MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
Student’s Printed Name:
CUID:
Instructor:
Section:
Instructions: You are not permitted to use a calculator on any portion of this test. You are not
allowed to use any textbook, notes, cell phone, laptop, PDA, or any technology on either portion
of this test. All devices must be turned o↵ while you are in the testing room.
During this test, any communication with any person (other than the instructor or a designated
proctor) in any form, including written, signed, verbal, or digital, is understood to be a violation
of academic integrity.
No part of this test may be removed from the examination room.
Read each question carefully. In order to receive full credit for the free response portion of the test,
you must:
1.
2.
3.
4.
Show legible and logical (relevant) justification that supports your final answer.
Use complete and correct mathematical notation.
Include proper units, if necessary.
Give exact numerical values whenever possible.
You have 90 minutes to complete the entire test.
On my honor, I have neither given nor received inappropriate or unauthorized information at any time before or during this test.
Student’s Signature:
Do not write below this line.
Free Response
Possible Points
Points Earned
Problem
1.
14
2.
14
3.
14
4.
10
5.
18
Free Response
70
Multiple Choice
30
Test Total
100
Version A - Page 1 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
This page intentionally left blank.
Version A - Page 2 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
Multiple Choice: There are 8 multiple choice questions. They do not all have the same
point value. Each question has one correct answer. The multiple choice problems will
count for 30% of the total grade. Use a number 2 pencil and bubble in the letter of
your response on the scantron sheet for problems 1 - 8. For your own record, also
circle your choice on your test since the scantron will not be returned to you. Only the
responses recorded on your scantron sheet will be graded. You are NOT permitted
to use a calculator on any portion of this test.
1. (4 pts.) Suppose f 00 is continuous and f and f 0 have the values given below. Evaluate
Z 3
xf 00 (x) dx.
1
f (x)
f 0 (x)
(a) 9
x=1
2
3
(b) 15
x=2
5
1
x=3
8
4
(c) 5
(d) 3
Answer: (d)
2. (4 pts.) Evaluate the integral
(a)
1
1
x+
sin(6x) + C
2
12
(b)
1
cos3 (3x) + C
9
Z
cos2 (3x) dx.
(c)
1
1
x + sin(3x) + C
2
6
(d)
1
x
2
1
sin(6x) + C
12
Answer: (a)
Version A - Page 3 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Z p
3. (4 pts.) When evaluating
9
p
9 25x2 and dx are replaced by
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
25x2 dx by trigonometric substitution, the expressions
(a)
p
9
25x2 = 3 sin ✓ and dx =
3
cos ✓d✓
5
(c)
p
9
25x2 = 3 sin ✓ and dx =
3
sin ✓d✓
5
(b)
p
9
25x2 = 3 cos ✓ and dx =
3
sin ✓d✓
5
(d)
p
9
25x2 = 3 cos ✓ and dx =
3
cos ✓d✓
5
Answer: (d)
4. (2 pts.) To derive the formula for integration by parts, one would use:
(a) The Mean Value Theorem
(c) The Product Rule
(b) The Chain Rule
(d) None of these
Answer: (c)
Version A - Page 4 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
5. (4 pts.) Which of the following integrals gives the volume of the solid whose base is bounded
by the graphs of y = x + 1 and y = x2 1 and whose cross sections perpendicular to the base
and parallel to the y-axis are isosceles triangles with height equal to twice the base? See figures
below.
(a)
(b)
Z
Z
2
1
1
x+1
2
(x
⇡ x+1
(x2
2
2
1)
1)
2
2
dx
dx
1
(c)
(d)
Z
Z
2
x+1
(x2
1)
2
dx
1
2
2⇡x x + 1
(x2
1) dx
1
Answer: (c)
6. (4 pts.) At what point does the function f (x) = 3x2
[0, 2]?
(a) 1
2
(b) p
3
(c) 2
2 equal its average value on the interval
(d)
p
2
Answer: (b)
Version A - Page 5 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
7. (4 pts.) A force of 12 pounds is required to stretch a spring 0.5 feet from its natural position.
Assuming Hooke’s Law applies, how much work is done in stretching the spring 3 feet from its
natural position?
(a) 54 ft-lb
(b) 72 ft-lb
(c) 36 ft-lb
(d) 108 ft-lb
Answer: (d)
8. (4 pts.) Find the area of the region bounded by the curves y = x3
1  x  2.
(a)
23
4
(b)
7
4
(c)
9
4
x2 and y =
(d)
x2 + 4x for
55
4
Answer: (a)
Version A - Page 6 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
Free Response. The Free Response questions will count for 70% of the total grade.
Read each question carefully. To receive full credit, you must show legible, logical, and
relevant justification which supports your final answer. Give answers as exact values.
You are NOT permitted to use a calculator or any other technology on any portion
of this test.
Z
1. (14 pts.) Evaluate the integral:
x2 e 3x dx.
