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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)   ∫( 3e^x+1/x+ln(3) )dx =  3e^x+ln|x|+x ln(3)+C


(b)   ∫ x  x^(1/3)  dx = ∫ x  x^(1/3) dx =∫   x^(4/3) dx =( x^(7/3))/(7/3)+C =(3x^(1/3)^7)/7+C


(c)   ∫5/(1 - x^2)^(1/2) dx =5 tan^(-1)(x)+C


(d)   ∫( e^(-x)+sin(π x) + sec(x)tan(x) )dx = -e^(-x)- 1/πcos(π x) + sec(x)+C


(e)   ∫(cos(3x))^4sin(3x) dx =
-1/3∫(cos(3x))^4(-1)sin(3x)3 dx =-1/3∫u^4 du =-u^5/15+C=-cos^5(3x)/15+C

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

(f)   ∫x(x - 3)^(1/2)dx =
∫(u+3)u^(1/2)du =∫(u^(3/2)+3u^(1/2))du =u^(5/2)/(5/2)+3u^(3/2)/(3/2)+C=
2/5u^(1/2)^5+2u^(1/2)^3+C=2/5(x - 3)^(1/2)^5+2(x - 3)^(1/2)^3+C

u = x - 3
du = dx

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

(a)   ∫__ (π/4)^(π/3)sec^2(x) dx =tan(π/3)-tan(π/4)=3^(1/2)-1


(b)    ∫__0^ln(2)3e^xdx =3e^ln(2)-3e^0=6-3=3


(c)  ∫_0^1(x^2+x-1)dx =[x^3/3 + x^2/2 - x] _0^1=1/3+1/2-1=-1/6


(d)  ∫__0^(π/2)^(1/2)cos (x^2) 2x  dx =∫__0^(π/2)cos (u)  du =sin (π/2)-sin (0)=1

u = x^2
du = 2x dx


(e)  ∫_ (-2)^( -1)x/( x^2 + 2 )^2dx =
   1/2∫_ (-2)^( -1)1/( x^2 + 2 )^22x dx =1/2∫_6^( 3) u^(-2) du =1/2[-1/u] _6^3=1/2(-1/3+1/6)=-1/12

u = x^2+2
du = 2x dx



(f)  ∫__0^e1/(x + e)dx = ∫__e^(2e)1/udu =[ ln | u | ] _e^(2e)=ln|2e|- ln|e|=ln(2)+ln(e)-ln(e)=ln(2)

u = x +e
du = dx

(3)  Consider the limit of Riemann sums  Underscript[lim , max Δx_k0]    Underoverscript[∑ , k = 1, arg3](( x_k^*)^2+1)Δ x_k,  where a=1 and b=2.
Find the value of this limit by writing it as a definite integral and evaluating.
  

∫_1^2(x^2+1)dx =[x^3/3 + x] _1^2=(2^3/3+2)-(1^3/3+1)=8/3+2-1/3-1=10/3


(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.

∫_0^4(t-1)dt =[t^2/2 - t] _0^4=4^2/2-4=4 units



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

∫_0^4|t-1|dt =∫_0^1|t-1|dt +∫_1^4|t-1|dt =∫_0^1(1-t)dt +∫_1^4(t-1)dt=1/2+9/2=5units


(5) This question concerns the function F(x)=∫__2^x  (9t - t^2 - 20)^(1/3)dt.

(a)   F '(x)= (9x - x^2 - 20)^(1/3)



(b)   Find the interval(s) on which  F(x)  increases.
F '(x)= (9x - x^2 - 20)^(1/3)=(-(x^2 - 9x + 20))^(1/3)=-((x - 5) (x - 4))^(1/3)
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-x^2.

y=1-x^2=(1-x)(1+x), so the x-intercepts are 1 and -1.

[Graphics:HTMLFiles/T1S05D_99.gif]


Area =∫_ (-1)^1(1-x^2)dx =[x - x^3/3] _ (-1)^1=(1-1^3/3)-(-1-(-1)^3/3)=2-2/3=4/3square units



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


(a)   Find   ∫_ (-3)^2f(x)dx .        (Suggestion:  Sketch the graph of f.)

[Graphics:HTMLFiles/T1S05D_108.gif]



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 ∫_ (-3)^xf(x)dx = 0

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