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MTHS 1080 Test 1-Version A Fall 2013
Student’s Printed Name: _____________________SOLUTIONS_________________ Instructor: _________________________________ Section # :_________ 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 off while you are in the testing room. During this test, any communication with any person (other than the instructor or his 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 very carefully. In order to receive full credit for the free response portion of the test, you must:
1. Show legible and logical (relevant) justification which supports your final answer. 2. Use complete and correct mathematical notation. 3. Include proper units, if necessary. 4. Give exact numerical values whenever possible. 5. Follow the directions given for the problem.
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
Problem Possible Points Points Earned
1 6
2 8
3 9
4 6
5 17
6 11
7. (Scantron) 1 Total Free Response 58 Multiple Choice 42
Test Total 100
MTHS 1080 Test 1-Version A-SOUTIONS Fall 2013
1
Multiple Choice: There are 14 multiple choice questions. They all have the same point value. Each question has one correct answer. The multiple choice problems will count 42% of the total grade. Use a number 2 pencil and bubble in the letter of your response on the scantron sheet for problems 1 – 14. 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. Which one of the following is equivalent to log2 10( )? (3 pts.)
a) ln5 + ln2
b) ***1+ ln5ln2
c) ln5
d) ln10 − ln2
2. Simplify tan cos−1 x⎡⎣ ⎤⎦ . (3 pts.)
a) x2 −1x
b) 1− x2
c) *** 1− x2
x
d) x1− x2
3. We can approximate the length of a curve with the sum of the lengths of line
segments, expressed as L ≈ Δx( )2 + Δyk( )2k=1
n
∑ . What theorem helps us rewrite
this sum as a Riemann sum, L ≈ 1+ ′f xk( )2k=1
n
∑ Δx ?
(3 pts.)
a) ***Mean Value Theorem b) Fundamental Theorem of Calculus
c) L’Hopital’s Rule
d) Intermediate Value Theorem
MTHS 1080 Test 1-Version A-SOUTIONS Fall 2013
2
4.
Evaluate the following integral:
g3(x)g 4(x)−1∫ ′g (x)dx .
(3 pts.)
a) ln g(x) − g
4 (x)4
+C b) ***
ln g 4(x)−14
+C
c) g
2 (x)2
− g4 (x)4
+C
d) ln g4 (x)−1 +C
5. Given that
tan x + y( ) = tan x + tan y
1− tan x tan y, evaluate
tan−1 1
2⎛⎝⎜
⎞⎠⎟+ tan−1 1
3⎛⎝⎜
⎞⎠⎟. (3 pts.)
a)
π2
b)
π3
c) ***
π4
d)
π6
6. The length of a curve is given by 1+ 4
1− 4x2dx
a
b
∫ . Which of the following could be
the equation of the curve?
(3 pts.)
a) f (x) = 4
1− 4x2 b)
f (x) = 32x
1− 4x2( )2
c) f (x) = 2arctan(2x)
d) *** f (x) = arcsin(2x)
MTHS 1080 Test 1-Version A-SOUTIONS Fall 2013
3
7.
The region bounded by y = cos x and the x-‐ and y-‐axes 0 ≤ x ≤ π / 2( ) is revolved around the x-‐axis. Using the shell method, the volume of the solid is given by which integral?
(3 pts.)
a) 2π xcos xdx0
π2
∫ b) π cos−1 y( )2 dy0
1
∫
c)*** 2π ycos−1 ydy
0
1
∫
d) cos2 xdx0
π2
∫
8. Evaluate ′f 0( ) given f x( ) = cos 2x( )
x2 + 6x +1.
(3 pts.)
a) ***−3 b) 0
c) 1
d) 3
9. Find the derivative of ywith respect to x for the function y = x
cos x . (3 pts.)
a) y = xcos x −sin xx
⎛⎝⎜
⎞⎠⎟
b)*** y = xcos x cos xx
− sin x ln x⎛⎝⎜
⎞⎠⎟
c) y = cos xx
− sin x ln x
d) y = cos x( )xcos x−1
MTHS 1080 Test 1-Version A-SOUTIONS Fall 2013
4
10.
