Math 1071 Spring 2016, Final Exam Review Solutions This review packet only covers the sections from Chapter 6 covered after Exam 2. The final exam will be cumulative. You should look at the review packet for Exams 1 and 2, as well as past assignments, quizzes, worksheets, and relevant sections of our text to remind yourself of earlier material. This review packet is not intended to be exhaustive. 1. For each of the answers below, we use C as the constant of integration. (a) R 5 dx =5x + C (b) R x 99 dx = 1 100 x 100 + C (c) R x -99 dx = - 1 98 x -98 + C (d) R 5 x 3 dy =5 R x -3 dx = 5(- 1 2 x -2 )+ C = - 5 2 x -2 + C (e) R y 3 2 p 2 dy = 1 p 2 R y 3 2 dy = 1 p 2 h⇣ 1 (5/2) ⌘ y 5/2 i + C = 1 p 2 ( 2 5 )y 5 2 + C = 2 5 p 2 y 5 2 + C (f) R 3 p u 2 du = R u 2 3 du = 1 (5/3) u 5 3 + C = 3 5 u 5 3 + C (g) R (6x 2 +4x) dx =6 R x 2 dx +4 R xdx =2x 3 +2x 2 + C (h) R( 3 t 2 - 6t 2 ) dt =3 R t -2 dt - 6 R t 2 dt = 3(-t -1 ) - 2t 3 + C = -3t -1 - 2t 3 + C (i) R( x + 1 x 3 ) dx = R x dx + R x -3 dx = 1 2 x 2 + - 1 2 x -2 + C (j) R( ⇡ + 1 x ) dx = R ⇡ dx + R 1 x dx = ⇡x + ln |x| + C (k) R t+1 p t dt = R ⇣ t p t + 1 p t ⌘ dt = R (t 1 2 + t - 1 2 ) dt = R t 1 2 dt + R t - 1 2 dt = 1 (3/2) t 3 2 + 1 (1/2) t 1 2 + C = 2 3 t 3 2 +2t 1 2 + C (l) R( e x - 1 x 2 ) dx = R e x dx - R x -2 dx = e x + x -1 + C (m) R (u 2 + 1)(3 - u) du = R (3u 2 - u 3 +3 - u) du =3 R u 2 du - R u 3 du +3 R du - R u du = u 3 - 1 4 u 4 +3u - 1 2 u 2 + C (n) R t( p t - t 4 ) dt = R (t 3 2 - t 5 ) dt = R t 3 2 dt - R t 5 dt = 1 5/2 t 5 2 - 1 6 t 6 + C = 2 5 t 5 2 - 1 6 t 6 + C 1
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Math 1071 Spring 2016, Final Exam Review Solutions
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Math 1071 Spring 2016, Final Exam Review Solutions
This review packet only covers the sections from Chapter 6 covered after Exam 2. The
final exam will be cumulative. You should look at the review packet for Exams 1 and 2,
as well as past assignments, quizzes, worksheets, and relevant sections of our text to remind
yourself of earlier material. This review packet is not intended to be exhaustive.
1. For each of the answers below, we use C as the constant of integration.
(a)R5 dx = 5x+ C
(b)Rx
99dx = 1
100x100 + C
(c)Rx
�99dx = � 1
98x�98 + C
(d)R
5x
3 dy = 5Rx
�3dx = 5(�1
2x�2) + C = �5
2x�2 + C
(e)R
y
32p2dy = 1p
2
Ry
32dy = 1p
2
h⇣1
(5/2)
⌘y
5/2i+ C = 1p
2(25)y
52 + C = 2
5p2y
52 + C
(f)R
3pu
2du =
Ru
23du = 1
(5/3)u53 + C = 3
5u53 + C
(g)R(6x2 + 4x) dx = 6
Rx
2dx+ 4
Rxdx = 2x3 + 2x2 + C
(h)R �
3t
2 � 6t2�dt = 3
Rt
�2dt� 6
Rt
2dt = 3(�t
�1)� 2t3 + C = �3t�1 � 2t3 + C
(i)R �
x+ 1x
3
�dx =
Rx dx+
Rx
�3dx = 1
2x2 +�1
2x�2 + C
(j)R �
⇡ + 1x
�dx =
R⇡ dx+
R1x
dx = ⇡x+ ln |x|+ C
(k)R
t+1pt
dt =R ⇣
tpt
+ 1pt
⌘dt =
R(t
12 + t
� 12 ) dt =
Rt
12dt+
Rt
� 12dt
= 1(3/2)t
32 + 1
(1/2)t12 + C = 2
3t32 + 2t
12 + C
(l)R �
e
x � 1x
2
�dx =
Re
x
dx�Rx
�2dx = e
x + x
�1 + C
(m)R(u2 +1)(3� u) du =
R(3u2 � u
3 +3� u) du = 3Ru
2du�
Ru
3du+3
Rdu�
Ru du
= u
3 � 14u
4 + 3u� 12u
2 + C
(n)Rt(pt� t
4) dt =R(t
32 � t
5) dt =Rt
32dt�
Rt
5dt = 1
5/2t52 � 1
6t6 +C = 2
5t52 � 1
6t6 +C
1
2. For each of the answers below, we use C as the constant of integration.
(a) Let u = 3x+ 1. Then du = 3dx, so dx = 13du.
ThenR6(3x+ 1)10 dx =
R2u10
du = 211u
11 + C = 211(3x+ 1)11 + C.
(b) Let u = 3� x
2. Then du = �2xdx and �12du = x dx.
ThenRx(3�x
2)6 dx = �12
Ru
7du = �1
2(17u
7)+C = � 114u
7+C = � 114(3�x
2)7+C
(c) Let u = x
4+8x+3. Then du = (4x3+8) dx = 4(x3+2) dx. Then 14 du = (x3+2) dx,
soR(x3 + 2) 3
px
4 + 8x+ 3 dx = 14
R3pu du = 1
4
Ru
13du = 1
4
⇣1
4/3u43
⌘+ C
= 316(x
4 + 8x+ 3) + C
(d) Let u = x+ 1. Then du = dx, soR2px+ 1 dx = 2
R pu du = 2
Ru
12du = 2
⇣1
3/2u32
⌘+ C = 4
3(x+ 1)32 + C
(e) Let u = x+ 1. Then du = dx, soR
3p
(x+ 1)2 dx =R(x+ 1)
23dx =
Ru
23du = 1
(5/3)u53du+ C = 3
5(x+ 1)53 + C
(f) Let x = ln 2x. Then du = 1x
dx soR
ln 2xx
dx =Ru du = 1
2u2 + C = 1
2(ln 2x)2 + C
(g) Let x = 2x+ 1. The du = 2 dx and 12 du = dx. So
R1
2x+1 dx = 12
R1u
du = 12 ln |u|+ C = 1
2 ln |2x+ 1|+ C
(h) Let u = e
�x + 1 and du = �e
�x
dx. ThenR
e
�x
e
�x+1 dx = �R
1u
du = � ln |u|+ C = � ln |e�x + 1|+ C = � ln(e�x + 1) + C
(since e
�x + 1 > 0 for all x)
(i) Let u = ln x. Then du = dx
x
andR
1x lnx
dx =R
1u
du = ln |u|+ C = ln | ln x|+ C
(j) Let u = ln x2. Then du = 2x
dx andR
1x lnx
2 dx = 12
R1u
du = 12 ln |u|+ C = 1
2 ln | ln x2|+ C
3. Use a left- and right-hand sum with rectangles of equal width for the given value of n to
approximate the integral. Round the answers to two decimal places.