REVIEW
TEST 4
1
1. Find the following integral
x
A. x 8 + c
C. 7
x6+
c
7
dx
B. x 8 /8 + c
D. x 6 /6 + c
E. None of the above
2
Sorry, that answer is incorrect.
Remember the power rule for integration –
n 1
x
 x dx  n  1  c
n
Please click here to try again.
3
Great, the correct answer is x 8 /8 + c, by substituting
the power 7 into the power rule.
n 1
x
n
 x dx  n  1  c
Please click here for the next question.
4
2. Find the following integral
1
 x 2 dx
A. x –3 /- 3 + c
B. x –1 /- 3 + c
x–1
D. - 2x –3 + c
C. -
+c
E. None of the above
5
Not quite, that answer is incorrect.
Remember the power rule for integration –
n 1
x
 x dx  n  1
n
Please click here to try again.
6
Hey great, the correct answer is - x – 1 + c, by
substituting the power - 2 into the power rule.
n 1
x
n
 x dx  n  1  c
Please click here for the next question.
7
3. Find the following integral
3
 3
  3 x  x 3  dx
A.
3 x4 3 x 4

c
4
4
4
C.
3x
9x 
4
2
c
B.
2
3
x
2
9x 
c
2
4
D. 3 x  3 x
4
2
2
c
E. None of the above
8
Too bad, that answer is incorrect.
Remember the power rule for integration –
n 1
x
 x dx  n  1
n
Please click here to try again.
9
4
OK, the correct answer is
2
3x
3x

4
2
c
Using the power rule twice; once with n = 3 and again
with n = - 3.
n 1
x
 x dx  n  1
n
Please click here for the next question.
10
4. Ready to step it up a notch?
e
x
dx
A. e x + c
B. x e x + c
ex 1
c
C.
x 1
D. 2e 2x + c
E. None of the above
11
Not quite, that answer is incorrect.
Remember the rule for exponential functions
for integration –
e
x
dx  e  c
x
Please click here to try again.
12
Terrrrriffffic, the correct answer is e x + c,
Using the rule for exponential expressions for
integration.
e
x
dx  e  c
x
Please click here for the next question.
13
5. Find the following integral
1
 x dx
A. ln x + c
B. e x + c
1
C. 2  c
x
D. x + c
E. None of the above
14
Not quite, that answer is incorrect.
Remember the rule for logarithmic functions
for integration –
1
 x dx  ln x  c
Please click here to try again.
15
Terrrrriffffic, the correct answer is ln x + c,
since
1
 x dx  ln  c by definition .
Please click here for the next question.
16
6. Now, let’s try something really interesting.
3
x
(

3
e
) dx
 x
A.
3
x
e c
2
x
C. 3ln x – 3ex + c
B. ln x – ex + c
D.
3e c
x
E. None of the above
17
That answer is incorrect. It is a hard one!
You must use both the rule for exponential
functions for integration and the rule for
logarithmic functions for integration as in the
previous two problems.
Please click here to try again.
18
Terrrrriffffic, the correct answer is 3ln x – 3ex + c
since
3
x
(

3
e
) dx 
 x
3
x
x
dx

3
e
dx

3
ln
x

3
e
c
x

Please click here for the next question.
19
Let’s go back to the power rule.
7.

1
A.
x
3
3
3x
C.
4
x dx
4
2
4
3
c
B.
x 3
c
3
1
3
c
D.
x 3
c
3
E. None of the above
20
That answer is incorrect. Try changing the
radical to exponential form!

3
x dx 

1
x 3 dx
Then use the power rule.
Please click here to try again.
21
4
Grrrrreat, the correct answer is

3
x dx 
3x
4
3
c
1

x 3 dx
And using the power rule for integration yields 4

x
3
x dx 
4
1
3
3
3 43
 x c
4
Please click here for the next question.
22
8. Find the following integral

1
dx
x
A. -2 x -1 / 2 + c
B. 2 x 1 / 2 + c
C. -2 x 1 / 2 + c
D. 2 x -1 / 2 + c
E. None of the above
23
That answer is incorrect. Try changing the
radical to exponential form!

1
dx 
x

1
x
2
dx
Then use the power rule.
Please click here to try again.
24
WOW, the correct answer is indeed 2x ½ + c.

1
dx 
x

12
1
x
2
x
dx 
1
 c  2x
1
2
c
2
Please click here for the next question.
25
9. Let’s complete that idea.