Evaluate 4 x44x
dx∫ . (3 pts.)
a) 44 x+1
ln 4 4x +1( ) +C
b) 44x
ln 4( )2 +C
c)*** 44
x
ln 4( )2+C
` d) 4
4x
ln 4+C
11. Which integral gives the area of the region in the first quadrant bounded by the axes, y = ex , x = ey , and the line x = 4 ?
(3 pts.)
a) ex dx0
e
∫ + ln xdxe
4
∫
b) ey − ln y( )dy0
ln4
∫
c) ex − ln x( )dx
0
4
∫ ` d)*** ex dx
0
1
∫ + ex − ln x( )dx1
4
∫
12. Which of the following integrals gives the volume of the solid whose base is the triangle in the xy-‐plane with vertices (0,0), (2,0), and (0,2) and whose cross sections perpendicular to the base and parallel to the y-‐axis are semicircles?
(3 pts.)
a) π 2 − y( )2 dy
0
2
∫ b) *** π82 − x( )2 dx
0
2
∫
c) π
42 − x( )2 dx
0
2
∫
d) 2π x 2 − x( )dx
0
2
∫
MTHS 1080 Test 1-Version A-SOUTIONS Fall 2013
5
13.
Which integral represents the arclength of the curve y = log10 x on the interval 1≤ x ≤ 5?
(3 pts.)
a) *** 1+ 1x ln10( )2
dx1
5
∫ b) 1− log10 xx
⎛⎝⎜
⎞⎠⎟2
dx1
5
∫
c) 1+ ln x2
ln10( )2dx
1
5
∫
d) 1+ log10 xx
⎛⎝⎜
⎞⎠⎟2
dx1
5
∫
14. Consider the region R shown below. Which of the following is NOT always true?
(3 pts.)
a)
f x( )− c⎡⎣ ⎤⎦a
b
∫ dx = b− g y( )⎡⎣ ⎤⎦c
d
∫ dy
b)
π f x( )⎡⎣ ⎤⎦
2− c2( )
a
b
∫ dx = 2π y b− g y( )⎡⎣ ⎤⎦c
d
∫ dy
c)
π b2 − g y( )⎡⎣ ⎤⎦
2( )dy =c
d
∫ 2π x f x( )− c⎡⎣ ⎤⎦a
b
∫ dx
d) ***
π f x( )⎡⎣ ⎤⎦
2− c2( )dx
a
b
∫ = π b2 − g y( )⎡⎣ ⎤⎦2( )dy
c
d
∫
MTHS 1080 Test 1-Version A-SOUTIONS Fall 2013
6
Free Response: The Free Response questions will count 57% of the total grade (plus 1 point for correctly filling out the scantron). Read each question carefully. In order to receive full credit you must show legible and logical (relevant) justification which supports your final answer. All limits of integration should be expressed as exact values not as decimal approximations (ie. π and not 3.14). Solutions should be simplified using exact values (i.e. π ) and radicals (if necessary, i.e. 2 ) and by applying properties in this unit. If a u-‐substitution is required for integration, clearly identify u and du. (This does not apply to problems for which only setting up an integral is required.) You are NOT permitted to use a calculator on any portion of this test. Evaluate the following integrals.
1. (6 pts) 1− 1− cos x( )2⎡⎣ ⎤⎦∫ sin xdx
Let u = 1− cos x. Then du = sin xdx.
1− 1− cos x( )2⎡⎣⎢
⎤⎦⎥sin x dx∫ = 1− u2⎡⎣ ⎤⎦∫ du = u − u3
3+C
⇒ 1− 1− cos x( )2⎡⎣⎢
⎤⎦⎥sin x dx∫ = 1− cos x( )− 1− cos x( )3
3+C
OR
Let u = cos x. Then du = −sin xdx.
1− 1− cos x( )2⎡⎣⎢
⎤⎦⎥sin x dx∫ = − 1− 1− u( )2⎡
⎣⎢⎤⎦⎥∫ du = − 1− 1− 2u + u2( )⎡
⎣⎤⎦∫ du = −2u + u2( )∫ du = −u2 + u3
3+C
⇒ 1− 1− cos x( )2⎡⎣⎢
⎤⎦⎥sin x dx∫ = −cos2 x + cos3 x
3+C
2. (8 pts) dx4x2 − 8x + 5∫
First you need to complete the square.
