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