2 

 3 x  3  dx
x

A. 2x 2/3 + 3x 3/2 + c
B. -6x -1/2 + 3x -4/3 + c
C. 2x 3/2 + 3x 2/3 + c
D. -2x 3/2 - 3x - 2/3 + c
E. None of the above
26
That answer is incorrect.
That answer is incorrect. Try changing the
radicals to exponential form! See the previous two
problems.
Please click here to try again.
27
Great, the correct answer is 2x 3/2 + 3x 2/3 + c, since

2 

12
1 3
3
x

dx

(
3
x

2
x
)
dx




3
x

3x 3 2
2 x2 3
32
23

 c  2x  3x  c
32
23
Please click here for the next question.
28
10. OK. Let’s try some involving substitution
techniques.
x
4
(x  3) dx
5
7
A.
( x 5  3) 8
c
5
( x 5  3) 8
c
8
B.
C.
( x 5  3) 8
c
40
D. 8 ( x5  3)8  c
E. None of the above
29
No that answer is incorrect.
4
5
7
x
(
x

3
)
dx

Try substitution. Let u = x 5 – 3 and find du and
make an adjustment to x 4 that is appropriate.
Please click here to try again.
30
x
4
(x  3) dx
5
7
( x 5  3) 8
c
Yes! Yes! Yes!, the correct answer is
40
by letting u = x 5 – 3 and finding du = 5x 4 dx
The problem then needs adjusting by multiplying and
dividing by 5.
1
1 7
4
5
7
 x (x  3) dx  5  5 x (x  3) dx  5  u du
8
1 u8
1 5



x  3  c
5 8
40
4
5
7
Please click here for the next question.
31
11.
4
(
2
x

4
x

7
)
(
5
x
 2) dx

5
5
5
6
(
2
x

4
x

7
)
A.
c
6
B. 5 (2x 5 – 4x + 7) 4 (20x 3 ) + c
1
C.
( 2x5  4x  7)6  c
12
5 ( 2x 5  4x  7 ) 4
D.
( 5x 4  2)  c
12
E. None of the above
32
That answer is incorrect. You must
substitute correctly!
4
(
2
x

4
x

7
)
(
5
x
 2) dx

5
5
Let u = 2x 5 – 4x + 7, then du = 10x 4 – 4 which is
2 (5x 4 – 2) . SO the original problem becomes
5
1
5
4
(
2
x

4
x

7
)
2
(
5
x
 2) dx

2
5
1
OR
u du

2
Please click here to try again.
33
1
Grrrrreat, the correct answer is
( 2x5  4x  7)6  c
12
Let u = 2x 5 – 4x + 7, then du = 10x 4 – 4 which is
2 (5x 4 – 2) . SO the original problem becomes
5
1
5
4
(
2
x

4
x

7
)
2
(
5
x
 2) dx 

2
5
1
1 u6
1
5
6
u
du


c

(
2
x

4
x

7
)
 c

2
2 6
12
Please click here for the next question.
34

12.
A. 2 (x 2 – 3)
-½
5x
x2  3
+c
C. 2 (x 2 – 3) – 1/2 + c
dx
B. 5 (x 2 – 3)
- 1/2
+c
D. 5 (x 2 – 3) ½ + c
E. None of the above
35
Too bad that answer is incorrect.
5x

x 3
2
First, rewrite the problem - 5  x x  3 
2
dx
1 2
dx
Let u = x 2 – 3, then du = 2x dx which means that
we need a factor of 2 in the problem or

5 x x  3
2

1 2
5
dx 
2
 2 x x
2
3

1 2
dx
Please click here to try again.
36

Yes, you are an integrating machine.

5 x x  3
First, rewrite the problem
2

5x
x 3
1 2
2
dx
dx
Let u = x 2 – 3, then du = 2x dx which means that
we need a factor of 2 in the problem or

5 x x  3
2

1 2
1
5
dx 
2
 2 x x
2
3


1 2
5
dx 
2
u
1 2
du

12
5 u 2
12
2

5u c5 x 3 c
2 12
Please click here for the next question.
37
13.
e
A.
2 x3
3
C.
3e
c
2 x3
c
 2x
2
e
2x3
dx
e
B.
2 x3
6
D. 2 x e
c
2 x3
c
E. None of the above
38
No that answer is incorrect. You
must substitute correctly!
 2x
2
e
2x3
dx
Let u = 2x 3 , then du = 6x 2 dx .
Please click here to try again.
39
Correctamundo, the answer ise
 2x
2 x3
3
2
e
2x3
c
dx
Let u = 2x 3 , then du = 6x 2 dx , then
 2x
2
e
2x3
1
1 u
2 2x3
dx   6 x e dx   e du 
3
3
1 u
1 2x3
 e c e c
3
3
Please click here for the next question.
40
14.
1
 2x  1 dx
A. 2 (2x – 1) + c
B. ½ (2x – 1) + c
C. ½ ln | 2x – 1| + c
D. 2x – 1 + c
E. None of the above
41
No that answer is incorrect.
No that answer is incorrect. You
must substitute correctly!
1
 2x  1 dx
Let u = 2x - 1 , then du = 2 dx .
Please click here to try again.
42
Yes, the answer is ½ ln | 2x – 1| + c
Let u = 2x - 1 , then du = 2 dx , then,
1
1
2
1 1
 2x  1 dx  2  2x  1 dx  2  u du
1
1
 ln | u |  c  ln | 2x  1|
2
2
Please click here for the next question.
43
15. Now that you have
indefinite integrals down
cold let’s try some
definite integrals.
A. 5
B. 6.67
C. – 8.67
D. 8.67

