Calculus II Test #3 April 16, 2002 Name____________________ R.  Hammack Score ______

(1) Evaluate the following integrals.

(a) uv = x ln x = x ln x = x ln x - x + C

u = ln x            dv = dx
du = 1/x dx          v = x

(b) uv = -1/3 x cos(3x)= -1/3 x cos(3x)+1/3
= -1/3 x cos(3x)+1/9 sin (3x) + C

u =  x            dv = sin(3 x) dx
du = dx          v = -1/3 cos(3x)

(c)

- 1/4 cot θ  + C = + C

x = 2sin θ
dx = 2cos θ dθ

(d) 5 ln|x-4| + 3 ln|x+2| + C

= +

= +

Set x = 4,    get 8(4)-2 =  6A, or 30 = 6A, so A=5.
Set x = -2,  get 8(-2)-2 = -6B, or -18=-6B, so B=3

(2) Evaluate the following improper integrals. Please show all of your work.

(a)     0 + 1/8 = 1/8

(b)
We are going to have to find an antiderivative of ln x, so let's start off doing that by integration by parts.

u = ln x         dv = dx
du = 1/x dx       v = x

Now notice that has an infinite discontinuity at 0. Therefore

[-1] +
+ 0 =
This last limit is an indetrerminate form ∞/∞, so applying L'Hopital's rule gives a final answer of
= [-l] = -1

(3)
This limit has indeterminate form 0/0, so we can use L'Hopital's rule to get 3 (0) = 3