Calculus II 
Final Exam

May 23, 2002

Name____________________ 
R. Hammack

Score______

(1) Evaluate the following integrals.
(a)
(b)
u = ln(x)
du = 1/x dx
(c)
du = 2x dx
(d)
(e)
(f)
(g)
(2) Evaluate the following definite integrals
(a)
(b)
(c) =
du = 2x dx
(3) Evaluate the following integrals.
(a)
(b)
Set x = 4. Get A = 1/5
Set x = 1. Get B = 1/5
(4) Evaluate the following definite integrals. (Notice that they are both improper.)
(a)
(b) = 1
(5) Consider the region between the graph of and the xaxis.
(6) Find the area enclosed between the curves , , , and x = 1.
(8) Decide if the following sequences converge or diverge. In the case of convergence, find the limit.
(a)
The sequence converges to 8.
(b)
(Using L'Hopital's Rule)
The sequence converges to 0.
(9) Decide if the following series converge or diverge.
(a)
Let's try the ratio test on this one (it works well with powers).
Thus, the series converges by the ratio test.
(b) Pseries with p = 1/2 < 1, so it diverges.
(c) Alternating series that meets conditions of the Alternating Series Test, so it converges.
(d)
The given series converges by the comparison by comparison with the convergent series whose kth term is 3/2^k.
(10) Both of the following series converge. Say what number they converge to.
(a) =
(Because it's a geometric series with a = 3 and r = 1/2)
(b) = e
(because it's the Maclaurin series for with x = 1.)
(11) The remaining problems on this exam are based on the function .
(a) Derive the Maclaurin series for the function . Show
all of your work.
Maclaurin Series:
(b) Find the interval of convergence of the Maclaurin series from part
(a). Show all of your work.
Let's use the ratio test to check for absolute convergence.
Thus, we get convergence provided that and
divergence when
Thus the series converges absolutely on the interval (1, 1) and diverges on
What about at x = 1? Then the series is
which is the convergent alternating harmonic series.
What about at x = 1? Then the series is
which is the divergent harmonic series.
Thus the interval of convergence is (1, 1].
(c) Use your answer to part (a) above to express ln(2) as an infinite
series.
Part (a) and (b) say
on the interval (1, 1]. Since 1 is on this interval, we have:
ln(2) =ln( 1+1) =
(d) Use your answer to part (a) above to find a power series representation
of the function .
=
(e) Write as
an infinite series. (You may just write out the first 5 or 6 terms.)