3
1
2
x dx
E. None of the above
44
Too bad that answer is incorrect.
To find the definite integral you have two options1. Calculate the indefinite integral and evaluate at
the top limit (3) and subtract the value at the
bottom limit (1)
OR
2. Use your calculator to do the work. Use the “Calc”
menu and integrate.
Please click here to try again.
45
Yes, the answer is 8.67
I used the “Calc”
menu and
“integrate”.
Please click here for the next question.
46
16.

3
1
ln ( 2x  1) dx
A. 3
B. 5.17
C. 2.02
D. - 3.5
E. None of the above
47
Too bad that answer is incorrect.
To find the definite integral you have two options1. Calculate the indefinite integral and evaluate at
the top limit (3) and subtract the value at the
bottom limit (1)
OR
2. Use your calculator to do the work. Use the “Calc”
menu and integrate.
Please click here to try again.
48
Yes, the answer is 2.02
I used the “Calc”
menu and
“integrate”.
Please click here for the next question.
49
17.

2
2x
e dx
1
A. 3
B. 5.17
C. 27.23
D. - 3.5
E. None of the above
50
Too bad that answer is incorrect.
To find the definite integral you have two options1. Calculate the indefinite integral and evaluate at
the top limit (2) and subtract the value at the
bottom limit (- 1)
OR
2. Use your calculator to do the work. Use the “Calc”
menu and integrate.
Please click here to try again.
51
Yes, the answer is 27.23
I used the “Calc”
menu and
“integrate”.
Please click here for the next question.
52
18. Find the area between y = 4x – x 2 and the
x-axis
A. 4
B. 10.67
C. 5.38
D. 7.33
E. None of the above
53
That answer is incorrect.
This area is the definite integral between the x-intercepts of
the curve. Graph the equation to find the intercepts and
then to find the definite integral you have two options1. Calculate the indefinite integral and evaluate at the top
limit (3) and subtract the value at the bottom limit (1)
OR
2. Use your calculator to do the work. Use the “Calc” menu
and integrate.
Please click here to try again.
54
WOW, the correct answer is indeed 10.67
I used the “Calc” menu and
“integrate”. Note the curve
crosses the x- axis at 0 and 4
the limits of your definite
integral.

4
0
4x  x 2 dx
Please click here for the next question.
55
19. Find the area between y = 3 – 2x 2 and the
x-axis
A. 4
B. 4.90
C. 5.89
D. 7.33
E. None of the above
56
That answer is incorrect.
This area is the definite integral between the x-intercepts (You
will need a calculator to find the x-intercepts.) of the curve.
Graph the equation to find the intercepts and then to find
the definite integral you have two options1. Calculate the indefinite integral and evaluate at the top
limit (3) and subtract the value at the bottom limit (1)
OR
2. Use your calculator to do the work. Use the “Calc” menu
and integrate.
Please click here to try again.
57
WOW, the correct answer is indeed 4.90. That is a
tough one because of the x-intercepts.
I used the “Calc” menu and
“integrate”. Note the curve
crosses the x- axis at – 1.225
and 1.225 the limits of your
definite integral.

1.225
 1.225
3  2 x 2 dx
Please click here for the next question.
58
20. Find the area between y = 3 – 2x 2 and
y = 2x 2 – 4x.
A. - 5.33
B. 6.67
C. 5.33
D. 4.50
E. None of the above
59
That answer is incorrect.
This area is the definite integral of the difference of the two
functions. You will need a calculator to find the xintercepts. Graph the function that is the difference of the
two equations to find the intercepts and then to find the
definite integral you have two options1. Calculate the indefinite integral and evaluate at the top
limit (3) and subtract the value at the bottom limit (1)
OR
2. Use your calculator to do the work. Use the “Calc” menu
and integrate.
Please click here to try again.
60
Grrrrreat, the correct answer is 5.33
I used the “Calc” menu and
“integrate”. Note the curve
crosses the x- axis at – 0.5
and 1.5 the limits of your
definite integral.

