Calculus II 
Quiz #3 
February 27, 2002
R.  Hammack 
Score ______

(1) Use the Fundamental Theorem of Calculus to evaluate the following integrals. Simpify your answer as much as possible.
(a)  [Graphics:Images/quiz3sol_gr_1.gif][Graphics:Images/quiz3sol_gr_2.gif]= 4+8-1+1 = 12

(b)   [Graphics:Images/quiz3sol_gr_3.gif]= 3(5) - 3 = 12

(c)  [Graphics:Images/quiz3sol_gr_4.gif][Graphics:Images/quiz3sol_gr_5.gif]= sin(π/2)-sin(π/4) = [Graphics:Images/quiz3sol_gr_6.gif]

(d) [Graphics:Images/quiz3sol_gr_7.gif]
Because the integrand has an absolute value, we need to analyze it in some detail. Think of the graph of [Graphics:Images/quiz3sol_gr_8.gif]It is a parabola that opens "up", and its x intercepts are 0 and 1. A graph based on this information is sketched on the right. As you can see, [Graphics:Images/quiz3sol_gr_10.gif] is positive on the interval (-1,0) and negative on (0,1). Thus we break up our integral as follows: [Graphics:Images/quiz3sol_gr_9.gif]
[Graphics:Images/quiz3sol_gr_16.gif]1/3 +1/2 -1/3 + 1/2 = 1

  Find the area of the region under the graph of   [Graphics:Images/quiz3sol_gr_17.gif]between [Graphics:Images/quiz3sol_gr_18.gif]  and   [Graphics:Images/quiz3sol_gr_19.gif].

(3) Suppose [Graphics:Images/quiz3sol_gr_22.gif]   Find the interval(s) on which F increases, and the interval(s) on which F decreases.
To answer this question we need to look at the sign of the derivative of F.  By F.T.C. II, [Graphics:Images/quiz3sol_gr_23.gif].
The denominator of this expression is always positive, so the sign of F '(x) is controlled by the numerator.
As you can see, the nunerator is positive when x > 3, and negative when x < 3.
Therefore F(x) increases on the interval (3,∞), and decreases on (-∞, 3),