Solution:
Use integration by parts. Let u = x2 and dv = e
Z
To integrate
Z
du = dx and v =
Therefore,
Z
3x
xe
e
x2 e
3x
1 2
x e
3
1 2
=
x e
3
dx =
3x
dx. Then du = 2x dx and v =
Z
3x
3x
+
2
3
3x
3
. So
3x
2xe
Z
e
dx
3
xe
3x
dx
dx, use integration by parts again. Let u = x and dv = e
3x
dx. Then
3x
3
. So
1 2
x e
3
3x
x2 e
dx =
3x
2
+
3
Z
xe
1 2
x e
3
3x
3x
1 2
dx =
x e
3
1 2
=
x e
3
1 2
=
x e
3
2
xe
9
3x

2
1
+
xe
3 3
2
3x
xe 3x +
9
2
3x
xe 3x
9
3x
2
e
27
3x
3x
Z
e
3x
3
Z
2
e 3x dx
9
2 3x
e
+C
27
+ C.
Work on Problem:
Points
Defines u, dv and finds du, v for first IBP
4 points (1 point each)
Rewrites integral using integration by parts formula
2 points
Defines u, dv and finds du, v for second IBP
4 points (1 point each)
Rewrites integral using integration by parts formula
⇣
⌘
R 3x
Finishes integrating integrates e 3 dx
2 points
Final answer with + C
dx
1 point
1 point
Notes:
• Deduct 0.5 points for notation errors.
Version A - Page 7 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
2. (14 pts.) Evaluate the integral:
Z
p
x2
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
x
dx.
+ 4x + 8
Solution:
First complete the square. x2 + 4x + 8 = (x + 2)2 + 4. Then use a trig substitution. Let
x + 2 = 2 tan ✓. Then dx = 2 sec2 ✓ d✓. So
Z
p
x
dx =
2
x + 4x + 8
Z
x
p
dx =
(x + 2)2 + 4
Z
2 tan ✓ 2
p
2 sec2 ✓ d✓
2
4 tan ✓ + 4
Z
tan ✓ 1
=4 p
sec2 ✓ d✓
2
4 sec ✓
Z
tan ✓ 1
=4
sec2 ✓ d✓
2 sec ✓
Z
= 2 (tan ✓ 1)(sec ✓) d✓
Z
= 2 (sec ✓ tan ✓ sec ✓) d✓
= 2(sec ✓
Now use that
x+2
ln | sec ✓ + tan ✓|) + C
x+2
= tan ✓ to construct the following triangle.
2
p
(x + 2)2 + 4
✓
2
p
(x + 2)2 + 4
The triangle gives sec ✓ =
. Therefore,
2
!
p
p
Z
(x + 2)2 + 4
(x + 2)2 + 4 x + 2
x
p
dx = 2
ln
+
+C
2
2
2
x2 + 4x + 8
Work on Problem:
Points
Completes the square
2 points
Defines x + 2 = 2 tan ✓
2 points
Finds dx
2 points
Rewrites the integral using the trig sub
Uses the trig identity 1 +
tan2 ✓
=
sec2 ✓
1 point
2 points
Simplifies the integral
1 points
Finds the antiderivatives of sec ✓ tan ✓ and sec ✓
2 points (1 point each)
Uses the triangle to write the final answer in terms of x with + C
2 points
Notes:
• Deduct 0.5 points for each notation error
(with a maximum deduction of 2 points for notation errors)
Version A - Page 8 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
3. (14 pts.) Evaluate the integral:
Solution (Method 1):
Z
⇡/4
tan3 x
sec2 x
0
Z
dx =
⇡/4
0
Z
tan3 x
dx.
