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.

(a)R 2

1 ln x dx, n = 3

Let f(x) = ln x. When n = 3, �x = 2�13 = 1

3 .

Left-hand sum:

2

Z 2

1

ln x dx ⇡ 1

3· f(1) + 1

3· f

✓4

3

◆+

1

3· f

✓5

3

◆

=1

3· ln(1) + 1

3· ln

✓4

3

◆+

1

3· ln

✓5

3

◆

⇡ 0.266169

Right-hand sum:

Z 2

1

ln x dx ⇡ 1

3· f

✓4

3

◆+

1

3· f

✓5

3

◆+

1

3f(2)

=1

3· ln

✓4

3

◆+

1

3· ln

✓5

3

◆+

1

3ln(2)

⇡ 0.497218

(b)R 2

0 xe

x

dx, n = 4

Let f(x) = xe

x

dx. When n = 4, �x = 2�04 = 1

2 .

Left-hand sum:

Z 2

0

xe

x

dx ⇡ 1

2· f(0) + 1

2· f

✓1

2

◆+

1

2· f(1) + 1

2· f

✓3

2

◆

=1

2· (0)e(0) + 1

2· 12e

(1/2) +1

2· (1)e(1) + 1

2· (32)e(3/2)

⇡ 5.13259

Right-hand sum:

Z 2

0

xe

x

dx

1

2· f

✓1

2

◆+

1

2· f(1) + 1

2· f

✓3

2

◆⇡ 1

2· f(2)

=1

2· 12e

(1/2) +1

2· (1)e(1) + 1

2· (32)e(3/2) +

1

2· (2)e(2)

⇡ 12.5216

(c)R 1

�13p3� x

2dx, n = 4

Let f(x) = 3p3� x

2. When n = 4, �x = 1�(�1)4 .

3

Left-hand sum:

Z 1

�1

3p3� x

2dx ⇡ 1

2· f(�1) +

1

2· f

✓�1

2

◆+

1

2· f(0) + 1

2· f

✓1

2

◆

=1

23p

3� (�1)2 +1

2· 3p

3� (�1/2)2 +1

2· 3p3� (0)2 +

1

2· 3p

3� (1/2)2

⇡ 2.7521

Right-hand sum:

Z 1

�1

3p3� x

2dx ⇡ 1

2· f

✓�1

2

◆+

1

2· f(0) + 1

2· f

✓1

2

◆+

1

2· f(1)

=1

2· 3p

3� (�1/2)2 +1

2· 3p

3� (0)2 +1

2· 3p3� (1/2)2 +

1

23p3� (1)2

⇡ 2.7521

4. Evaluate the following definite integrals.

(a)R 2

1 4x3dx = x

4|21 = 24 � 14 = 15

(b)R �2

�2 3x4dx = 0

(c)R 0

�1(9x2 � 1) dx = 3x3 � x|0�1 = (3(0)3 � 0)� (3(�1)3 � (�1)) = 2

(d)R 2

1 (x�2 + 3x�4) dx =

�� 1

x

� 1x

3

���21=

��1

2 �123

��

��1

1 �113

�= 11

8

(e)R 0

�1 e�x

dx = (�e

�x)|0�1 = �e

0 + e

1 = �1 + e

(f)R �1

�2 e

2xdx =

�12e

2x����1

�2= 1

2e�2 � 1

2e�4

(g)R 4

23x

dx = (3 ln |x|)|42 = 3 ln 4� 3 ln 2

(h)R 0

�1(1 + 2x)5 dx = 12 ((1 + 2x)6)|0�1 =

12(1 + 0)6 � 1

2(1� 2)6 = 12 �

12 = 0

(i)R 1

�1 xex

2+1dx =

⇣12e

x

2+1⌘���

1

�1= 1

2e2 � 1

2e2 = 0

(j)R 2

11

2x+1 dx =�12 ln |2x+ 1|

���21= 1

2 ln 5�12 ln 3

(k)R 0

�2x

x

2+1 dx =�12 ln(x

2 + 1)���0

�2= 1

2 ln 1�12 ln 5 = �1

2 ln 5

(l)R 4

1e

px

px

dx =�2e

px

���41= 2e2 � 2e

5. The graph of f is shown. Evaluate each integral by interpreting it in terms of areas.

4

0 2 4 6 8 10 120

2

4

6

8

10

For the following solutions, the areas below the graph have been labeled.

You may find the following formulas helpful:

• Area of a triangle = 12(base)(height)

• Area of a rectangle = (base)(height)

• Area of a circle = ⇡(radius)2

• Area of a trapezoid = 12(height1 + height2)(base)

(a)R 4

0 f(x) dx = A1 + A2 =12(2 + 4)(2) + (2)(4) = 14

(b)R 8

2 f(x) dx = A2 + A3 + A4 = (2)(4) + (4)(4) + 12⇡(2)

2 = 24 + 2⇡

(c)R 12

10 f(x) dx = A5 =122(4 + 8) = 12

6. (a) If f(6) = 15, f 0 is continuous, andR 10

6 f

0(x) dx = 23, what is the value of f(10)?

