1 review test 4. 2 1. find the following integral a. x 8 + c b. x 8 /8 + c c. 7 x 6 + c d. x 6 /6 +...

1

REVIEWTEST 4

2

1. Find the following integral

A. x 8 + c B. x 8 /8 + c

C. 7 x 6 + c D. x 6 /6 + c

E. None of the above

dxx7

3

Sorry, that answer is incorrect.

Remember the power rule for integration –

Please click here to try again.

c1n

xdxx

1nn

4

Great, the correct answer is x 8 /8 + c, by substituting the power 7 into the power rule.

Please click here for the next question.

c1n

xdxx

1nn

5


A. x –3 /- 3 + c B. x –1 /- 3 + c

C. - x – 1 + c D. - 2x –3 + c


dxx

12

6

Not quite, that answer is incorrect.



1n

xdxx

1nn

7

Hey great, the correct answer is - x – 1 + c, by substituting the power - 2 into the power rule.


c1n

xdxx

1nn

8


A. B.

C.


dxx

3x3

33

c2

x3

4

x3 24

c4

x3

4

x3 44

c2

x3x9

22

c4

x3x9

42

D.

9

Too bad, that answer is incorrect.



1n

xdxx

1nn

10

OK, the correct answer is

Using the power rule twice; once with n = 3 and again with n = - 3.


c2

x3

4

x3 24

1n

xdxx

1nn

11

4. Ready to step it up a notch?

A. e x + c B. x e x + c

C. D. 2e 2x + c


dxex

c1x

e 1x

12


Remember the rule for exponential functions for integration –


cedxe xx

13

Terrrrriffffic, the correct answer is e x + c,

Using the rule for exponential expressions for integration.


cedxe xx

14

A. ln x + c B. e x + c

C. D. x + c


dxx

1


cx

12

15


Remember the rule for logarithmic functions for integration –


cxlndxx

1

16

Terrrrriffffic, the correct answer is ln x + c,

since


.definitionbyclndxx

1

17

6. Now, let’s try something really interesting.

dx)e3x

3( x

cex

3 x2 A. B. ln x – ex + c

C. 3ln x – 3ex + c D.


ce3 x

18

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.


19

Terrrrriffffic, the correct answer is 3ln x – 3ex + c

since


ce3xln3dxe3dxx

3dx)e3

x

3( xxx

20

7. Let’s go back to the power rule.

A. B.

D.


dxx3

c4

x3 34

cx3

1 32

c3

x 34

c3

x 31

C.

21

That answer is incorrect. Try changing the radical to exponential form!


dxxdxx 31

3

Then use the power rule.

22

Grrrrreat, the correct answer is


c4

x3 34

dxxdxx 31

3

cx4

3

34x

dxx 343

4

31

And using the power rule for integration yields -

23


dxx

1

A. -2 x -1 / 2 + c B. 2 x 1 / 2 + c

C. -2 x 1 / 2 + c D. 2 x -1 / 2 + c


24

That answer is incorrect. Try changing the radical to exponential form!


dxxdxx

12

1

Then use the power rule.

25

WOW, the correct answer is indeed 2x ½ + c.


cx2c

21x

dxxdxx

12

1212

1

26

9. Let’s complete that idea.

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


dx

x

2x3

3

27

That answer is incorrect.

That answer is incorrect. Try changing the radicals to exponential form! See the previous two

problems.


28

Great, the correct answer is 2x 3/2 + 3x 2/3 + c, since


dx)x2x3(dx

x

2x3 3121

3

cx3x2c32

x2

23

x3 32233223

29

10. OK. Let’s try some involving substitution techniques.

A. B.

D.


dx)3x(x 754

c40

)3x( 85

c5

)3x( 85

c8

)3x( 85

c)3x(8 85 C.

30

No that answer is incorrect.

Try substitution. Let u = x 5 – 3 and find du and make an adjustment to x 4 that is appropriate.


dx)3x(x 754

31

Yes! Yes! Yes!, the correct answer is

by letting u = x 5 – 3 and finding du = 5x 4 dx


The problem then needs adjusting by multiplying and dividing by 5.

c40

)3x( 85

dx)3x(x 754

duu5

1dx)3x(x5

5

1dx)3x(x 7754754

c3x40

1

8

u

5

1 858

32

11.

A.

B. 5 (2x 5 – 4x + 7) 4 (20x 3 ) + c

D.


dx)2x5()7x4x2( 455

c)7x4x2(12

1 65

c6

)7x4x2( 65

c)2x5(12

)7x4x2(5 445

C.

33

That answer is incorrect. You must substitute correctly!

Let u = 2x 5 – 4x + 7, then du = 10x 4 – 4 which is

2 (5x 4 – 2) . SO the original problem becomes


dx)2x5()7x4x2( 455

dx)2x5(2)7x4x2(2

1 455

duu2

1OR

5

34

Grrrrreat, the correct answer is


c)7x4x2(12

1 65

Let u = 2x 5 – 4x + 7, then du = 10x 4 – 4 which is

2 (5x 4 – 2) . SO the original problem becomes

c)7x4x2(12

1c

6

u

2

1duu

2

1 656

5

dx)2x5(2)7x4x2(2

1 455

35

12.

A. 2 (x 2 – 3) - ½ + c B. 5 (x 2 – 3) - 1/2 + c

C. 2 (x 2 – 3) – 1/2 + c D. 5 (x 2 – 3) ½ + c


dx3x

x52

36

Too bad that answer is incorrect.


