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)...

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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.

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!

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

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

...

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

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

...

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

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

...

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

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

...

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

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

...

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

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

...

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 i^th interval as our x_i: 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[x^3/2]^π/2₀ −1

2[cos x]^π/2₀

= 2π^3/2 3 · 2^3/2 +1

2

= 2π^3/2 3 ·√

4 · 2+ 1 2

= 2π^3/2 3 · 2 ·√

2+ 1 2

= π^3/2 3√

2 +1 2

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(a) π^3/2 3√

2 +1

2 (b) 2π^3/2

3√

2 (c) π^3/2

3√ 2 (d) 2π^3/2

3√ 2 +1

2 (e) π^3/2

3√ 2− 1

2

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

-2 -1 1 2 3^x

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 20^y

(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) = 4t^1/2 + C for some constant C. We are also given that the initial velocity is 1 so that v(0) = C = 1. Thus, v(t) = 4^1/2+ 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 (y^1/3) =R1

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

−f (y^1/3) = 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(y^1/3) ∗ 1/3y^−2/3=p

1 + sin(y)

⇒ f⁰(y^1/3) = −3y^2/3p

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(x^2/3)

√3

x dx.

(a) 3

2cos(x^2/3) + C (b) −2

3cos(x^2/3) + C

(c) − cos(x^2/3) + C (d) −3

2cos(x^2/3) + C (e) cos(x^2/3) + C

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

3x^−1/3 and Z

sin(u) du = 2 3

Z sin(x^2/3)

√3

x dx

⇒ − cos(u) + C = 2 3

Z sin(x^2/3)

√3

x dx

⇒

Z sin(x^2/3)

√3

x dx = −3

2cos(x^2/3) + 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(x^1/2).

We see that F⁰(x) = sec²(x^1/2)

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