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


(1) Evaluate the following integrals.

(a) [Graphics:Images/1S02Dsol_gr_17.gif] uv [Graphics:Images/1S02Dsol_gr_18.gif]= x ln x [Graphics:Images/1S02Dsol_gr_19.gif]= x ln x [Graphics:Images/1S02Dsol_gr_20.gif]= x ln x - x + C

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


(b) [Graphics:Images/1S02Dsol_gr_21.gif]uv [Graphics:Images/1S02Dsol_gr_22.gif]= -1/3 x cos(3x)[Graphics:Images/1S02Dsol_gr_23.gif]= -1/3 x cos(3x)+1/3[Graphics:Images/1S02Dsol_gr_24.gif]
= -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) [Graphics:Images/1S02Dsol_gr_25.gif][Graphics:Images/1S02Dsol_gr_26.gif][Graphics:Images/1S02Dsol_gr_27.gif][Graphics:Images/1S02Dsol_gr_28.gif][Graphics:Images/1S02Dsol_gr_29.gif]

[Graphics:Images/1S02Dsol_gr_30.gif][Graphics:Images/1S02Dsol_gr_31.gif][Graphics:Images/1S02Dsol_gr_32.gif][Graphics:Images/1S02Dsol_gr_33.gif]- 1/4 cot θ  + C = [Graphics:Images/1S02Dsol_gr_34.gif] + C

x = 2sin θ
dx = 2cos θ dθ


(d) [Graphics:Images/1S02Dsol_gr_35.gif] [Graphics:Images/1S02Dsol_gr_36.gif][Graphics:Images/1S02Dsol_gr_37.gif]5 ln|x-4| + 3 ln|x+2| + C

[Graphics:Images/1S02Dsol_gr_38.gif]= [Graphics:Images/1S02Dsol_gr_39.gif]+[Graphics:Images/1S02Dsol_gr_40.gif]

[Graphics:Images/1S02Dsol_gr_41.gif]= [Graphics:Images/1S02Dsol_gr_42.gif]+[Graphics:Images/1S02Dsol_gr_43.gif]

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)    [Graphics:Images/quiz7sol_gr_1.gif][Graphics:Images/quiz7sol_gr_2.gif][Graphics:Images/quiz7sol_gr_3.gif][Graphics:Images/quiz7sol_gr_4.gif][Graphics:Images/quiz7sol_gr_5.gif] 0 + 1/8 = 1/8


(b)    [Graphics:Images/quiz7sol_gr_6.gif]
We are going to have to find an antiderivative of ln x, so let's start off doing that by integration by parts.

[Graphics:Images/quiz7sol_gr_7.gif][Graphics:Images/quiz7sol_gr_8.gif][Graphics:Images/quiz7sol_gr_9.gif][Graphics:Images/quiz7sol_gr_10.gif][Graphics:Images/quiz7sol_gr_11.gif]
  u = ln x         dv = dx
du = 1/x dx       v = x

Now notice that [Graphics:Images/quiz7sol_gr_12.gif]has an infinite discontinuity at 0. Therefore
[Graphics:Images/quiz7sol_gr_13.gif][Graphics:Images/quiz7sol_gr_14.gif][Graphics:Images/quiz7sol_gr_15.gif][Graphics:Images/quiz7sol_gr_16.gif][Graphics:Images/quiz7sol_gr_17.gif][Graphics:Images/quiz7sol_gr_18.gif][Graphics:Images/quiz7sol_gr_19.gif]
[Graphics:Images/quiz7sol_gr_20.gif][Graphics:Images/quiz7sol_gr_21.gif][Graphics:Images/quiz7sol_gr_22.gif][-1][Graphics:Images/quiz7sol_gr_23.gif][Graphics:Images/quiz7sol_gr_24.gif][Graphics:Images/quiz7sol_gr_25.gif] +[Graphics:Images/quiz7sol_gr_26.gif][Graphics:Images/quiz7sol_gr_27.gif]
[Graphics:Images/quiz7sol_gr_28.gif][Graphics:Images/quiz7sol_gr_29.gif][Graphics:Images/quiz7sol_gr_30.gif] + 0 = [Graphics:Images/quiz7sol_gr_31.gif][Graphics:Images/quiz7sol_gr_32.gif]
This last limit is an indetrerminate form ∞/∞, so applying L'Hopital's rule gives a final answer of
[Graphics:Images/quiz7sol_gr_33.gif][Graphics:Images/quiz7sol_gr_34.gif]= [Graphics:Images/quiz7sol_gr_35.gif][-l] = -1

(3)  
This limit has indeterminate form 0/0, so we can use L'Hopital's rule to get [Graphics:Images/quiz7sol_gr_36.gif][Graphics:Images/quiz7sol_gr_37.gif][Graphics:Images/quiz7sol_gr_38.gif][Graphics:Images/quiz7sol_gr_39.gif]3 [Graphics:Images/quiz7sol_gr_40.gif](0) = 3