Holy Cross College, Spring Semester, 2005 MATH 132, Section 01, Final Exam Solutions May 11

1. The following graph shows the function y = f(x).

a. [10 points] Let F (x) be the antiderivative of f(x) with F (0) = 0. Compute the entries in the following table of values: Solution:

\begin{matrix}x & 0 & 1 & 2 & 3 & 4 \\ F (x) & 0 & −1 & 0 & .5 & −.5 \end{matrix}

For example, the area between the graph and the x-axis from x = 0 to x = 1 (counted with a negative sign) is −1. So by the Fundamental Theorem of Calculus:

−1 = ∫_{0}^{1} f(x) dx = F (1) − F (0) = F (1) − 0 ⇒ F (1) = −1

The others are obtained similarly.

b. [10 points] Sketch the graph y = F (x) on the interval 0 ≤ x ≤ 4:

2. Integrate each of the following. You must show all work and cite any table entries you use for full credit.

MATH 132-01, Final Exam Solutions 2

a. [5 points] ∫ \frac{x^{2} + 3x + 5}{x^{3/2}} dx

Solution: Divide x^{3/2} into each term on the top, then integrate:

∫ \frac{x^{2} + 3x + 5}{x^{3/2}} dx = ∫ x^{1/2} + 3x^{−1/2} + 5x^{−3/2}
= \frac{2}{3}x^{3/2} + 6x^{1/2} − 10x^{−1/2} + C

b. [10 points] ∫ \frac{e^{t} + 1}{(e^{t} + t)^{2}} dt

Solution: Let u = e^{t} + t, so du = (e^{t} + 1) dt. Then the form is

∫ \frac{du}{u^{2}} = \frac{−1}{u} + C = \frac{−1}{e^{t} + t} + C.

c. [10 points] ∫ \frac{dx}{(9 − x^{2})^{3/2}}

Solution: We use the trigonometric substitution x = 3 sin θ, so dx = 3 cos θ and

∫ \frac{dx}{(9 − x^{2})^{3/2}} = ∫ \frac{3 cos θ dθ}{27 cos^{3} θ}
= \frac{1}{9} ∫ \frac{dθ}{cos^{2} θ}
= \frac{1}{9} tan θ + C
= \frac{1}{9} \frac{sin θ}{cos θ} + C
= \frac{1}{9} \frac{x}{\sqrt{9 − x^{2}}} + C

d. [10 points] ∫_{0}^{1} x^{2}e^{2x} dx Solution: Use integration by parts twice, or # 14 in the table of integrals:

∫ x^{2}e^{2x} dx = (\frac{x^{2}}{2} − \frac{x}{2} + \frac{1}{4}) e^{2x} + C

Then for the definite integral

∫ x^{2}e^{2x} dx = (\frac{x^{2}}{2} − \frac{x}{2} + \frac{1}{4}) e^{2x}|_{0}^{1} = \frac{1}{4}(e^{2} − 1) ≐ 1.597

e. [10 points] ∫ \frac{(x − 1) dx}{x^{3} + 25x}

Solution: Use partial fractions based on the factorization x^{3} + 25x = x(x^{2} + 25). We have

\frac{x − 1}{x(x^{2} + 25)} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 25} .

Clearing denominators,

x − 1 = A(x^{2} + 25) + (Bx + C)x = (A + B)x^{2} + Cx + 25A

Equating coefficients of like powers of x: (A + B) = 0, C = 1, and 25A = −1. So A = \frac{−1}{25} and B = \frac{−1}{25}. Then using # 25 in the table

∫ \frac{\frac{−1}{25}}{x} + \frac{\frac{1}{25}x + 1}{x^{2} + 25} = \frac{−1}{25} ln |x| + \frac{1}{50} ln |x^{2} + 25| + \frac{1}{5} arctan (\frac{x}{5}) + C

3. All parts of this problem refer to the region in the plane bounded by y = 2 + cos(x), y = 0, x = 0, and x = π, shown below:

a. [10 points] Find the volume of the solid with base R whose cross-sections by planes perpendicular to the x-axis are squares.