1.5
 0.5
( 3  2 x )  ( 2x  4x) dx
2
2
Please click here for the next question.
61
APPLICATIONS
Now, let’s try some application problems
62
21. The Lorenz curve for the distribution of
income for students at York College is given
by f (x) = x 1.5.
A. Find the index of income concentration.
A. 0.10
B. 0.20
C. 0.25
D. 0.35
E. None of the above
63
Sorry that answer is incorrect.
The index of income concentration is 2 0 [x  f ( x) ] dx
1
where f (x) is the Lorenz curve.
Please click here to try again.
64
Yes, the answer 0.2 using the definition of the index
of income concentration and integrating on your
calculator.
2  [x  f ( x) ] dx
1
0
2  [x  x1.5 ] dx  2 (0.10)  0.20
1
0
This answer will be used in
the next problem.
Please click here for the next question.
65
22. The Lorenz curve for the distribution of
income for students at York College is given
by f (x) = x 1.5.
B. Interpret the results of the previous
problem.
After you have written a response click here to
check your answer.
66
Answers will vary. Remember an index of 0.0
indicates complete equality of income distribution
while and index of 1.00 indicates complete inequality.
This index of 0.20 show a fairly equitable distribution
of income.
Please click here for the next question.
67
23. The “Screaming Green Machine” t-shirts have the
following revenue and average cost equations.
R(x) = 20x – 0.002x 2 and C(x) = (100 + 5x)/x
A. Find the total cost function.
A. C (x) = 100 + 5x
B. C (x) = 20 – 0.002x
C. C (x) = 100x + 5 x 2
D. C (x) = 20x 2 – 0.002x 3
E. None of the above
68
Too bad that answer is incorrect.
Remember the average cost equation.
C
C
x
Solve for C.
Please click here to try again.
69
Yes, the answer is
C (x) = 100 + 5x
Multiply the average cost by x.
Please click here for the next question.
70
24. The “Screaming Green Machine” t-shirts have the
following revenue and average cost equations.
R(x) = 20x – 0.002x 2 and C(x) = (100 + 5x)/x
A. Find the total profit function.
A. P (x) = 100 + 5x
B. P (x) = 15x – 0.002 x 2 – 100
C. P (x) = 25x – 0.002 x 2
D. P (x) = 25x – 0.002 x 2 - 100
E. None of the above
71
Too bad that answer is incorrect.
P=R –C
Please click here to try again.
72
Yes, the answer is
P (x) = 15x – 0.002 x 2 – 100, since
P (x) = R - C
= (20 x – 0.002 x 2 ) – (100 + 5x)
= 15x – 0.002 x 2 – 100
You will need this answer for the next question.
Please click here for the next question.
73
25. The “Screaming Green Machine” t-shirts have the
following revenue and average cost equations.
R(x) = 20x – 0.002x 2 and C(x) = (100 + 5x)/x
for 0 < x < 7000
B. Find the maximum profit.
A. $27,500
B. $28,025
C. $28,575
D. $29,525
E. None of the above
74
No that answer is incorrect.
P=R –C
P = 15x – 0.002 x 2 - 100
Now find the maximum profit.
Please click here to try again.
75
Yesaroonie, the answer is
P (x) = 15x – 0.002 x 2 – 100, since
P (x) = R - C
= 20 x – 0.002x 2 – (100 + 5x)
= 15x – 0.002 x 2 – 100
Now graph it to find the max.
max P = $28,025
You will need this information
for the next problem.
0 < x < 7000
Please click here for the next question.
76
26. The “Screaming Green Machine” t-shirts have the
following revenue and average cost equations.
R(x) = 20x – 0.002x 2 and C(x) = (100 + 5x)/x
C. Find the price that yields the maximum profit.
A. $11.00
B. $11.50
C. $12.00
D. $12.50
E. None of the above
77
No that answer is incorrect.
Use the x value (sales giving max profit) from the last
problem and substitute it into the price equation. Do
you remember how to find the price equation. Think
REVENUE!
Please click here to try again.
78
Perfect, the answer is $12.50
One way to get this is to plug the x value from the
maximum profit (previous problem) into the price
equation which comes from the revenue equation.
x = 3750
And from R(x) = 20x – 0.002x 2 the price equation is
p (x) = R (x) / x = 20 – 0.002x
p (3750) = 20 – (0.002) (3750) = 12.50
Please click here for the next question.
79
27. “Rolling Stones” t-shirts have the following
profit and cost equations.
P(x) = 15x – 0.003x 2 - 100 and
C(x) = 100 + 5x for 0 < x < 7000
A. Find the average cost function.
A. C = x (100 + 5x)
B. C = (100 + 5x) / x
C. C = 100x + 5x
D. C = 100x + 5x - 15x – 0.003x 2 - 150
E. None of the above
80
Sorry that answer is incorrect.
Remember the average cost equation.
C
C
x
Please click here to try again.
81
Shazzammm!
The correct answer is C = (100 + 5x) / x
since,
C
C
x
Please click here for the next question.
82
28. “Rolling Stones” t-shirts have the
following profit and cost equations.
P(x) = 15x – 0.003x 2 - 100 and
C(x) = 100 + 5x
for 0 < x < 7000
B. Find the revenue function.
A. R(x) = x (100 + 5x)
B. R(x) = 20x – 0.003x 2
C. R(x) = 10x – 0.003x 2
D. R(x) = 20x – 0.003x 2 - 200
E. None of the above
83
No that answer is incorrect.
Profit = Revenue – Cost.
How would you find Revenue?
Please click here to try again.
84
Hot Dog!
The correct answer is R = 10x – 0.003x 2
since, Profit = Revenue – Cost and hence
Revenue = Profit + Cost
= (15x – 0.003x 2 - 100) + (100 + 5x)
= 20x – 0.003x 2
You will need this for the next question.
Please click here for the next question.
85
29. “Rolling Stones” t-shirts have the following
profit and cost equations.
P(x) = 15x – 0.003x 2 - 100 and
C(x) = 100 + 5x for 0 < x < 7000
B. Find the price-demand function.
A. p (x) = - 10 + 0.003x
B. p (x) = 20x – 0.003x 2
C. p (x) = 20 – 0.003x
D. p (x) = (100 + 5x) / x
E. None of the above
86
Not the correct answer.
Remember that R (x) = xp
And use R from the previous problem.
Please click here to try again.
87
Great work.
The correct answer is p (x) = 20 – 0.003x
since R = px and then p = R/x
and using R from the previous problem yields
p (x) = 20 – 0.003x.
Please click here for the next question.
88
30. A water supply is treated with a bactericide. The rate
of increase in harmful bacteria t days after the
treatment is given by the following where N is the
dN
1800 t
number of bacteria per milliliter.