sec2 x
⇡/4
0
sin3 x
Z ⇡/4
sin3 x
cos3 x dx =
cos2 x dx
3x
cos
1
0
cos2 x
Z ⇡/4
sin3 x
=
dx
cos x
0
Z ⇡/4
sin2 x sin x
=
dx
cos x
0
Z ⇡/4
(1 cos2 x) sin x
=
dx
cos x
0
Now use u-substitution. Let u = cos x. Then du =
Z
⇡/4
(1
0
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
cos2 x) sin x
dx =
cos x
=
Z
sin x dx. So
Z
p
1/ 2
1
1
p
1/ 2
1
u
u2 )
du
u
◆
u du
◆1
u2
2 1/p2
✓
◆ ✓ ✓
◆
1
1
p
= 0
ln
2
2
✓
◆
1
1
= ln p
4
2
=
So
Z
⇡/4
0
tan3 x
dx =
sec2 x
ln
✓
1
p
2
◆
✓
✓
(1
ln |u|
1
4
◆
1
4
Version A - Page 9 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
Work on Problem:
Points
Rewrites in terms of sines and cosines
2 points
Simplifies
1 point
Uses an appropriate Pythagorean identity
3 points
• No credit is given for a wrong identity
Defines u and finds du
2 points (1 point each)
Rewrites the integral in terms of u with u-limits
3 points
• 2 points for writing in terms of u/du, 1 point for limits
Finds the antiderivative of each term
2 points
Evaluates
1 point
Notes:
• Deduct 0.5 points for notation errors
• Deduct 0.5 points for leaving ln 1 in answer
• Deduct 0.5 points for mixing x and u in integral
• Deduct 0.5 points for additional incorrect trig identities
Version A - Page 10 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
Solution (Method 2):
Z
⇡/4
0
tan3 x
dx =
sec2 x
Z
⇡/4
0
tan2 x tan x
dx =
sec2 x
=
=
=
Z
Z
Z
Z
⇡/4
0
(sec2 1) tan x
dx
sec2 x
⇡/4
tan x
dx
sec2 x
tan x
0
Z
⇡/4
tan x dx
0
Z
⇡/4
tan x dx
0
⇡/4
0
sin x
cos x dx
1
cos2 x
⇡/4
sin x cos x dx
0
Now use u-substitution. Let u = sin x. Then du = cos x dx. So
Z
⇡/4
tan x dx
0
Z
⇡/4
sin x cos x dx =
0
=
Z
⇡/4
tan x dx
0
⇡/4
ln | sec x||0
p
= ln( 2)
So
Z
⇡/4
0
p
tan3 x
dx
=
ln(
2)
sec2 x
dx
Z
p
1/ 2
u du
0
u2
2
p
1/ 2
0
1
4
1
4
Version A - Page 11 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
Solution (Method 3):
Z
⇡/4
0
tan3 x
dx =
sec2 x
Z
⇡/4
3
tan x sec
2
x dx =
0
=
Z
Z
⇡/4
tan x sec x(tan2 x) sec
3
x dx
0
⇡/4
tan x sec x(sec2 x
1) sec
3
x dx
0
Now use u-substitution. Let u = sec x. Then du = sec x tan x dx. So
Z
⇡/4
tan x sec x(sec2 x
1) sec
3
x dx =
0
=
=
=
Z
Z
p
1
1
✓
✓
p
2
(u2
2✓
ln |u| +
p
= ln( 2)
So
Z
⇡/4
0
p
tan3 x
dx = ln( 2)
2
sec x
1
u
ln |u|
3
1)u
u
u
2
2
1
2u2
1
4
3
du
◆
du
◆p 2
1
◆p 2
1
1
4
Version A - Page 12 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
4. (10 pts.) A 15-ft chain with a 100-pound load attached to it hangs from a rod that is 15 ft
above the ground. Set up, but do not evaluate or simplify, the integral that represents the
work done in winding up the entire chain (with the load attached) onto the rod if the chain
weighs 3 pounds per foot.
Solution:
Let x = the distance from the bottom of the chain to the ground,
WT = total work done,
WC = work to lift the chain,
WL = work to lift the load.
Then WL = (100 lbs)(15 ft) = 1500 ft-lbs and
WC =
Z
So
WT = 1500 +
15
3(15
x) dx
0
Z
15
3(15
x) dx ft-lbs
0
Work on Problem:
Points
Finds the work to lift the load
3 points
Sets up an integral to find the work to lift the chain
6 points
Sums up both parts to express the total work done
1 point
Notes:
• Deduct 0.5 points notation errors.
• Do not deduct for missing units.
Version A - Page 13 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
5. (18 pts.) Let R be the region bounded by y = e2x , y = 4, and the y-axis.
(a) Set up, but do not evaluate or simplify, the integral that gives the volume of the solid
obtained by rotating the region R around the y-axis using the disk/washer method.
Solution:
V =
Z
4
⇡
1
✓
ln y
2
◆2
dy
(b) Set up, but do not evaluate or simplify, the integral that gives the volume of the solid
obtained by rotating the region R around the line y = 1 using the disk/washer method.
Solution:
V =
Z
ln 2
0
h
⇡ 52
e2x + 1
2
i
dx
(c) Set up, but do not evaluate or simplify, the integral that gives the volume of the solid
obtained by rotating the region R around the line x = 2 using the shell method.
Solution:
V =
Z
ln 2
2⇡(2
x)(4
e2x ) dx
0
Work on each Problem:
Points
Limits of integration
1 point
Appropriate constant (⇡ on (a), (b); 2⇡ on (c))
1 point
Rest of integrand
4 points
Notes:
• Deduct 0.5 points for each notation error
(with maximum deduction of 1 point for notation errors)
Version A - Page 14 of 15
MATH 1080
Calculus of One Variable II
Test 1 - Answer Key
Version A
Spring 2016
Sections 6.1 - 6.5, 7.1 - 7.3
Scantron: Check to make sure your Scantron form meets the following criteria:
My Scantron:
• is bubbled with firm marks so that the form can be machine read;
• is not damaged and has no stray marks (the form can be machine read);
• has 8 bubbled in answers;
• has MATH 1080 and my Section number written at the top;
• has my Instructor’s last name written at the top;
• has Test No. 1 written at the top;
• has the correct test version written at the top and bubbled in below my XID;
• shows my correct XID both written and bubbled in.
**Bubble a zero for the leading C in your XID**.
Version A - Page 15 of 15