By the Fundamental Theorem of Calculus, part 2,R 10

6 f

0(x) dx = f(10) � f(6) so

23 = f(10)� 15. Then f(10) = 38.

(b) If f(4) = 20, f 0 is continuous, andR 4

1 f

0(x) dx = 6, what is the value of f(1)?

By the Fundamental Theorem of Calculus, part 2,R 4

1 f

0(x) dx = f(4) � f(1), so

6 = 20� f(1). Then f(1) = 14.

7. Using the graph of f 0(x), given below, and the fact that f(0) = 12, compute the following

values.

5

AA

A

A A1 2 3

4

5

0 2 4 6 8 100

2

4

6

8

(a) We can see graphically thatR 2

0 f

0(x) dx = 12(2)(6) = 6. By the Fundamental Theo-

rem of Calculus, part 2, we also knowR 2

0 f

0(x) dx = f(2)�f(0). Then 6 = f(2)�12.

Thus f(2) = 18

(b) We can see graphically thatR 5

0 f

0(x) dx = 12(2)(6)+(3)(6) = 24. By the Fundamental

Theorem of Calculus, part 2, we also knowR 5

0 f

0(x) dx = f(5) � f(0). Then 24 =

f(5)� 12. Thus f(5) = 36

(c) We can see graphically thatR 9

0 f

0(x) dx = 12(2)(6)+(4)(6)+ 1

2(2)(6+4)+(1)(9) = 49.

By the Fundamental Theorem of Calculus, part 2, we also knowR 9

0 f

0(x) dxf(9) �f(0). Then 49 = f(9)� 12, so f(9) = 61.

8. Suppose the rate of sales of an item is given by

S

0(t) = �3t2 + 36t

where t is the number of weeks after an advertising campaign has begun. How many

items were sold during the third week?

The number of items sold during the third week is given by

Z 3

2

S

0(t) dt =

Z 3

2

(�3t2 + 36t) dt

=��t

3 + 18t2���3

2

=��(3)3 + 18(3)2

����(2)3 + 18(2)2

�

= 71

6

9. A tank holding 10,000 gallons of a polluting chemical breaks at the bottom and spills out

at the rate given by

f

0(t) = 400e�0.01t,

where t is measured in hours. How much spills during the first day?

The amount of oil lost spilled during the first day is

Z 24

0

f

0(t) dt =

Z 24

0

400e�0.01tdt

=��40000e0.01t

���240

=��40000e0.01(24)

����40000e0.01(0)

�

⇡ 8534.89 gallons

10. For the following solutions, we denote the average value of f(x) by f(x)

(a) f(x) = 12�(�2)

R 2

�2 2x dx = 14 x

2|2�2 =142

2 � (�2)2 = 0

(b) f(x) = 13�0

R 3

0 x

3dx = 1

3 ·14x

4��30= 1

3

⇥14(3)

4 � 14(0)

4⇤= 81

12

(c) f(x) = 1ln 2�0

R ln 2

0 e

x

dx = 1ln 2 e

x|ln 20 = 1

ln 2

⇥e

ln 2 � e

0⇤= 1

ln 2 (2� 1) = 1ln 2

(d) f(x) = 1e

2�1

Re

2

14x

dx = 1e

2�1 · 4 ln |x||e

2

1 = 1e

2�1 [4 ln |e2|� 4 ln |1|] = 1

e

2�1(8�0) = 8e

2�1

(e) f(x) = 12�0

R 2

0 x(x� 1) dx = 12

R 2

0 (x2 � x) dx = 1

2

⇣x

3

3 � x

2

2

⌘���2

0

= 12

h⇣(2)3

3 � (2)2

2

⌘�

⇣(0)3

3 � (0)2

2

⌘i= 1

3

11. The population in a certain city is projected to be given by

P (t) = 100000e0.05t,

where t is given in years from now. Find the average population over the next 10 years.

The average population over the next 10 years is

1

10� 0

Z 10

0

P (t) dt =1

10

Z 10

0

100, 000e0.05t dt

=1

10

�2000000et/20

���101

⇡ 129744

7

12. The profit in millions of dollars of a certain firm is given by

P (t) = 6(t� 1)(t� 2),

where t is measured in years. Find the average profit per year over the period of time

[0, 4].

