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Calculus  II                                                          Test #1                                                  March 4, 2005

Name____________________                     R.  Hammack                                                Score ______
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(1) Find the following antiderivatives.

(a)   ( 3++ln(3) )dx =  3+ln|x|+x ln(3)+C

(b)   x    dx = x   dx =    dx =+C =+C

(c)    dx =5 (x)+C

(d)   ( +sin(π x) + sec(x)tan(x) )dx = -- cos(π x) + sec(x)+C

(e)   sin(3x) dx =
-(-1)sin(3x)3 dx =- du =-+C=-+C

u = cos(3x) dx
du = -sin(3x)3 dx

(f)   xdx =
(u+3)du =(+3)du =+3+C=
+2+C=+2+C

u = x - 3
du = dx

(2) Find the following definite integrals. Simplify your answer as much as possible.

(a)   (x) dx =tan(π/3)-tan(π/4)=-1

(b)    3dx =3-3=6-3=3

(c)  (+x-1)dx ==+-1=-

(d)  cos () 2x  dx =cos (u)  du =sin (π/2)-sin (0)=1

u =
du = 2x dx

(e)  dx =
2x dx = du ==(-+)=-

u = +2
du = 2x dx

(f)  dx = du ==ln|2e|- ln|e|=ln(2)+ln(e)-ln(e)=ln(2)

u = x +e
du = dx

(3)  Consider the limit of Riemann sums      (+1)Δ ,  where a=1 and b=2.
Find the value of this limit by writing it as a definite integral and evaluating.

(+1)dx ==(+2)-(+1)=+2--1=

(4)  Suppose a particle moving on the number line has a velocity of v(t)=t-1 units per second at time t.

(a)   Find the object's displacement between times t = 0 and t = 4.

(t-1)dt ==-4=4 units

(b)     Find the total distance the object has traveled between times t = 0 and t = 4.

|t-1|dt =|t-1|dt +|t-1|dt =(1-t)dt +(t-1)dt=+=5units

(5) This question concerns the function F(x)=  dt.

(a)   F '(x)=

(b)   Find the interval(s) on which  F(x)  increases.
F '(x)= ==-
From this you can see the derivative is positive on the interval [4,5], so that is where F increases.

(6)   Find the area of the region contained between the x-axis and the graph of y=1-.

y=1-=(1-x)(1+x), so the x-intercepts are 1 and -1.

Area =(1-)dx ==(1-)-(-1-)=2-=square units

 (7) Suppose f(x) ={ 2              if x <0 2-x     if x≥0

(a)   Find   f(x)dx .        (Suggestion:  Sketch the graph of f.)

Integral equals area of a 3 by 2 rectangle plus area of triangle of height and base 2.
Thus its value is (3)(2)+1/2(2)(2) = 8

(b)    Find a value of x, different from -3, for which f(x)dx = 0

Notice that from the graph, if x = 6, the net signed area is 0. Hence the answer is x = 6.