Calculus II 
Test #1
March 5, 2002
Name____________________    
R.  Hammack 
Score ______

(1) Evaluate the following integrals.
(a) [Graphics:Images/1S02Dsol_gr_1.gif][Graphics:Images/1S02Dsol_gr_2.gif]


(b) [Graphics:Images/1S02Dsol_gr_3.gif] 2[Graphics:Images/1S02Dsol_gr_4.gif] 2[Graphics:Images/1S02Dsol_gr_5.gif]2tan(u) +C = 2 tan([Graphics:Images/1S02Dsol_gr_6.gif])+C

[Graphics:Images/1S02Dsol_gr_7.gif]
[Graphics:Images/1S02Dsol_gr_8.gif]


(c) [Graphics:Images/1S02Dsol_gr_9.gif][Graphics:Images/1S02Dsol_gr_10.gif][Graphics:Images/1S02Dsol_gr_11.gif]-2/3 [Graphics:Images/1S02Dsol_gr_12.gif]+C = - [Graphics:Images/1S02Dsol_gr_13.gif]+C

u = cos x
du = -sin x dx


(d) [Graphics:Images/1S02Dsol_gr_14.gif][Graphics:Images/1S02Dsol_gr_15.gif][Graphics:Images/1S02Dsol_gr_16.gif]4/2 - 1/2 = 3/2

u = ln x
du = 1/x dx




(2) A object moving on a straight line has a (nonconstant) acceleration of  6t  feet per second per second at time t. It travels a distance of 33 feet between times t = 0 and t = 3.  Find the object's velocity function v(t).
a(t) = 6t
v(t) =
[Graphics:Images/1S02Dsol_gr_47.gif] = 3[Graphics:Images/1S02Dsol_gr_48.gif]+C        (We just need to find C)
s(t) = [Graphics:Images/1S02Dsol_gr_49.gif] = [Graphics:Images/1S02Dsol_gr_50.gif]+Ct +K
From s(0) = 0 we get K = 0, so  s(t) = [Graphics:Images/1S02Dsol_gr_51.gif]+Ct Now, 33 = s(3) = [Graphics:Images/1S02Dsol_gr_52.gif] gives the equation 33 = 27 + 3C, or  3C = 6, so C = 2.
Thus v(t) =  3[Graphics:Images/1S02Dsol_gr_53.gif]+2   


(3) [Graphics:Images/1S02Dsol_gr_54.gif][Graphics:Images/1S02Dsol_gr_55.gif]

(4) Find the average value of the function [Graphics:Images/1S02Dsol_gr_67.gif]on the interval [0, ln(2)].
Using the formula for average value, the average value is
[Graphics:Images/1S02Dsol_gr_68.gif]= [Graphics:Images/1S02Dsol_gr_69.gif] =[Graphics:Images/1S02Dsol_gr_70.gif]= [Graphics:Images/1S02Dsol_gr_71.gif][Graphics:Images/1S02Dsol_gr_72.gif][Graphics:Images/1S02Dsol_gr_73.gif][Graphics:Images/1S02Dsol_gr_74.gif][Graphics:Images/1S02Dsol_gr_75.gif]


(5) [Graphics:Images/1S02Dsol_gr_76.gif]20
Over the interval [6,2], the graph of y = |x| forms two triangles. Adding their areas, we get the answer of
(1/2)(2)(2) + (1/2)(6)(6) = 2 + 18 = 20.


(6)  The graph of a function f(x) has slope [Graphics:Images/quiz1solved_gr_14.gif]  at each point (x, f(x)).
Also, the graph of  f(x) passes through the point (1, 5).  Find the function f(x).

The information says slope = f '(x) = [Graphics:Images/quiz1solved_gr_15.gif].
This means  f(x) =[Graphics:Images/quiz1solved_gr_16.gif]
[Graphics:Images/quiz1solved_gr_17.gif].
Thus f(x) = [Graphics:Images/quiz1solved_gr_18.gif]
To determine f completely, we just need to find C. To do this, use the fact that f(1)=5.
5=f(1)= [Graphics:Images/quiz1solved_gr_19.gif]
5= 0 + 3/4 +C
Therefore C = 17/4, and  the answer to the question is f(x) = [Graphics:Images/quiz1solved_gr_20.gif]

 


(7)  The expression   [Graphics:Images/quiz2_gr_10.gif][Graphics:Images/quiz2_gr_11.gif]   represents a definite integral over the interval [3, 5].   Write the definite integral. (You do not need to find its value.)    ANSWER:[Graphics:Images/quiz2_gr_12.gif]