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Math 116 — Practice for Exam 1

Generated January 31, 2022 Name: SOLUTIONS

Instructor: Section Number:

1. This exam has 9 questions. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck.

2. Do not separate the pages of the exam. If any pages do become separated, write your name on them and point them out to your instructor when you hand in the exam.

3. Please read the instructions for each individual exercise carefully. One of the skills being tested on this exam is your ability to interpret questions, so instructors will not answer questions about exam problems during the exam.

4. Show an appropriate amount of work (including appropriate explanation) for each exercise so that the graders can see not only the answer but also how you obtained it. Include units in your answers where appropriate.

5. You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad).

However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a 3′′× 5′′note card.

6. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of the graph, and to write out the entries of the table that you use.

7. You must use the methods learned in this course to solve all problems.

Semester Exam Problem Name Points Score

Winter 2011 1 7 paint truck 11

Fall 2011 1 7 AC current 13

Winter 2018 1 1 14

Fall 2017 1 8 10

Winter 2016 1 1 catapult 16

Winter 2018 1 6 7

Winter 2010 1 10 15

Winter 2015 1 2 dog bowl 13

Fall 2018 1 8 pond 12

Total 111

Recommended time (based on points): 100 minutes

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Math 116 / Exam 1 (February 2011) page 9

7. [11 points]

A truck carrying a large tank of paint leaves a garage at 9AM. The tank starts to leak in such a way that x miles from the garage, the density of paint on the road is e−x2/5000 gallons per mile. At 10AM, a cleaning crew leaves from the same garage and follows the path of the truck, scrubbing the paint from the road as it travels until it catches up to the leaking truck. At t hours after 10AM, the leaking truck is 50 ln(t + 2) miles from the garage, and the cleanup crew is 35t miles from the garage. You may use your calculator to evaluate any definite integrals for this problem.

a. [4 points] Calculate the total amount of paint that has leaked from the truck by 11AM.

Solution: The truck at 11 AM is at 50 ln 3 = 54.9306 miles from the garage. Total amount of paint leaked from the truck is

Z 50 ln 3 0

e−x2/5000dx= 45.6246 gallons.

b. [2 points] At time t hours after 10AM, what interval I of the road is still covered in paint?

(you may assume that t represents a time before the trucks meet)

Solution: The truck is at 50 ln(t + 2) miles from the garage and the crew is at 35t miles from the garage. I = [35t, 50 ln(t + 2)].

c. [3 points] Let P (t) represent the amount of paint in gallons on the road t hours after 10 AM. Find a formula (which may include a definite integral) for P (t).

Solution:

P(t) =

Z 50 ln(t+2) 35t

e−x2/5000dx

d. [2 points] Calculate P(1).

Solution:

P(t) = e−(50 ln(t+2))2/5000

 50 t+ 2



− 35



e−(35t)2/5000



P(1) = e−(50 ln(3))2/5000 50 3



− 35



e−(35)2/5000

= −18.2795 gal/hr

University of Michigan Department of Mathematics Winter, 2011 Math 116 Exam 1 Problem 7 (paint truck) Solution

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Math 116 / Exam 1 (October 12, 2011) page 9

7. [13 points] Household electricity in the United States is supplied in the form of an alternating current that varies sinusoidally with a frequency of 60 cycles per second (Hz). The voltage is given by the equation

E(t) = 170 sin(120πt), where t is given in seconds and E is in volts.

a. [7 points] Using integration by parts, find R sin2θdθ. Show all work to receive full credit.

(Hint: sin2θ+ cos2θ= 1.)

Solution: We first note that R sin2θdθ = R sin θ(sin θ)dθ, so that we may take u = sin θ, dv = sin θdθ (and du = cos θdθ, v = − cos θ). Then integration by parts gives

Z

sin2θdθ = sin θ(− cos θ) − Z

− cos θ(cos θ)dθ

= − sin θ cos θ + Z

cos2θdθ.

Using the trig. identity given in the hint, we obtain Z

sin2θdθ= − sin θ cos θ + Z

1 − sin2θ dθ = − sin θ cos θ + Z

dθ − Z

sin2θ.

The integral on the far right also appears on the left, so combining like terms, we get 2

Z

sin2θdθ = − sin θ cos θ + Z

dθ Z

sin2θdθ = 1 2



− sin θ cos θ + Z



= −1

2sin θ cos θ +θ 2 + C.

b. [6 points] Voltmeters read the root-mean-square (RMS) voltage, which is defined to be the square root of the average value of [E(t)]2 over one cycle. Find the exact RMS voltage of household current.