dt
1  t2
A. Find the minimum value of dN/dt. 0 ≤ x ≤ 18
A. 900
B. 875
C. - 875
D. - 900
E. None of the above
89
Sorry that answer is wrong.
Graph dN/dt, the given equation, on your calculator
and find the minimum (y-value).
Please click here to try again.
90
WOW, yes a tough one without a calculator!
The correct answer is - 900
Please click here for the next question.
91
31. A water supply is treated with a bactericide. The rate
of increase in harmful bacteria t days after the
treatment is given by the following where N is the
number of bacteria per milliliter.
dN
1800 t
dt

1  t2
B. If the initial count was 6000 bacteria per
milliliter, find the equation for N (t).
A. N = 1 / (1 + t 2 ) + 6000
B. N = - 900 ln (1 + t 2 ) + 6000
C. N = 6000 / (1 + t 2 )
D. N = - 1800 ln (1 + t 2 ) + 6000
E. None of the above
92
Too bad, that is incorrect.
You need to integrate the initial equation dN/dt to
find the equation for N.
Don’t forget the initial condition of 6000 bacteria per
milliliter.
Please click here to try again.
93
Well done!
The correct answer is N = - 900 ln (1 + t 2 ) + 6000
Integrate the original equation and use the
point (0, 6000) to get the equation for N.
t
2t
2
 1800 
dt


900
dt


900
ln
(
1

t
)c
2
2

1t
1t
Now substitute the point.
You will need this answer for the next problem.
Please click here for the next question.
94
32. A water supply is treated with a bactericide. The
rate of increase in harmful bacteria t days after the
treatment is given by the following where N is the
number of bacteria per milliliter.
C. Find the bacteria count after 7 days.
A. 6,000
B. 4,335
C. 2,479
D. 1,981
E. None of the above
95
Sorry that is incorrect.
Plug t = 7 into the equation for N that you found in
the previous problem.
N (t) = - 900 ln (1 + t 2 ) + 6000
Please click here to try again.
96
Well done!
The correct answer 2,479 bacteria.
N (t) = - 900 ln (1 + 7 2 ) + 6000
= - (900) (3.912) + 6000
= - 3521 + 6000
= 2479
Please click here for the next question.
97
That was a lot of review. I hope you found it
helpful. Good luck on the test!
98