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 x-axis.

(a) This region is rotated around the y-axis. What is the volume of the resulting solid?

Volume by shells:
cubic units

(b) This region is rotated around the x-axis. What is the volume of the resulting solid?

Volume by slicing:
cubic units

(6) Find the area enclosed between the curves ,   ,   ,  and  x = 1.

(7) Find the derivative of the function

(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)   P-series 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.)