Math 136Final Exam Sample Solutions

December 10, 2016

You may use your calculator and integral table. Show all your work. If you use a formula fromthe tables, state which formula and the values of any constants in the formula.

Problem Points Score

1 25

2 20

3 20

4 20

5 20

6 25

7 20

8 10

9 20

10 20

Total 200

1

1. (25 pts.) The graph of a function y = f(x) isshown in Figure 1(a). Define a new functionF by

F (x) =

∫ x

−2f(t) dt

(a) On which intervals is F (x) increasing?On which intervals is F (x) decreasing?Explain your answer.

By the First Fundamental The-orem of Calculus, F ′ = f .It follows that F is increasingwhere f is positive and decreas-ing where f is negative. So F isincreasing for (approximately)−2 < x < −0.4 and 0.6 < x < 2(in the domain of the graph.)

(b) On which intervals is F (x) concaveup? On which intervals if F (x) concavedown? Explain your answer.

F is concave up where f ′′ > 0and concave down where f ′′ <0. Since F ′′ = (F ′)′ = f ′, this iswhere f is increasing and whereit is decreasing. So F is concaveup on (approximately) the in-tervals −2 < x < −1 and 0.1 <x < 1.5, and concave down onthe intervals −1 < x < 0.1 and1.5 < x < 2.

(c) Is F (2) positive, negative, or zero. Ex-plain your answer.

F (2) is the net area between -2and 2 under the graph of f . Itappears that there is more areaabove the x-axis than below it.We conclude that F (2) > 0.

(d) Use your answers to above to sketch thegraph of F (x) on the axes in Figure 1(b).

(a)

(b)

Figure 1:

2

2. (20 pts.) Figure 2 shows portions of the curves y = 2 cos(π2x), y = x + 1, y = 1, andy = sin3(πx).

(a) Label the curves and their points of in-tersection on the plot.

The cosine curve is the topcurve of the region. The otherthree are the curves at the bot-tom of the region reading fromleft to right. The points of in-tersection of top and bottomcurves occure at (−1, 0) and(1, 0). The bottom curves inter-sect at (0, 1) and (12 , 1).

(b) Compute the area of the region S thatthe curves enclose.

Figure 2:

We need to compute four integrals, one for each curve, and subtract the integralsfor the bottom curves on their respective domains from that of the top curve. Sincey = x+ 1 and y = 1 are lines, we know their areas from geometry and do not haveto evaluate an integral to calculate them. These areas are both equal to 1

2 , so theirsum is 1. Therefore the area A is:

A =

∫ 1

−12 cos(

π

2x) dx− 1−

∫ 1

12

sin3(πx) dx

=4

π

[sin(

π

2x)

]1−1− 1−

∫ 1

12

(1− cos2(πx))(sin(πx) dx

=4

π(1− (−1))− 1−

∫ 1

12

sin(πx)− cos2(πx) sin(πx) dx

=8

π− 1−

[− 1

πcos(πx) +

1

3πcos3(πx)

]112

=8

π− 1− 1

π+

1

3π=

22

3π− 1 ≈ 1.334.

3

3. (20 pts.) Find the location of the centroid ofthe region under the graph of f(x) = e−x/2

between x = 0 and x = 2. (See Figure 3.)

Figure 3:

We need to calculate the following integrals:

M =

∫ 2

0f(x), dx, My =

∫ 2

0xf(x), dx, My =

1

2

∫ 2

0f(x)2, dx

M =

∫ 2

0e−x/2 dx

= −2[e−x/2

]20

= 2(1− e−1) ≈ 1.264

My =

∫ 2

0xe−x/2 dx

=

[−2xe−x/2 +

∫2e−x/2 dx

]20

=[−2xe−x/2 − 4e−x/2

]20

= −8e−1 + 4 ≈ 1.057

Mx =1

2

∫ 2

0e−x, dx

=1

2

[−e−x

]20

=1

2(1− e−2) ≈ 0.432.

Then the centroid is

(x, y) =My

M,Mx

M) ≈ (0.836, 0.342).

4

(a) (b)

Figure 4:

4. (20 pts.) The region R in the first quadrant lies below the graph of f(x) = 1 + ln(x) andabove the graph of g(x) = 2− x for x ≤ 2.