University of Michigan Department of Mathematics Fall, 2011 Math 116 Exam 1 Problem 7 (AC current) Solution

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Math 116 / Exam 1 (October 12, 2011) page 10

Solution: Since the frequency of the current is 60 cycles per second, one cycle is com- pleted every 601 seconds. Thus

RMS voltage = v u u

t 1

1 60− 0

Z 601

0

E(t)2dt

= s

60 Z 601

0

1702sin2(120πt) dt.

Substituting w = 120πt, dw = 120πdt, we get

RMS voltage = v u u t60

Z w(601) w(0)

1702sin2(w) · 1 120πdw

= s

1702

Z 0

sin2(w) dw.

Using the antiderivative we found in part (a) with C = 0, the Fundamental Theorem of Calculus gives

RMS voltage =

s1702



−1

2sin θ cos θ + θ 2



0

= s

1702

 (−1

2sin (2π) cos (2π) +2π

2 ) − (−1

2sin(0) cos(0) + 0 2)



= 170

√2 Volts.

Note that due to the periodicity of the sine function, the average value over one cycle could also have been computed over 0 ≤ t ≤ 1 (or any other number of periods).

University of Michigan Department of Mathematics Fall, 2011 Math 116 Exam 1 Problem 7 (AC current) Solution

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Math 116 / Exam 1 (February 5, 2018) page 2

1. [14 points] Let f (x) be a twice-differentiable function. Use the table to compute the following expressions. Show your work.

x 0 1 2 3 4 5 6 7 8 9

f(x) 1 2 4 11 1 3 5 4 2 3

f(x) 2 3 7 4 −5 2 1 −2 −3 1

a. [3 points]

Z 8 1

f(√3 x)

3

x2 dx

Solution: Use u-substitution with u =√3

x to get u=√3

x du= 1

3 1

3

x2dx.

Then the antiderivative is Z f(√3

x)

3

x2 dx= 3 Z

f(u) du = 3f (√3 x).

Hence we have Z 8

1

f(√3 x)

3

x2 dx= 3(f (2) − f(1)) = 6.

Answer: a. 6

b. [3 points]

Z 9

7

12f(x) (f (x))2 dx

Solution: Use u-substitution with u = f (x) to get u= f (x) du= f(x)dx.

Then the antiderivative is

Z 12f(x)

(f (x))2 dx= 12 Z 1

u2 du= − 12 f(x). Hence we have

Z 9

7

12f(x)

(f (x))2 dx= − 12

f(9)+ 12

f(7) = −12 3 +12

4 = −1.

Answer: b. −1

University of Michigan Department of Mathematics Winter, 2018 Math 116 Exam 1 Problem 1 Solution

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Math 116 / Exam 1 (February 5, 2018) page 3

c. [3 points]

Z 3 0

xf′′(x) dx

Solution: Use integration by parts with u = x and dv = f′′(x)dx to get

u = x v = f(x)

du = dx dv = f′′(x)dx.

Then the antiderivative is Z

xf′′(x) dx = xf(x) − Z

f(x) dx = xf(x) − f(x).

Z 3 0

xf′′(x) dx = (3f(3) − f(3)) − (0f(0) − f(0)) = (3 · 4 − 11) − (0 · 2 − 1) = 2.

Answer: c. 2

d. [5 points] The average value of 2f(x)

(f (x))2+ f (x) on [4, 6].

Solution: By definition, this average is given by the integral 1

6 − 4 Z 6

4

2f(x)

(f (x))2+ f (x)dx= Z 6

4

f(x)

(f (x))2+ f (x)dx.

To compute this antiderivative, first use the u-substitution u = f (x) to get Z f(x)

(f (x))2+ f (x)dx=

Z 1

u2+ udu.

Using the method of partial fractions we find the identity 1

u2+ u = 1 u − 1

u+ 1.

Thus Z

1

u2+ udu= ln |u| − ln |u + 1|, and changing variables back to x leads to

Z f(x)

(f (x))2+ f (x)dx= ln |f(x)| − ln |f(x) + 1|.