The average profit in millions of dollars is given by

1

4� 0

Z 4

0

P (t) dt =1

4

Z 4

0

6(t� 1)(t� 2) dt

=3

2

Z 4

0

(t2 � 3t+ 2) dt

=3

2

✓t

3

3� 3t2

2+ 2t

◆����4

0

=3

2

✓43

3� 3(4)2

2+ 2(4)

◆�

✓(0)3

3� 3(0)2

2+ 2(0)

◆�

=3

2

✓16

3

◆

= 8

13. Find the area of the region enclosed by the given curves.

(a) y = x

3, y = 0, x = 1, x = 2

A =R 2

1 x

3dx = x

4

4

���2

1= 16

4 � 14 = 15

4

(b) y = e

�x

, y = 3, x = �1, x = 0

8

A =R 0

�1(3� e

�x) dx = (3x+ e

�x)|0�1 =�3(0) + e

�(0)���3(�1) + e

�(�1)�= 4� e

(c) y =px, y = x, x = 0, x = 1

A =R 1

0 (px� x) dx =

⇣23x

3/2 � x

2

2

⌘���1

0= 1

6

(d) y = x

2 � 2x+ 1, y = x+ 1

To find the bounds for integration, we determine where the two given curves intersect:

x

2 � 2x+ 1 = x+ 1 ) x

2 � 3x = 0 ) x(x� 3) = 0 ) x = 0, x = 3

Then

A =R 3

0 [(x+ 1)� (x2 � 2x+ 1)] dx =R 3

0 (�x

2 + 3x) dx =⇣

�x

3

3 + 3x2

2

⌘���3

0

9

=⇣

�(3)3

3 + 3(3)2

2

⌘�

⇣�(0)3

3 + 3(0)2

2

⌘= 9

2

(e) y = x

2, y = 8� x

2

To find the bounds for integration, determine where the two given curves intersect:

x

2 = 8� x

2 ) 2x2 � 8 = 0 ) 2(x2 � 4) = 0 ) x = �2, x = 2

Then

A =R 2

�2[(8� x

2)� x

2] dx =R 2

�2(8� 2x2) dx =⇣8x� 2x3

3

⌘���2

�2

=⇣8(2)� 2(2)3

3

⌘�

⇣8(�2)� 2(�2)3

3

⌘= 64

3

(f) y = e

2x, y = e

�2x, x = �1, x = 2

Observe that the roles of “top function” and “bottom function” switch when the two

functions intersect, so in order to find the area between the curves, we must find the

two areas separately and then add them together.

The graphs intersect when e

2x = e

�2x ) 2x = �2x ) 4x = 0 ) x = 0.

Then A =R 0

�1(e�2x � e

2x) dx+R 2

0 (e2x � e

�2x) dx

=��1

2e�2x � 1

2e2x���0

�1+

�12e

2x + 12e

�2x���2

0

=⇥��1

2 �12

��

��1

2e�2(�1) � 1

2e2(1)

�⇤+⇥�

12e

2(2) + 12e

�2(2)��

�12 +

12

�⇤

= 12e

2 + 12e

�2 + 12e

4 + 12e

�4 � 2

10

(g) y = x

3 � 3x, y = 2x2

As in the example above, the roles of the “top function” and “bottom function”

switch when the two functions intersect, so in order to find the area between the

curves, we must find two areas separately and then add them together. We also

need to find the intersection in order to find the bounds for our integrations.

The graphs intersect when

x

3 � 3x = 2x2 ) x

3 � 2x2 � 3x = 0 ) x(x+ 1)(x� 3) = 0 ) x = 0, x = �1, x = 3.

Then A =R 0

�1[(x3 � 3x)� 2x2] dx+

R 3

0 [2x2 � (x3 � 3x)] dx

=⇣

x

4

4 � 3x2

2 � 2x3

3

⌘���0

�1+

⇣2x3

3 � x

4

4 + 3x2

2

⌘���3

0= 7

12 +454 = 71

6

(h) y = x

3 � 3x, y = �2x3

Again, the roles of the “top function” and “bottom function” switch when the two

functions intersect, so in order to find the area between the curves, we must find two

areas separately and then add them together. We also need to find the intersection

in order to find the bounds for our integrations.

The graphs intersect when

x

3 � 3x = �2x3 ) 3x3 � 3x = 0 ) 3x(x2 � 1) = 0 ) x = �1, x = 0, x = 1

11

Then A =R 0

�1[(x3 � 3x)� (�2x3)] dx+

R 1

0 [�2x3 � (x3 � 3x)) dx

=R 0

�1[3x3 � 3x] dx+

R 1

0 [�3x3 + 3x] dx

=⇣

3x4

4 � 3x2

2

⌘���0

�1+

⇣�3x4

4 + 3x2

2

⌘���1

0

=�0�

�34 �

32

��+���3

4 + 32

�� 0

�= 3

2

(i) y = 3px, y = x, x = �8, x = 1

The bounds for x are given, but we still need to note that the “top function” and

“bottom function” switch roles when the functions intersect. In this case they inter-

sect at 3 points:

The graphs intersect when 3px = x ) x = x

3 ) x

3 � x = 0 ) x(x2 � 1) = 0

) x = 0, x = 1, x = �1.

Then A =R �1

�8 (3px� x) dx+

R 0

�1(x� 3px) dx+

R 1

0 (3px� x) dx

=⇣

34x

43 � x

2

2

⌘����1

�8+

⇣x

2

2 � 34x

43

⌘���0

�1+

⇣34x

43 � x

2

2

⌘���1

0= 81

4 + 14 +

14 = 83

4

12