(a) Sketch the region R and the solid of revolution obtained by rotating R around the yaxis.

(b) Find the volume of the solid of revolution in part (a).

Use the method of shells.

2π

∫ 2

1x(1 + ln(x)− (2− x)) dx = 2π

∫ 2

1x ln(x)− x+ x2)) dx

= 2π

[1

2x2 ln(x)− 1

4x2 − 1

2x2 +

1

3x3]21

= 2π

[1

2x2 ln(x)− 3

4x2 +

1

3x3]21

= 2π(1

122 ln(2)) ≈ 9.234.

5

5. (20 pts.) Paperback college calculus texts don’t last forever. In fact, it’s not clear they lastas long as they should. At one small college, where all the calculus courses used the samepaperback text, the average time to failure was determined to be 1.5 semesters, where failurewas defined to be at least 50 pages had fallen out of the book. For the rest of the problem,assume that the time to failure of a calculus text can be modeled by an exponential densityfunction.

(a) What is the formula for the exponential probability density function for this college’scalculus textbook time to failure?

Use one semester as the unit of time. Then the average time to failure is 1.5.The formula is

p(t) =1

1.5e−t/1.5.

(b) If a student plans to take Calculus 1 and Calculus 2 (so 2 semesters of calculus) whatare the chances the student’s text book will fail before they finish Calculus 2?

Evaluate the integral of p from t = 0 to t = 2.

∫ 2

0

1

1.5e−t/1.5 dt =

[−e−t/1.5

]20

= 1− e−2/1.5 ≈ 0.736

Approximately three quarters of the books will fail.

(c) The bookstore at the college will buy back books if they haven not failed. If 100 studentswho take Calculus 1 in the fall decide not to continue in Calculus 2 in the spring, howmany of them would be able to sell their textbooks back to the bookstore at the end ofthe semester?

Compute the integral of p from 0 to 1 and subtract the result from 1 or integratep from 1 to ∞. Multiply this fraction by 100.

1−∫ 1

0

1

1.5e−t/1.5 dt = 1−

[−e−t/1.5

]10

= 1− (1− e−1/1.5) ≈ 0.513.

Fifty-one of the students would be able to sell there books back to the bookstore.

6

Figure 5: ’A’a lava flow, Kilauea Volcano, Hawai‘i, courtesy of the US Geological Survey. ’A’a isthe Hawaiian word for this type of lava and has been adopted by geologists.

6. (25 pts.) ’A’a lava is typically 1000 C◦ when it emerges from a volcano and can take a verylong time to cool depending on its thickness. Assume an ’a’a flow cools according to Newton’slaw of cooling, so that the rate of change of temperature y satisfies

dy

dt= −k(y − T0),

where Ts is the temperature of the surrounding air.

(a) Solve the differential equation for T by separation of variables.

Cross multiply to separate y and dt, then integrate both sides, then solve for y.

∫dy

y − T0=

∫k dt

ln(|y − T0|) = −kt+ C

|y − T0| = ekt+C = Ce−kt

⇒ y = T0 + Ce−kt

(b) Express your solution in terms of the initial temperature y0 = 1000 C◦ of the lava andTs = 25 C◦, the average air temperature.

Since y(0) = y0 = 1000 we see that C = 1000− T0 = 975. The solution is then

y(t) = 25 + 975e−kt

(c) If an ’a’a flow takes 30 days to cool to 200 C◦, what is the value of k?

Solve 200 = 25+975e−k·30 for k. We see that 175975 = e−k·30, so k = − ln(175/975)/30 ≈

0.057.

(d) What will be the temperature of this flow after six months?

Evaluate y(180) = 25 + 975e−0.057·180 ≈ 25.034.

7

7. (20 pts.) A chemistry lab requires the accurate measurement of 100 mg of a compound thatis used in the experiment. In order for the experiment to succeed, the measurement has to beaccurate within 0.5 mg. From double checking the students’ measurements on a very accuratescale, she knows that the errors in measurement will be normally distributed with a mean of102 mg and a standard deviation of 1.2 mg.

The normal density with mean µ and standard deviation σ is

p(x) =1

σ√

2πe−

(x−µ)2

2σ2

(a) Sketch the normal density for this problem on the axes in Figure 7. Shade the regionunder the graph of p that would lead to a successful experiment.

