Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Practice Exam 2 Solutions - Calculus II - Spring 2008 | MATH 2202, Exams of Calculus

Material Type: Exam; Professor: Ellermeyer; Class: Calculus II; University: Kennesaw State University; Term: Spring 2008;

Typology: Exams

2010/2011

Uploaded on 06/03/2011

koofers-user-a5f
koofers-user-a5f 🇺🇸

10 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
MATH 2202 –Exam 3 (Version 2) Solutions
April 2, 2008
S. F. Ellermeyer Name
Instructions. Your solution to each problem must contain su¢ cient detail so that it is clear
to the reader what your reasoning process is. Full credit will not be given for solutions that
are lacking in detail (such as omission of important steps in computing an integral, etc.)
State your conclusions and other important steps in your reasoning in complete sentences.
You may use a calculator on this exam but you may not use any books or notes.
1. Use L’Hospital’s Rule to show that
lim
x!1
(ln (x))2
x= 0.
You must include all details of your reasoning, use correct notation, and write in
complete sentences.
Solution: First note that limx!1 x=1which means that we can attempt to use
L’Hospital’s Rule. Now consider the problem
lim
x!1
d
dx (ln (x))2
d
dx (x)= lim
x!1
2 (ln (x)) 1
x
1= lim
x!1
2 ln (x)
x.
Since limx!1 x=1, we can again try L’Hospital’s Rule. Since
lim
x!1
d
dx (2 ln (x))
d
dx (x)= lim
x!1
2
x
1= 0,
then
lim
x!1
(ln (x))2
x= 0
by L’Hospital’s Rule.
2. Show that Z1
8
(ln (x))2
xdx =1.
You must include all details of your reasoning (steps in computing integrals, etc.).
Solution: To evaluate the indenite integral
Z(ln (x))2
xdx,
we use the simple substitution u= ln (x),du =1
xdx. This gives us
Z(ln (x))2
xdx =Zu2du =1
3u3+C=1
3(ln (x))3+C.
1
pf3

Partial preview of the text

Download Practice Exam 2 Solutions - Calculus II - Spring 2008 | MATH 2202 and more Exams Calculus in PDF only on Docsity!

MATH 2202 ñExam 3 (Version 2) Solutions April 2, 2008 S. F. Ellermeyer Name

Instructions. Your solution to each problem must contain su¢ cient detail so that it is clear to the reader what your reasoning process is. Full credit will not be given for solutions that are lacking in detail (such as omission of important steps in computing an integral, etc.) State your conclusions and other important steps in your reasoning in complete sentences. You may use a calculator on this exam but you may not use any books or notes.

  1. Use LíHospitalís Rule to show that

lim x!

(ln (x))^2 x

You must include all details of your reasoning, use correct notation, and write in complete sentences. Solution: First note that limx!1 x = 1 which means that we can attempt to use LíHospitalís Rule. Now consider the problem

lim x!

d dx

(ln (x))^2

d dx (x)^

= lim x!

2 (ln (x))

x

= lim x!

2 ln (x) x

Since limx!1 x = 1 , we can again try LíHospitalís Rule. Since

lim x!

d dx (2 ln (x)) d dx (x)^

= lim x!

2 x 1

then lim x!

(ln (x))^2 x

by LíHospitalís Rule.

  1. Show that (^) Z 1

8

(ln (x))^2 x

dx = 1.

You must include all details of your reasoning (steps in computing integrals, etc.). Solution: To evaluate the indeÖnite integral Z (ln (x))^2 x

dx,

we use the simple substitution u = ln (x), du = (^) x^1 dx. This gives us Z (ln (x))^2 x

dx =

Z

u^2 du =

u^3 + C =

(ln (x))^3 + C.

Thus we have Z (^1)

8

(ln (x))^2 x

dx = lim t!

Z (^) t

8

(ln (x))^2 x

dx

= lim t!

(ln (t))^3

(ln (8))^3

(because limt!1 ln (t) = 1 ).

  1. Explain why the series X^1

k=

(ln (k))^2 k

diverges. In doing this, you may refer to the result of problem 2. You must include all details of your reasoning, use correct notation, and write in complete sentences that do not contain the word ìitî. (Hint: Use the Integral Test.) Solution: Let f be the function f (x) = (ln (x))^2 =x and note that

f 0 (x) =

2 ln (x) (ln (x))^2 x^2

ln (x) (2 ln (x)) x^2 from which we see that f 0 (x) < 0 throughout the interval (e^2 ; 1 ). This means that f is decreasing on the interval [8; 1 ). (Note that f is also positive and continuous on this interval.) In problem 2, we showed that the improper integral Z (^1)

8

(ln (x))^2 x

dx

diverges. Thus the series X^1

k=

(ln (k))^2 k

diverges by the Integral Test.

  1. Suppose that an and bn are sequences such that 0  an  bn for all n  1. Decide whether each of the following statements must be true or is not necessarily true. (Circle the correct choice.)

(a) If limn!1 an = 0, then

P 1

n=1 an^ converges. (must be true / is not necessarily true ). (b) If

P 1

n=1 bn^ diverges, then^

P 1

n=1 an^ also diverges. (must be true / is not necessarily true ). (c) If limn!1 an = 3, then

P 1

n=1 an^ and^

P 1

n=1 bn^ both diverge. ( must be true / is not necessarily true). (d) If limn!1 a bnn = 1, then either

P 1

n=1 an^ and^

P 1

n=1 bn^ both converge or both diverge. ( must be true / is not necessarily true).