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

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 x_{1} = π

2, what is x_{2}?
(a) 3π

4 (b) π

2 (c) π

4 (d) 0 (e) π

Solution: Take f (x) = cos(x) − x. Using that f^{0}(x) = − sin(x) − 1 and the formula
x_{2} = x_{1}− f (x_{1})

f^{0}(x_{1}), we have

x_{2} = 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
x^{3} − x^{2}+ 3x + 1 = 0.

If x_{1} = −1, find x_{2}.

Solution: First compute the derivative:

3x^{2}− 2x + 3.

Then

x_{2} = x_{1}+x^{3}_{1}− x^{2}_{1}+ 3x_{1} + 1

3x^{2}_{1}− 2x_{1}+ 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

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}: x_{1} = 1/2, x_{2} = 1, x_{3} = 3/2, and x_{4} = 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:

x_{1}f (x_{1}) + x_{2}f (x_{2}) + x_{3}f (x_{3}) + x_{4}f (x_{4}) = 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 − x^{2}dx.

Solution: Notice that points (x, y) satisfying y^{2} + x^{2} = 25 lie on the circle of radius 5.

Then y =√

25 − x^{2} defines the points above the x-axis on the circle (and y = −√

25 − x^{2}
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

5.(6 pts.) If f (x) =
Z x^{2}

0

cos(t^{2})dt, find f^{0}(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., G^{0}(x) = g(x). Let

h(x) = Z x

0

cos(t^{2}) dt.

Then by what we just said, h^{0}(x) = cos(x^{2}). Also, f (x) = h(x^{2}). Hence, by the chain
rule,

f^{0}(x) = 2xh^{0}(x^{2}) = 2x cos((x^{2})^{2}) = 2x cos(x^{4}).

(a) 2x cos(x^{2}) (b) −2x sin(x^{2}) (c) cos(x^{4})
(d) −2x^{2}sin(x^{4}) (e) 2x cos(x^{4})

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/2}_{0} −1

2[cos x]^{π/2}_{0}

= 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

(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

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*

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/sec^{2}. 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.

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 − x^{2} 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

4π2^{2} = 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.

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

x^{3}

p1 + sin(t) dt, then f^{0}(x) =

(a) p1 + sin(x^{3}) (b) −p1 + sin(x^{3}) (c) −3x^{2}p1 + sin(x^{3})
(d) p1 + sin(x) (e) 3x^{2}p1 + sin(x^{3})

Solution: Setting y = x^{3}, 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

−f^{0}(y^{1/3}) ∗ 1/3y^{−2/3}=p

1 + sin(y)

⇒ f^{0}(y^{1/3}) = −3y^{2/3}p

1 + sin(y)

⇒ f^{0}(x) = −3x^{2}p

1 + sin(x^{3})

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 −x^{2}

2 ]^{2}_{0}+ [x^{2}

2 − 2x]^{4}_{2} = (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.

13.(6 pts.) Evaluate

Z 2 1

sec^{2}(√

√ 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 F^{0}(x) = sec^{2}(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).

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/sec^{2}.

(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) = 3t^{2}+ C for some constant C.

The initial velocity is 12, so we know that v(0) = 12 = C. Thus, v(t) = 3t^{2} + 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

(3t^{2}+ 12)dx = [t^{3}+ 12t]^{1}_{0} = 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) = 3t^{2} + 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

20 = Z 2

0

v(t)dt = Z 2

0

3t^{2}+ Cdt = [t^{3} + Ct]^{2}_{0} = 8 + 2C.

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

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 ≈ R_{4} = f (x_{1})∆x + f (x_{2})∆x + f (x_{3})∆x + f (x_{4})∆x

= 1 2

1 4+ 1

2+ 3 4+ 1

= 5 4

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) (•) (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