Solution: We have, using # 18 in the table

V = ∫_{0}^{π}(2 + cos x)^{2} dx
= ∫_{0}^{π} 4 + 4 cos x + cos^{2} x dx
= (4x + 4 sin x + \frac{1}{2}x + \frac{1}{2} sin x cos x|_{0}^{π}
= \frac{9π}{2}

b. [10 points] True or False: The solid obtained by rotating R about the x-axis has volume π times the volume from part a. (For full credit, give the integral that would compute the volume of this solid of revolution, and answer the question.)

Solution: This is TRUE, since the volume of the solid of revolution is

∫_{0}^{π} π(2 + cos x)^{2} dx = π ∫_{0}^{π}(2 + cos x)^{2} dx.

c. [15 points] A thin metal plate has the shape of the region R (units in meters) and constant mass density 1kg/m^{2}. Find the x-coordinate of its center of mass.

Solution: We have, integrating by parts on the top

\overline{x} = \frac{∫_{0}^{π} x(2 + cos x) dx}{∫_{0}^{π}(2 + cos x) dx}
= \frac{x^{2} + x sin x + cos x|_{0}^{π}}{2x + sin x|_{0}^{π}}
= \frac{π^{2} − 2}{π}

d. [10 points] Set up, but do not evaluate the integral giving the arclength of the top edge of R.

Solution: The arclength integral is L = ∫_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^{2}} dx. Here, since y = 2 + cos x, \frac{dy}{dx} = − sin x and

L = ∫_{0}^{π} \sqrt{1 + sin^{2} x} dx.

4. [5 points each] The following graph shows a probability density function (pdf) for a quantity x:

a. Which of the following two plots shows the corresponding cumulative distribution for x:

,

Solution: Graph B is the correct cdf for the given density function. We can see this because the given density should be the derivative of the cdf. Plot B has two maxima for the slope, corresponding to the two “peaks”.

b. Using the appropriate graph, estimate the median of x.

Solution: The median is the T -value where ∫_{−∞}^{T} p(x) dx = \frac{1}{2}. Drawing in a horizontal line in plot B, it appears that the line y = 1/2 intersects the graph at about x = 1. That is the median value.

5. [5 points each] Does each of the following series converge? For full credit, you must justify your answer completely by showing how the indicated test leads to your stated conclusion.

a. ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{1/3}} – alternating series test.

MATH 132-01, Final Exam Solutions 6 Solution: lim_{n→∞} \frac{1}{n^{1/3}} = 0 and for all n ≥ 1, \frac{1}{n^{1/3}} > \frac{1}{(n+1)^{1/3}} . Hence the alternating series test (see p. 420 in the textbook) implies that this series converges. b. ∑_{n=1}^{∞} \frac{1}{n^{9/10}} – integral test Solution: The improper integral

∫_{1}^{∞} \frac{dx}{x^{9/10}} = lim_{b→∞} 10x^{1/10}|1^{b}

does not converge, because as b → ∞, so does b^{1/10}. Hence by the integral test, this series diverges. c. ∑_{n=1}^{∞} \frac{1}{ne^{n}} – any applicable method

Solution: Method 1: Since n ≥ 1, \frac{1}{ne^{n}} < \frac{1}{e^{n}}. The series ∑_{n=1}^{∞} \frac{1}{e^{n}} is a geometric series with ratio \frac{1}{e} < 1. Hence it converges. By the comparison test, ∑_{n=1}^{∞} \frac{1}{ne^{n}} converges too. Method 2: Apply the ratio test:

lim_{n→∞} |\frac{1}{(n + 1)e^{n+1}} · \frac{ne^{n}}{1} | = lim_{n→∞} \frac{n}{n + 1} · \frac{1}{e} = \frac{1}{e}

Since \frac{1}{e} < 1, the ratio test implies the series converges.