The region from 99.5 to 100.5 should be shaded.

(b) For what fraction of students will their measurements lead to a successful experiment?

Integrate p from 99.5 to 100.5 using a calculator:∫ 100.5

99.5p(x) dx ≈ 0.087.

The fraction is 0.087 of the students.

(c) For what fraction of students will their measurements be too high to lead to a successfulexperiment?

Integrate p from 100.5 to ∞ using a calculator:∫ ∞100.5

p(x) dx ≈ 0.894.

The fraction is 0.894 of the students.

(d) The lab supervisor usually lets students whose experiments do not work repeat the lab.If there were 32 students in the lab section, how many should the lab supervisor expectin the repeat lab?

The fraction whose experiments do not work is approximately 0.913. Multiply-ing by 32, we see that 29 or 30 students will want to repeat the lab.

Figure 6:

8

8. (10 pts)

(a) Does the following sequence converge or diverge? If it converges, what is its limit?

limn→∞

sin(n2)

n2

The numerator is between -1 and 1 and the denominator goes to∞, so the limitis 0.

(b) Does the following series converge or diverge? If it converges, what is its limit?

3

π2− 9

π3+

27

π4− 81

π5+ · · ·

This is a geometric series with ratio r = − 3π . Since this is between -1 and 1,

the series converges to

3/π2

1− (−3/π)=

3

π(π + 3)

9. (20 pts.) Determine whether the following series converge or diverge as indicated.

(a) Use the integral test to determine whether the following series converges or diverges.

∞∑n=1

2

n(n+ 1).

Use the method of partial fractions to evaluate the integral∫1

x(x+ 1)dx

Solve1

x(x+ 1)=A

x+

B

x+ 1.

Cross multiplying, we must have 1 = A(x + 1) + Bx. Setting x = 0, we see1 = A. Setting x = −1, we see that B = −1. Therefore∫

1

x(x+ 1)dx =

∫1

x− 1

x+ 1dx = ln |x| − ln |x+ 1|+ C = ln |x/(x+ 1)|+ C.

Use this to evaluate the corresponding improper integral:

limb→∞

∫ b

1

1

x(x+ 1)dx = lim

b→∞[ln |x/(x+ 1)|]b1 = lim

b→∞ln |b/(b+1)|−ln(

1

2= ln(1)+ln(2) = ln(2).

Since the limit exists and is finite, the improper integral converges. We concludethe series converges.

(b) Use the comparison test to determine whether the following series converges or diverges.

∞∑n=2

1

ln(n2).

9

Notice that ln(n2) = 2 ln(n), so that the series is

1

2

∞∑n=2

1

ln(n).

It suffices to consider the series without the factor of 12 . Since ln(n) < n,

1ln(n) >

1n . This shows that term by term the series is larger than the harmonic

series. Since the harmonic series diverges, the original series diverges.

10. (20 pts.)

(a) Find the radius and interval of convergence of the following series:

∞∑n=0

n2x2n

4n

Evaluate the limit of the ratio of consecutive terms:

limn→∞

∣∣∣∣∣∣(n+1)2x2(n+1)

4n+1

n2x2n

4n

∣∣∣∣∣∣ = limn→∞

∣∣∣∣∣(n+ 1)2x2

4n2

∣∣∣∣∣ = 4x2 limn→∞

∣∣∣∣∣(n+ 1)2

n2

∣∣∣∣∣ = 4x2 · 1 = 4x2.

Since this limit must be less than 1 in order for there to be convergence, wehave 4x2 < 1 or −2 < x < 2.

(b) i. What is the Taylor series for sin(x) centered at a = 0?

∞∑0

(−1)n+1 x2n+1

(2n+ 1)!

ii. Plot sin(x) and the Taylor polynomials for sin(x) up to degree 5 on the axes below.

These polynomials are T1(x) = x, T3(x) = x − x3/3!, T5(x) = x − x3/3! +x5/5!.

iii. What does this indicate about the Taylor polynomials and the Taylor series?

This indicates that as n increases the Taylor polynomials become betterapproximations to sin(x) on larger and larger intervals.

10

Figure 7:

11