First, rewrite the problem -

dx3x

x52

Let u = x 2 – 3, then du = 2x dx which means that we need a factor of 2 in the problem or

dx3xx5212

dx3xx22

5dx3xx5

212212

37

Yes, you are an integrating machine.


First, rewrite the problem

dx3x

x52

Let u = x 2 – 3, then du = 2x dx which means that we need a factor of 2 in the problem or

dx3xx5212

duu2

5dx3xx2

2

5dx3xx5

21212212

c3x5cu521

u

2

5 212212

1

38

13.

B.

C. D.


dxex23x22

c3

e3x2

c6

e3x2

cex23x2 ce3

3x2

A.

39

No that answer is incorrect. You must substitute correctly!


dxex23x22

Let u = 2x 3 , then du = 6x 2 dx .

40

Correctamundo, the answer is


c3

e3x2

Let u = 2x 3 , then du = 6x 2 dx , then

dxex23x22

due3

1dxex6

3

1dxex2 ux22x22 33

ce3

1ce

3

1 3x2u

41

14.

A. 2 (2x – 1) + c B. ½ (2x – 1) + c

C. ½ ln | 2x – 1| + c D. 2x – 1 + c


dx

1x2

1

42



No that answer is incorrect. You must substitute correctly!

Let u = 2x - 1 , then du = 2 dx .

dx

1x2

1

43

Yes, the answer is ½ ln | 2x – 1| + c


duu

1

2

1dx

1x2

2

2

1dx

1x2

1

Let u = 2x - 1 , then du = 2 dx , then,

|1x2|ln2

1c|u|ln

2

1

44

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


dxx3

1

2

45



To find the definite integral you have two options-

1. 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.

46

Yes, the answer is 8.67


I used the “Calc” menu and “integrate”.

47

16.

A. 3 B. 5.17

C. 2.02 D. - 3.5


dx)1x2(ln3

1

48




OR



49




50

17.

A. 3 B. 5.17

C. 27.23 D. - 3.5


dxe2

1

x2

51



1. Calculate the indefinite integral and evaluate at the top limit (2) and subtract the value at the bottom limit (- 1)

OR



52




53

18. Find the area between y = 4x – x 2 and the x-axis

A. 4 B. 10.67

C. 5.38 D. 7.33


54



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 options-


OR


55

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.

dxxx44

0

2

56

19. Find the area between y = 3 – 2x 2 and the x-axis

A. 4 B. 4.90

C. 5.89 D. 7.33


57



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 options-


OR


58

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.

dxx23225.1

225.1

2

59

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


60



This area is the definite integral of the difference of the two functions. You will need a calculator to find the x-intercepts. 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 options-


OR


61

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.

dx)x4x2()x23( 25.1

5.0

2

62

APPLICATIONS

Now, let’s try some application problems

63

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


64

Sorry that answer is incorrect.

The index of income concentration is

where f (x) is the Lorenz curve.


1

0dx])x(fx[2

65

Yes, the answer 0.2 using the definition of the index of income concentration and integrating on your calculator.


20.0)10.0(2dx]xx[21

0

5.1

1

0dx])x(fx[2

This answer will be used in the next problem.

66

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.

67

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.


68

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


69


Remember the average cost equation.


x

CC

Solve for C.

70

Yes, the answer is

C (x) = 100 + 5x

Multiply the average cost by x.


71


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


72



P = R – C

73

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.

74


for 0 < x < 7000B. Find the maximum profit.

A. $27,500

B. $28,025

C. $28,575

D. $29,525


75


P = R – C

P = 15x – 0.002 x 2 - 100


Now find the maximum profit.

76

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


0 < x < 7000

You will need this information for the next problem.

77


C. Find the price that yields the maximum profit.

A. $11.00

B. $11.50

C. $12.00

D. $12.50


78


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!


79

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


80

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


81

Sorry that answer is incorrect.


x

CC

Remember the average cost equation.

82

Shazzammm!

The correct answer is C = (100 + 5x) / x

since,


x

CC

83


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


84


Profit = Revenue – Cost.

How would you find Revenue?


85

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.

86


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


87

Not the correct answer.

Remember that R (x) = xp

And use R from the previous problem.


88

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.


89

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 number of bacteria per milliliter.

A. Find the minimum value of dN/dt. 0 ≤ x ≤ 18

A. 900

B. 875

C. - 875

D. - 900


2t1

t1800

dt

dN

90

Sorry that answer is wrong.

Graph dN/dt, the given equation, on your calculator and find the minimum (y-value).


91

WOW, yes a tough one without a calculator!

The correct answer is - 900


92


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


2t1

t1800

dt

dN

93

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.


94

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.

c)t1(ln900dtt1

t2900dt

t1

t1800 2

22

Now substitute the point.

You will need this answer for the next problem.

95


C. Find the bacteria count after 7 days.

A. 6,000

B. 4,335

C. 2,479

D. 1,981


96

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

97

Well done!

The correct answer 2,479 bacteria.


N (t) = - 900 ln (1 + 7 2 ) + 6000= - (900) (3.912) + 6000= - 3521 + 6000= 2479

98

That was a lot of review. I hope you found it helpful. Good luck on the test!

1 review test 4. 2 1. find the following integral a. x 8 + c b. x 8 /8 + c c. 7 x 6 + c d. x 6 /6 +...

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