# PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 2. (a) (b) (c) (d) (e) (a) (b) (c) (d) (e) 4. (a) (b) (c) (d) (e)...

## Full text

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Math 10550, Practice EXAM IV November 18, 3016

• The Honor Code is in effect for this examination. All work is to be your own.

• No calculators.

• The exam lasts for 1 hour and 15 min.

• Be sure that your name is on every page in case pages become detached.

• Be sure that you have all 14 pages of the test.

1. (a) (b) (c) (d) (e)

2. (a) (b) (c) (d) (e)

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3. (a) (b) (c) (d) (e)

4. (a) (b) (c) (d) (e)

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5. (a) (b) (c) (d) (e)

6. (a) (b) (c) (d) (e)

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7. (a) (b) (c) (d) (e)

8. (a) (b) (c) (d) (e)

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9. (a) (b) (c) (d) (e)

10. (a) (b) (c) (d) (e)

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11. (a) (b) (c) (d) (e)

12. (a) (b) (c) (d) (e)

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13. (a) (b) (c) (d) (e)

Please do NOT write in this box.

Multiple Choice

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Multiple Choice

1.(6 pts.) If we want to use Newton’s method to find an approximate solution to cos(x) − x = 0

with initial approximation x1 = π

2, what is x2? (a) 3π

4 (b) π

2 (c) π

4 (d) 0 (e) π

Solution: Take f (x) = cos(x) − x. Using that f0(x) = − sin(x) − 1 and the formula x2 = x1− f (x1)

f0(x1), we have

x2 = π

2 − −π/2

−2

= π/2 − π/4

= π/4

2.(6 pts.) Newton’s method is used to find a root of the equation x3 − x2+ 3x + 1 = 0.

If x1 = −1, find x2.

Solution: First compute the derivative:

3x2− 2x + 3.

Then

x2 = x1+x31− x21+ 3x1 + 1

3x21− 2x1+ 3 = −1 + −1 − 1 − 3 + 1

3 + 2 + 3 = 1 + −4 8 = 1

2 (a) 1

2 (b) −3

2 (c) 1

(d) −1

2 (e) −3

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3.(6 pts.) Find the right endpoint approximation to the definite integral Z 2

0

10 1 + xdx

using four approximating rectangles of equal base width.

Solution: First we divide our integral into four subintervals, [0, 1/2], [1/2, 1], [1, 3/2], [3/2, 2]. Then because we’re doing right approximations we choose the rightmost endpoint of the ith interval as our xi: x1 = 1/2, x2 = 1, x3 = 3/2, and x4 = 1. Now we compute f (xi) for each i: f (x1) = 10

3/2 = 20

3 , f (x2) = 5 = 15

3 , f (x3) = 10 5/2 = 20

5 = 4 = 12 3 , and f (x4) = 10

3 . Finally, we sum the areas of each rectangle:

x1f (x1) + x2f (x2) + x3f (x3) + x4f (x4) = 1 2

20 3 +1

2 15

3 +1 2

12 3 +1

2 10

3

= 1 2

 57 3



= 19 2

(a) 47

6 (b) 19

2 (c) 47

3

(d) 19 (e) 125

6

4.(6 pts.) Compute

Z 0

−5

25 − x2dx.

Solution: Notice that points (x, y) satisfying y2 + x2 = 25 lie on the circle of radius 5.

Then y =√

25 − x2 defines the points above the x-axis on the circle (and y = −√

25 − x2 defines points below it). Hence, the integral from −5 to 0 is the area of a quarter of a circle of radius 5. That is 25π

4 .

(a) 25π (b) 25π

2 (c) 25π

4

(d) 5 (e) 25

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5.(6 pts.) If f (x) = Z x2

0

cos(t2)dt, find f0(x).

Solution: Recall the fundamental theorem of calculus:

G(x) = Z x

a

g(t) dt,

where g is continuous on [a, b] and differentiable on (a, b), and G is the antiderivative of g, i.e., G0(x) = g(x). Let

h(x) = Z x

0

cos(t2) dt.

Then by what we just said, h0(x) = cos(x2). Also, f (x) = h(x2). Hence, by the chain rule,

f0(x) = 2xh0(x2) = 2x cos((x2)2) = 2x cos(x4).

(a) 2x cos(x2) (b) −2x sin(x2) (c) cos(x4) (d) −2x2sin(x4) (e) 2x cos(x4)

6.(6 pts.) Calculate the following definite integral Z π/2

0

√x + 1

2sin(x) dx.