Therfore, Z 6

4

f(x)

(f (x))2+ f (x)dx= (ln |f(6)| − ln |f(6) + 1|) − (ln |f(4)| − ln |f(4) + 1|)

= ln(5) − ln(6) + ln(2)

≈ 0.510825

Answer: d. ln(5)−ln(6)+ln(2) ≈ 0.510825

University of Michigan Department of Mathematics Winter, 2018 Math 116 Exam 1 Problem 1 Solution

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Math 116 / Exam 1 (October 9, 2017) page 9

8. [10 points] A portion of the graph of a function h is shown below. The domain of h(x) includes the interval −1 ≤ x ≤ 5.

Note the following:

• h(x) is linear on each of the intervals [1, 2], [2, 3], and [4, 5].

• The portion of the graph of y = h(x) for

−1 < x < 1 is symmetric across the y- axis.

• The area of shaded region A is 4/3.

• The area of shaded region B is 13/3.

−1 1 2 3 4 5

−2

−1 1

A

B

y = h(x)

x y

Throughout this problem, the function H is the antiderivative of h satisfying H(1) = 2.

a. [2 points] For each of the following, compute the exact value. Show your work.

i. H(−1) Solution:

H(−1) = H(1) + Z −1

1

h(x) dx = H(1) − Z 1

−1

h(x) dx = 2 − 4 3 = 2

3 .

ii. H(2) Solution:

H(2) = H(1) + Z 2

1

h(x) dx = 2 + (−1) = 1 .

b. [8 points] Use the axes below to carefully sketch a graph of y = H(x) for −1 ≤ x ≤ 5.

•Clearly label the coordinates of the points on your graph at x = 0, 3, and 5.

•Be sure that local extrema and concavity are clear.

•If there are features of this function that are difficult for you to draw, indicate these on your graph.

−1 1 2 3 4 5

−4

−3

−2

−1 1 2 (0,43)

(3, −1)

(5, −176) x y

University of Michigan Department of Mathematics Fall, 2017 Math 116 Exam 1 Problem 8 Solution

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Math 116 / Exam 1 (February 8, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 2 1. [16 points] At a time t seconds after a catapult throws a rock, the rock has horizontal velocity v(t) m/s. Assume v(t) is monotonic between the values given in the table and does not change concavity.

t 0 1 2 3 4 5 6 7 8

v(t) 47 34 24 16 10 6 3 1 0

a. [4 points] Estimate the average horizontal velocity of the rock between t = 2 and t = 5 us- ing the trapezoid rule with 3 subdivisions. Write all the terms in your sum. Include units.

Solution:

R5

2 v(t) dt

5 − 2 = Lef t(3) + Right(3)

2 · 3 = (v(2) + v(3) + v(4)) + (v(3) + v(4) + v(5))

6 =

= 24 + 16 + 10 + 16 + 10 + 6

6 = 82

6 = 41 3 The average horizontal velocity of the rock is 41/3 m/s.

b. [4 points] Estimate the total horizontal distance the rock traveled using a left Riemann sum with 8 subdivisions. Write all the terms in your sum. Include units.

Solution:

Z 8

0

v(t) dt = Lef t(8) = v(0) + v(1) + v(2) + v(3) + v(4) + v(5) + v(6) + v(7) =

= 47 + 34 + 24 + 16 + 10 + 6 + 3 + 1 = 141

The total horizontal distance the rock traveled is approximately 141 meters.

University of Michigan Department of Mathematics Winter, 2016 Math 116 Exam 1 Problem 1 (catapult) Solution

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Math 116 / Exam 1 (February 8, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 3 c. [4 points] Estimate the total horizontal distance the rock traveled using the midpoint rule

with 4 subdivisions. Write all the terms in your sum. Include units.

Solution: Z 8 0

v(t) dt = M id(4) = 2(v(1) + v(3) + v(5) + v(7)) =

= 2(34 + 16 + 6 + 1) = 114

The total horizontal distance the rock traveled is approximately 114 meters.

d. [4 points] A second rock thrown by the catapult traveled horizontally 125 meters. Deter- mine whether the first rock or the second rock traveled farther, or if there is not enough information to decide. Circle your answer. Justify your answer.

Solution:

the first rock the second rock not enough information The function v(t) is concave up since for example

−10 = v(2) − v(1)

2 − 1 > v(1) − v(0)

1 − 0 = −13

The trapezoid rule gives T rap(4) = 121 (or T rap(8) = 117.5). Since v(t) is concave up, this is an overestimate which means that the first rock traveled at most 121 meters, that is less than the second.