6. [10 points] Using the ratio test, determine the radius of convergence of the following power series

∑_{n=1}^{∞} \frac{(x + 2)^{n}}{n^{2}4^{n}}

Solution:

lim_{n→∞} |\frac{(x + 2)^{n+1}}{(n + 1)^{2}4^{n+1}} · \frac{n^{2}4^{n}}{(x + 2)^{n}} | = lim_{n→∞} \frac{n^{2}}{(n + 1)^{2}} \frac{1}{4}|x + 2| = \frac{1}{4}|x + 2|

The radius of convergence is 1/(1/4) = 4.

7. All parts of this problem refer to f(x) = cos (\frac{x}{2}). a. [15 points] Using the definition of Taylor polynomials, compute the Taylor polynomial of degree n = 6 for f(x) at a = 0.

Solution: The problem said to use the definition of Taylor polynomials, so we compute f′(x) = \frac{−1}{2} sin (\frac{x}{2}), etc. The 6th degree polynomial is

p_{6}(x) = ∑_{i=0}^{6} \frac{f^{(i)}(0)}{i!}x^{i} = 1 − \frac{1}{8}x^{2} + \frac{1}{384}x^{4} − \frac{1}{46080}x^{6}

b. [5 points] Use your answer from part a to approximate cos(1.2).

Solution: Since f(x) = cos (\frac{x}{2}), to approximate cos(1.2), we substitute x = 2.4 into p_{6}(x) from part a. (Note: p_{6}(x) is supposed to approximate cos (\frac{x}{2}), so to get 1.2 “inside” the cosine, we want x = 2.4. We get

p_{6}(2.4) = 1 − \frac{1}{8}(2.4)^{2} + \frac{1}{384}(2.4)^{4} − \frac{1}{46080}(2.4)^{6} ≐ 1 − .72_{.}0864 − .0041472 ≐ .3622528

c. [10 points] Using the error bound for Taylor approximations, estimate how big n must be to guarantee 4 decimal place accuracy in approximations to f(x) for all x between 0 and 2.

Solution: The (n + 1)st derivative of f(x) is \frac{1}{2^{n+1}} times either ± sin x or ± cos x. Hence we can use M = \frac{1}{2^{n+1}} in the error bound, and we get for x between 0 and 2:

|f(x) − p_{n}(x)| ≤ \frac{\frac{1}{2^{n+1}}}{(n + 1)!}2^{n+1} = \frac{1}{(n + 1)!}.

We have \frac{1}{8!} ≐ .000025 so n = 7 (or n = 6, since p_{6} = p_{7} for cos) is large enough.

8. When a course ends, it’s an unfortunate fact of life that students start to forget the material they have learned. One mathematical model for this process states that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant a < 1. Let y be the fraction of the original material remembered t weeks after the course has ended. Then y(0) = 1, and the model assumes y(t) > a for all t.

a. [5 points] Which of the following differential equations is the translation of the model described above:

I : \frac{dy}{dt} = −k(y − a) II : \frac{dy}{dt} = \frac{−k}{y − a}

Solution: I is the correct equation. b. [10 points] Solve the equation you selected in part a by separation of variables.

Solution: Dividing by y − a separates so

∫ \frac{dy}{y − a} = ∫ −k dt
ln |y − a| = −kt + c
y = a + de^{−kt}

where d = ±e^{c}. Note that it is given in the problem that y(0) = 1. This says 1 = y(0) = a + d · 1, so d = 1 − a.

c. [10 points] Suppose a = .2, and after 3 weeks the student only remembers 80% of the material from the course (that is, y(3) = .8). How long will it be before the student only remembers one half of the material?

Solution: Using this last comment, y(t) = .2+.8e^{−kt}. From the given information y(3) = .8, so

.8 = .2 + .8e^{−3k} ⇒ k = \frac{ln(.6/.8)}{−3} ≐ .0959

Then we want to solve for t:

.5 = .2 + .8e^{(−.0959)t} ⇒ t = \frac{ln(.3/.8)}{−.0959} ≐ 10.23

That is, the student will only remember half of the material 10.23 weeks after the end of the class.