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Material Type: Exam; Professor: Ellermeyer; Class: Calculus II; University: Kennesaw State University; Term: Spring 2008;
Typology: Exams
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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.
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.
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 =
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 ).
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.
(a) If limn!1 an = 0, then
n=1 an^ converges. (must be true / is not necessarily true ). (b) If
n=1 bn^ diverges, then^
n=1 an^ also diverges. (must be true / is not necessarily true ). (c) If limn!1 an = 3, then
n=1 an^ and^
n=1 bn^ both diverge. ( must be true / is not necessarily true). (d) If limn!1 a bnn = 1, then either
n=1 an^ and^
n=1 bn^ both converge or both diverge. ( must be true / is not necessarily true).