Final Exam Spring 2008

8/2/2019 Final Exam Spring 2008

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1. Find f(1) if f (x) = sin x +3

x2 and f(0) = 3

We would want to integrate this function because of the 2nd Fundamental theorem of calculus

ba

f(x) dx = F (b) F (a)

Where f(x) is the derivative of F (x)

f (x) dx =

sin x +

3

x2 dx

We would have to use this formula:xn dx =

xn+1

n + 1+ C to evaluate:

sin x +

3

x2 dx =

cos(x) +

35

x3

5

+ C

f(0) = 3

We have to solve C

3 = cos (0) + 0 + C3 = 1 + C

C = 4

Now, we just substitute f(1)

f(1) = cos(1) + 35

1

5+ 4

f(1) =3

5+

20

5 cos (1)

f(1) =23

5 cos (1)

Correct answer is option 4

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2. Evaluate

3et1 et dt

This integral would be ugly if we just integrate it as it is which is why we need to use substitution to make this

manegable

Let

u = 1 etdu = et dt

Lets shuffle it up

3et1 et dt = 3

du

udx

We will use again the formula discussed above this to evaluate this integral

3

du

udx = 3 u 1

2+ 2

2

= 6

u

3et1 et dt = 6

1 et + C


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4/9

31

x2 dx +

2 x2 dx =

13 x33

1

+

2x 13 x33

1

dx

=

9 13 +

(6 9)

2 13

=

26

3

+

14

3

=40

3

Add them all up, we get:

40

3+

4

3=

44

3


Remember to graph these problems, it will help you in the long run!

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5. Find the volume of the solid generated by revolving the region in the first quadrant bounded by y = 3 x2, y =2x, and y = 0 (the x axis) around the x axis

First, we want to picture the values they have given us. Since these are not complicated functions, lets draw

it out.

After you have drawn it out, make sure to think about the axis you are revolving it in and another question

you want to consider is what methods would be the most appropriate with this problem, here is a list with a

brief explanation:

(a) Disk Method:

i. Disk method is excellent in revolving simple functions that do not have gaps. There are two formulas,

depending on which axis you are revolving the region:

A. Horizontal axis of revolution, revolving on the x axis:

V =

ba

[R (x)]2

dx

B. Vertical axis of revolution, revolving on the y axis:

V =

ba

[R (y)]2

dy

(b) Washer Method:

i. Good method to use if you are dealing with holes in the middle

ii. It is the same thing as the Disk method but instead you will be revolving a region that has a hole

inside it like a washer

iii. Formula is

V =

ba

[R (x)]

2 [r (x)]2

dx

iv. Think of the formula as subtracting the outer circle with the inner circle

(c) Shell method:

i. We will be using strips in this method

ii. Here is a graphical representation of this method:

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iii. The formula is if we rotate around the y-axis:

V =

ba

2rh dx

iv. If we rotate the x axis

V =

ba

2rh dy

Now that we are equipped with all these methods, lets go back to the problem. We know based on the graph

that there is a little triangular gap that we have to account for. I personally like to use the washer method

because it is simple and clean.

First, we have to figure out the bounds that we are revolving this function:

3 x2 = 2x3 x2 2x = 0

x2 + 2x

3 = (x 3) (x 1)

x = 3, 1

Since we are only dealing in the first quadrant, we throw away -3. So our integral is now

10

outer circle inner circle dx = V

10

3 x22 3 x2 2x2 dx = V

Alright, so above is what you would get if you used the washer method. But since option 4 supposedly used

washer method, since it doesnt match with our answer, we throw that out of the window. Option 1 is wrong

because that would mean that we would rotate our region around the y-axis and option 3 looks crazy because

it doesnt correctly account the right inner radius function. So the right answer must be 2.

Websites used: http://library.thinkquest.org/3616/Calc/S3/shell3.jpg

and http://media.wiley.com/Lux/04/180004.image1.jpg

Hope that helps!

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6. Which integral finds the Moment with respect to the x axis, Mx, of the region in the first quadrant bounded byy = x3 and y = x.

First thing to do is to graph the function:

Even though you do not have an access to a graphing calculator, you can still find the intersection between x3

and x by setting them equal to each other.

x3 = x

0 = x

x2 10 = x (x + 1) (x 1)x = 0, 1, 1

Since we are asked to find the region in the first quadrant, we will disregard -1.

Lets discuss a little bit about center of mass. Recall in physics, center of mass deals with some point of an

object that when this object is thrown into the air and if it exhibits rotational motion, it would rotate of its

center of mass. There are two ways of calculating the center of mass, using either horizontal or vertical strips.

They would lead the exact same answer but different method. Since the problem asks us to use the horizontal

strips. This would be the approach taken to solve this problem

Here are the steps to find the center of mass:

1. Graph it out

2. Find the mass:

A = b

a

f(x)

g (x) dx

3 & 4. Find My and Mx

Mx =1

2

ba

[f(x)]2 [g (x)]2 dx

My =

ba

x [f(x)] [g (x)] dx

6. Find M

7 & 8. Find x and y

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x =My

M=

ba

x (f(x) g (x)) dxbx

f(x) g (x) dx=

1

A

ba

x (f(x) g (x)) dx

y =Mx

M=

ba

1

2

[f(x)]

2 [g (x)]2

dxbx

f(x) g (x) dx=

1

A

ba

1

2

[f(x)]

2 [g (x)]2

dx

A =ba

f(x) g (x) dx

But since we are asked to find only Mx, we use the formula above and we get:

1

2

10

x2 x6dx

Which is option 1

7. What integral is obtained from

x

4x2

4 dx by an appropriate trig substitution?First thing that you should notice is

x2 4. Recall the following substitutions:

a2 + x2 = a |sec |a2 x2 = a |cos |x2 a2 = a |tan |

There are very extremely important to remember. There is a figure in the book, page 444 that shows a nice

picture showing the reasoning

sin2 x + cos2 x = 1

Let:

x = 2 sec

dx = 2sec tan d

x

4x2

4 dx = (2sec )4sec2 4sec tan d=

16

sec4

(2tan )(2sec tan ) d

= 64

sec5 tan2 d

Answer would be option 2. Remember the trig integrals and derivatives of their counterpart

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8. Evaluate the integral

x2 + 2

x2 (x2 + 1)dx

We would need to use partial fractions for this problem:

x2 + 2

x2 (x2 + 1)=

A

x2+

Bx + C

x2 + 1

x2 + 2 = A

x2 + 1

+ (Bx + C)

x2

x2 + 2 = Ax2 + A + Bx3 + Cx2

x2 + 2 = (A + C) x2 + Bx3 + A

1 = A + C

0 = B

2 = A

1 = C

Since we now know what the coefficients are, we substitute them to find the integral

2x2 +

1

x2 + 1 dx = 1

x tan1

x + C

The answer is option 1

9. A tank is in the shape of an inverted cone with height 14 feet and radius on the ground 7 ft. If the tank is filled to

within 2 feet of the top with water weighing 62.4lb

ft3. How much work does it take to empty the tank through an

outlet at the top of the tank?

10. Determine the behavior of the limits

A = limx1

ln x

(x 1)2B = lim

x

(x 1)2ln x

11. Evaluate20

x cos(3x) dx

12. Evaluate

10

1

x2 + 10x + 25dx

13. Evaluate

sin2 (5y) dy

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Final Exam Spring 2008