University of Michigan Department of Mathematics Winter, 2016 Math 116 Exam 1 Problem 1 (catapult) Solution

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Math 116 / Exam 1 (February 5, 2018) page 8

6. [7 points] For each of the questions below, circle all of the available correct answers. You must circle at least one choice to receive any credit. No credit will be awarded for unclear markings.

No justification is necessary.

a. [3 points] Which F (x) are antiderivatives of f (x) = ex2 with F (3) = 5 for x > 0?

Note: due to a typo in the original exam (corrected here), a student’s answer to option IV did not impact their score.

I. F (x) = Z x2

0

eudu+ 5 II. F (x) =

Z x

3

5eu2du III. F (x) = 1

x2ex2 + 5 IV. F (x) = Z 9

x2− 1

2√ueudu+ 5

V. F (x) = Z x

3

eu2du+ 5 VI. F (x) = ex2 2x −

e9 6 + 5

b. [2 points] Suppose f (x) is an odd function. Which values of b make the following equation

true? Z b

π

sin(f (x)) dx = 0

I. b = −π II. b = 0 III. b = π IV. b = 2 V. b = 2π

c. [2 points] Which of the following could be the graph of f (x) = Z x3

x

e√u3 du?

I. 0 1 II. 0 1

III. 0 1

IV.

0 1

V.

0 1

University of Michigan Department of Mathematics Winter, 2018 Math 116 Exam 1 Problem 6 Solution

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Math 116 / Exam 1 (February 8, 2010) page 10

10. [15 points] Consider the area between the curves y = x2 and y = x4 in the positive quadrant as shown in the graph below. Use this area to answer the following questions.

x y

y = x2

y=x4

a. [5 points] Set up, but do not evaluate, a definite integral that describes the area described above. Write your final answer on the space provided.

R1

0(x2− x

4)dx or R1

0(y1/4− y

1/2)dy

b. [5 points] Set up, but do not evaluate, a definite integral that describes the volume of the solid generated by revolving the area described above about the line y = 2. Write your final answer on the space provided.

R1

0 π((2 − x4)2− (2 − x

2)2)dx

c. [5 points] Set up, but do not evaluate, a definite integral that describes the volume of the solid whose base is the area described above and whose cross-sections perpendicular to the x-axis are squares.

R1

0(x2− x

4)2dx

University of Michigan Department of Mathematics Winter, 2010 Math 116 Exam 1 Problem 10 Solution

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Math 116 / Exam 1 (February 9, 2015) page 4

2. [13 points] Fred is designing a plastic bowl for his dog, Fido. Fred makes the bowl in the shape of a solid formed by rotating a region in the xy-plane around the y-axis. The region, shaded in the figure below, is bounded by the x-axis, the y-axis, the line y = 1 for 0 ≤ x ≤ 4, and the curve y = −(x − 5)4+ 2 for 4 ≤ x ≤ 21/4+ 5. Assume the units of x and y are inches.

x y

a. [7 points] Write an expression involving one or more integrals which gives the volume of plastic needed to make Fido’s bowl. What are the units of your expression?

Solution: Using the cylindrical shell method, we have that the volume of plastic needed to make Fido’s bowl is given by

Z 4 0

2πxdx + Z 5+214

4

2πx(2 − (x − 5)4)dx.

Using the washer method, we have that the volume of plastic needed to make Fido’s bowl is given by π

Z 1 0

(5 + (2 − y)1/4)2dy+ π Z 2

1

(5 + (2 − y)1/4)2−(5 − (2 − y)1/4)2dy.

The units for either expression are in3.

b. [6 points] Fred wants to wrap a ribbon around the bowl before he gives it to Fido as a gift. The figure below depicts the cross section of the bowl obtained by cutting it in half across its diameter. The thick solid curve is the ribbon running around this cross section, and the dotted curve is the outline of the cross section which is not in contact with the ribbon. Write an expression involving one or more integrals which gives the length of the thick solid curve in the figure (the length of ribbon Fred needs to wrap the bowl).

Solution: The length of ribbon Fred needs to wrap the bowl is given by 10 + 2(214 + 5) + 2

Z 5+214

5

p1 + 16(x − 5)6dx.

x y

University of Michigan Department of Mathematics Winter, 2015 Math 116 Exam 1 Problem 2 (dog bowl) Solution

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© 2018 Univ. of Michigan Dept. of Mathematics (Creative Commons By−NC−SA license) Fall, 2018 Math 116 Exam 1 Problem 8 (pond) Solution

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