Solution:

Z π/2 0

√x dx + Z π/2

0

1

2sin(x) dx = 2

3[x3/2]π/20 −1

2[cos x]π/20

= 2π3/2 3 · 23/2 +1

2

= 2π3/2 3 ·√

4 · 2+ 1 2

= 2π3/2 3 · 2 ·√

2+ 1 2

= π3/2 3√

2 +1 2

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(a) π3/2 3√

2 +1

2 (b) 2π3/2

3√

2 (c) π3/2

3√ 2 (d) 2π3/2

3√ 2 +1

2 (e) π3/2

3√ 2− 1

2

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7.(6 pts.) The graph of f (x) is shown below:

-2 -1 1 2 3x

5 10 15 20

y

Which of the following gives the graph of an antiderivative for the function f (x)?

Solution: Notice that our answer can vary by adding a constant, i.e., a vertical shift.

There are a couple ways of thinking about this problem. If we think of the integral as area, then from −2 to 1 and 1 to 3 we can see that we’re gaining signed area, so we should expect our function to be increasing on these intervals. The only graph that does this is a (and notice it slightly decreases between 1 and 2 corresponding to losing signed area).

Alternatively, by the fundamental theorem of calculus, we can think of this problem as the graph of the derivative of F (x). Then the derivative is positive from −2 to 1 and 2 to 3, so F (x) is increasing on these intervals.

(a)

-2 -1 1 2 3 x

-10 -5 5 10 15 20y

(b) -2 -1 1 2 3 x

-40 -20 20

y

(c)

-2 -1 1 2 3 x

-50 -40 -30 -20 -10

y

(d)

-2 -1 1 2 3 x

-10 10 20 30

y

(e) -2 -1 1 2 3

x

-20 -15 -10 -5 5 10

y

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8.(6 pts.) A bug being chased by a kitten (both moving in a straight line) enters a kitchen with velocity 1 ft/sec, and accelerates at 2

√tft/sec2. How fast is the bug moving 9 seconds later.

(a) 4 ft/sec (b) 5 ft/sec (c) 37 ft/sec

(d) 13 ft/sec (e) 7 ft/sec

Solution: Since acceleration is given by a(t) = 2t−1/2 and the derivative of velocity is acceleration, we know that after integrating acceleration, the velocity is given by v(t) = 4t1/2 + C for some constant C. We are also given that the initial velocity is 1 so that v(0) = C = 1. Thus, v(t) = 41/2+ 1. Thus, v(9) = 12 + 1 = 13.

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9.(6 pts.) The graph shown below is that of f (x) for −1 ≤ x ≤ 4 where

f (x) =





2 if − 1 ≤ x ≤ 0

√4 − x2 if0 < x ≤ 2 4 − 2x if2 ≤ x ≤ 4 Which of the following equals

Z 4

−1

f (x)dx?

-1 1 2 3 4

-4 -3 -2 -1 1 2

y = f(x)

(a) 0 (b) π − 2 (c) 2π − 2

(d) π (e) 6 + π

Solution: From −1 to 0, the function is constant with output 2. So the area under the curve is given by the area of a rectangle with base 1 and height 2. The area of such a rectangle is 2. From 0 to 2, we see that the function traces out the upper left portion of a circle with radius 2 centered at (0, 0) and hence, we have that the area is given by

1

4π22 = π. From 2 to 4, we see that the curve traces the hypotenuse of a right triangle with base 2 and height 4. The area of such a triangle is 1

2(2)(4) = 4. However, since function is under the x−axis from 2 to 4, we need to account for this by subtracting the area of the triangle. So the integral is evaluated to be 2 + π − 4 = π − 2.

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10.(6 pts.) If f (x) = Z 1

x3

p1 + sin(t) dt, then f0(x) =

(a) p1 + sin(x3) (b) −p1 + sin(x3) (c) −3x2p1 + sin(x3) (d) p1 + sin(x) (e) 3x2p1 + sin(x3)

Solution: Setting y = x3, we rewrite as f (y1/3) =R1

y p1 + sin(t) dt. Then we have

−f (y1/3) = Z y

1

p1 + sin(t) dt.

Taking the derivative of both sides, we see via the chain rule and the Fund. Thm. of Calc. that

−f0(y1/3) ∗ 1/3y−2/3=p

1 + sin(y)

⇒ f0(y1/3) = −3y2/3p

1 + sin(y)

⇒ f0(x) = −3x2p

1 + sin(x3)

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11.(6 pts.) Evaluate Z 4

0

|x − 2|dx.

(a) 0 (b) 2 (c) 4 (d) −2 (e) 6

Solution: We see that x − 2 is positive when x > 2 and x − 2 is negative when x < 2.

