766 = 41.59
d. Error bound = ( 52.759 − 32 ) ∆t
Math 131 Fall 2012
e. ∆t =
CHAPTER 5 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
5
10 − 0
10
so we need ( 52.759 − 32 ) < 0.05 or n > 4151.8
n
n
f. Each term in the sum is the amount of oil consumed in a two year span assuming the rate
of oil consumption at the beginning of each two year span does not change for the next
two years.
Problem 7. The NASA Q36 Robotic Lunar Rover can travel up to 3 hours on a single charge
and after t hours of traveling, its speed is v(t ) miles per hour given by the function
v (t ) = sin 9 − t 2 .
a. Write an integral that expresses the distance traveled by the Q36 during its first two hours
of operation.
b. How many terms would be required in a Riemann sum for your integral in part a to be
accurate to within one foot, i.e., 1/5280 miles?
Solution.
a.
∫
2
0
v(t ) dt
b. Since v is increasing on [ 0, 2] , the left-hand sums will give underestimates, right-hand
sums will give overestimates, and the error will be bound by the difference between the
two. Since the two sums will contain all of the same terms except the largest term from
the overestimate v (2) ∆t and the smallest term from the underestimate v (0) ∆t , the error
bound will actually be ( v(2) − v(0) ) ∆t = 0.64563∆t . Setting this smaller than the desired
2
1
= 0.64563∆t ≤
or n ≥ 6817.85 . Thus
n
5280
we would need to use at least 6818 terms to ensure our left-hand sum is accurate to
within one foot.
1/5280 miles, we see that we need 0.64563 ⋅
Problem 8. Use your calculator to draw a very accurate graph of w( s) = s s on the interval ( 0,3] .
∫
a. Explain how to use your graph (not a calculator at this point) to estimate the value of
3
1
s s ds . Try to get close to the correct answer, but your explanation is the most important
part of your answer.
∫
b. Now use your calculator to evaluate a Riemann sum that is an underestimate but within
0.5 of the actual value of
3
1
s s ds . Explain how you know your sum is an underestimate
and that it is sufficiently accurate.
Math 131 Fall 2012
CHAPTER 5 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
6
Solution.
a. The graph of w( s ) is shown to below.
Each rectangle in the grid is one unit wide and two units high, so has area 2. There are
many ways to try to count these rectangles, and the graph to the right represents one such
attempt. Here we try to combine partial rectangles that appear to add to an whole
rectangle of area 2. These pairs are numbered with a and b indicating the two parts. For
example, the region labeled 6a appears to be slightly larger than half a rectangle. To
make up for this, we look for a region such as 6b that appears to be equal in size to the
remaining part of that rectangle. When we reach region 7, we are out of area to consider,
and this appears to be slightly smaller than a full rectangle of area 2. Thus we might
estimate that we have approximately 6.8 rectangles of area 2, or a total area of around
13.6. Of course, our visual inspection could be off by a fair amount, so we expect this to
only be a ballpark estimate.
b. w is increasing on the interval [1, 3] so left-hand sums will give underestimates, righthand sums will give overestimates, and the error will be bound by the difference between
the two. Since the two sums will contain all of the same terms except the largest term
from the overestimate w(3)∆s and the smallest term from the underestimate w(1)∆s , the
error bound will actually be ( w(3) − w(1) ) ∆s = 26∆s . Setting this smaller than the desired
2 ⋅ 26
= 26∆s ≤ 0.5 or n ≥ 104 . Choosing n = 104 for
n
example, we can enter the left-hand sum into the calculator as*
bound, we see that we need
sum(seq(x^x*2/104,x,1,3-1/104,2/104))=13.477
the right-hand sum as
sum(seq(x^x*2/104,x,1+2/104,3+1/104,2/104))=13.977
∫
and as we predicted, these are within 0.5 of each other. Thus we can say that 13.477 is an
underestimate for the integral
*
3
1
s s ds accurate to within 0.5.
Note that since ∆s = 2 104 has a repeating decimal expression, we have potential for a
rounding error to cause the sum(seq(…)) command to miss the last term. This will in
fact happen on the TI-83, for example. This is why we increased the ending s-value by
half of ∆s , using 3-1/104 and 3+1/104 instead of 3-2/104 and 3.
Math 131 Fall 2012
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CHAPTER 5 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
Problem 9. Find the average value of f ( x) =
1
between x = 1 and x = 10 .
x
1 10 1
1
dx =
( ln10 − ln1) = 0.2558
∫
1
10 − 1 x
10 − 1
Solution.
Problem 10. Let f and g be continuous functions for all real numbers.
∫ ( 2 f ( x) + 3) dx = 17 , find ∫
a. If
5
5
2
2
b. If f is odd and
c. If f is even and
∫
3
f ( x)dx .
f ( x )dx = 30 , find
−2
∫
3
2
f ( x )dx .
∫ ( f ( x) − 3) dx = 8 , find ∫
2
2
−2
0
f ( x)dx .
d. If the average value of f on the interval 2 ≤ x ≤ 5 is 4, find
∫ ( 3 f ( x) + 2 ) dx .
5
2
Solution.
a.
∫ ( 2 f ( x) + 3) dx = ∫ 2 f ( x)dx + ∫ 3dx = 2∫ f ( x)dx + 9 = 17 , which implies
2 ∫ f ( x)dx = 8 . This provides us with the solution ∫ f ( x)dx = 4
5
5
5
5
2
2
2
2
5
5
2
2
b. We have
∫
3
−2
implies that
c.
d.
f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx = 30 . Since f is odd,
∫
3
2
2
3
−2
2
2
2
−2
2
−2
2
2
−2
0
.
∫
5
f ( x)dx = 12 . Therefore,
∫ ( 3 f ( x) + 2 ) dx = 3∫ f ( x)dx + ∫ 2dx
∫ ( 3 f ( x) + 2 ) dx = 3 ⋅12 + 3 ⋅ 2 = 42 .
2
5
5
5
2
2
2
2
2
−2
f ( x)dx = 0 , which
f ( x)dx − ∫ 3dx = 2∫ f ( x)dx − 12 = 8 , which implies
1 5
f ( x )dx = 4 , so
5 − 2 ∫2
5
∫
f ( x)dx = 30 .
∫ ( f ( x) − 3) dx = ∫
∫ f ( x)dx = 10
0
.
by linearity of integrals. So,