4x2 −8x +5= 4 x2 − 2x( ) +5= 4 x2 − 2x +1( )− 4+5= 4 x −1( )2
+1 Then
dx4x2 − 8x + 5∫ = dx
4 x −1( )2 +1∫ .
Let u = 2 x −1( ). Then du = 2dx and du2
= dx
dx
4 x −1( )2+1
∫ .= 12
duu2 +1∫ = 1
2tan−1 u +C = 1
2tan−1 2x − 2( ) +C
⇒ dx4x2 −8x +5∫ = 1
2tan−1 2x − 2( ) +C
MTHS 1080 Test 1-Version A-SOUTIONS Fall 2013
7
3. (9 pts) e2x
1− e4 xdx
− ln22
− ln24
∫
Let u = e2x . Then du = 2e2xdx and du2
= e2xdx. If x = − ln22
, then u = e2 − ln2
2⎛⎝⎜
⎞⎠⎟ = e− ln2 = eln2−1
= 12
.
If x = − ln24
, then u = e− ln2
2 = eln2−1/2
= 12
.
e2x
1− e4xdx
− ln22
− ln24
∫ = 12
du
1− u212
12
∫
= 12
sin−1 u12
12
= 12
sin−1 12
⎛⎝⎜
⎞⎠⎟− sin−1 1
2⎡
⎣⎢
⎤
⎦⎥
= 12
π4− π
6⎛⎝⎜
⎞⎠⎟= π
24
⇒ e2x
1− e4xdx
− ln22
− ln24
∫ = π24
4. (6 pts) x2 + x( ) x +1∫ dx
Let u = x +1. Then du = dx and x = u −1.
x2 + x( ) x +1dx∫ = u −1( )2+ u −1( )⎡
⎣⎢⎤⎦⎥∫ udu
= u2 − u( )u1/2 du∫
= u5/2 − u3/2( )du∫
= 27
u7/2 − 25
u5/2 +C
⇒ x2 + x( ) x +1dx∫ = 27
x +1( )7/2− 2
5x +1( )5/2
+C
MTHS 1080 Test 1-Version A-SOUTIONS Fall 2013
8
5. Let R be the region in the first quadrant bounded by the graphs y = ln x ,
x = 1, and y = 1 shown below.
a. (4 pts) 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 x-‐axis using the disk/washer method.
V = π 12 − ln x( )2⎡
⎣⎢⎤⎦⎥
1
e
∫ dx
b. (4 pts) 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 shell method.
V = 2πx 1− ln x( )
1
e
∫ dx
c. (4.5 pts) 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 . State the name of the method you use. Method:_____________________________________
Shell: V = 2π 1− y( ) ey −1( )
0
1
∫ dy
Disk/Washer: V = π 1− ln x( )2
1
e
∫ dx
d. (4.5 pts) 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 = 1 . State the name of the method you use. Method:_____________________________________
Shell: V = 2π x −1( ) 1− ln x( )
1
e
∫ dx
Disk/Washer: V = π ey −1( )2
0
1
∫ dy
MTHS 1080 Test 1-Version A-SOUTIONS Fall 2013
9
6. Let R be the region below the graph of y =1
x2/3 1+ x1/3( ) in the first quadrant bounded by x = 1 and x = 8 as shown below.
a. (3 pts) Set up, but do not evaluate an integral(s) that represents the area of the region R .
A = dx
x2/3 1+ x1/3( )1
8
∫
b. (8 pts) Evaluate the integral you found in part a. giving your answer as an exact value and simplified. To receive full credit for part b., you must have a correct integral in part a.
A = dxx2/3 1+ x1/3( )1
8
∫
Let u = 1+ x1/3. Then du = 13
x−2/3dx and 3du = dxx2/3 . If x = 1, then u = 2. If x = 8, then u = 3.
dxx2/3 1+ x1/3( )1
8
∫ = 3 duu2
3
∫ = 3ln2
3
u 3ln 32
A = 3ln 32
square units