We compute Z 4

0

|x − 2|dx = Z 2

0

(2 − x)dx + Z 4

2

(x − 2)dx = [2x −x2

2 ]20+ [x2

2 − 2x]42 = (4 − 2) + (8 − 8) − (2 − 4) = 4.

12.(6 pts.) Evaluate

Z sin(x2/3)

3

x dx.

(a) 3

2cos(x2/3) + C (b) −2

3cos(x2/3) + C

(c) − cos(x2/3) + C (d) −3

2cos(x2/3) + C (e) cos(x2/3) + C

Solution: Setting u = x2/3, we have du = 2

3x−1/3 and Z

sin(u) du = 2 3

Z sin(x2/3)

3

x dx

⇒ − cos(u) + C = 2 3

Z sin(x2/3)

3

x dx

Z sin(x2/3)

3

x dx = −3

2cos(x2/3) + C.

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13.(6 pts.) Evaluate

Z 2 1

sec2(√

√ x)

x dx.

(a) tan(2)

2 − tan(1)

2 (b) 2 tan(√

2) − 2 tan(1)

(c) tan(√ 2)

2 − tan(1)

2 (d) tan(√

2) − tan(1) (e) 2 tan(2) − 2 tan(1)

Solution: First we look for an antiderivative. We try the function F (x) = tan(x1/2).

We see that F0(x) = sec2(x1/2)

2x1/2 . Thus, our initial guess is incorrect, but we see that F (x) = 2tan(x1/2) is an antiderivative for the integrand. Applying the FTC, we see that the answer is F (2) − F (1) = 2tan(21/2) − 2tan(1).

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Partial Credit

You must show your work on the partial credit problems to receive credit!

14.(13 pts.) A particle traveling on a line accelerates at a rate of 6t ft/sec2.

(a) If the initial velocity is known to be 12 ft/sec, find the distance traveled by the particle in the first second.

Solution: We are given that the acceleration a(t) = 6t. Since the derivative of velocity is acceleration, we know that velocity is of the form v(t) = 3t2+ C for some constant C.

The initial velocity is 12, so we know that v(0) = 12 = C. Thus, v(t) = 3t2 + 12. By inspection, we see that the velocity is always positive, and hence the displacement of the particle is the same as the distance traveled. Hence, the total distance traveled is given by

Z 1 0

(3t2+ 12)dx = [t3+ 12t]10 = 13f t.

(b) If the initial velocity is positive but not known, and the particle is known to travel 20 ft in the first 2 seconds, then find the initial velocity.

(Hint: A definite integral might be useful).

Solution: As on the actual exam, we assume that the initial velocity is nonnegative.

We know that the velocity v(t) = 3t2 + C where C is the initial velocity. Notice that the velocity is never negative. Hence, the displacement is the same as the total distance traveled. We also know that the displacement after 2 seconds is 20 ft and so we know that

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20 = Z 2

0

v(t)dt = Z 2

0

3t2+ Cdt = [t3 + Ct]20 = 8 + 2C.

Thus, we solve for C in the equation 8 + 2C = 20 and obtain C = 6.

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15.(5 pts.) Evaluate the right endpoint approximation, with 4 approximating rectangles, of the definite integral

Z 2 0

x 2 dx

Solution: We compute ∆x = (2 − 0)/4 = 2/4 = 1/2 and

x0 = 0, x1 = 1/2, x2 = 1, x3 = 3/2, x4 = 2.

Then

Z 2 0

x

2 dx ≈ R4 = f (x1)∆x + f (x2)∆x + f (x3)∆x + f (x4)∆x

= 1 2

 1 4+ 1

2+ 3 4+ 1



= 5 4

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Math 10550, Practice EXAM IV November 18, 3016

• The Honor Code is in effect for this examination. All work is to be your own.

• No calculators.

• The exam lasts for 1 hour and 15 min.

• Be sure that your name is on every page in case pages become detached.

• Be sure that you have all 14 pages of the test.

1. (a) (b) (•) (d) (e)

2. (a) (b) (c) (•) (e)

...

3. (a) (•) (c) (d) (e)

4. (a) (b) (•) (d) (e)

...

5. (a) (b) (c) (d) (•)

6. (•) (b) (c) (d) (e)

...

7. (a) (b) (c) (d) (•)

8. (a) (b) (c) (•) (e)

...

9. (a) (•) (c) (d) (e)

10. (a) (b) (•) (d) (e)

...

11. (a) (b) (•) (d) (e)

12. (a) (b) (c) (•) (e)

...

13. (a) (•) (c) (d) (e)

Please do NOT write in this box.

Multiple Choice

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