• Mar 16th 2009, 02:30 PM
azn989
I will be truly grateful for any help.
1) S (sinx - 3cotxsinx) dx =
A. cosx +3cscx + C
B. cosx + 3sinx + C
C. -cosx -3sinx+C
D. -cosx +3cscx+C
E. none of the above
im thinking none of the above since the answer i got, had -cosx but did not have -3sinx or 3cscx
2. S (4x-8)/(X^2-4X+7)^6 dx =
A. (-4)/(x^2-4x+7)^5 + C
B. (-4)/(5)(x^2-4x+7)^5 + C
C. -2/(x^2-4x+7)^5 + C
D. -2/(5)(x^2-4x+7)^5 + C
E. none of these
I have done this problem two times and still can't get an answer even remotely close to any of the 4 answers... so im guessing it is none of these?
3) Solve the differential equation dy/dx = 3x^2 + sinx+2, if y=2 when x=0.
A. y=x^3-cosx+2x+3
B. y=x^3-cosx+2x+2
C. y=x^3+cosx+2x+1
D. y=x^3-cosx+2x+C
E. y=X^3+cosx+2x
I think it is D but not sure. I am just guessing since it should definitely be an equation with a -cosx and then I just guessed it was D out of A,B, and D.
4) S (sinx-3cotxsinx)dx=
A. cosx+3cscx + C
B. cosx + 3sinx + C
C. -cosx -3sinx + C
D. -cosx + 3cscx + C
E. none of above
I am guessing none of the above cuz the answer I found did not have 3cscx, 3sinx or -3sinx. but I am really prone to making mistakes and do not trust my answer.
• Mar 16th 2009, 03:15 PM
skeeter
Quote:

Originally Posted by azn989
1) Think of estimating the area under the curve of radical(4-x^2) on [-2,0] using right-hand end points. As you use larger and larger values for n, the approximated area will approach:
A. Pi
B. 2pi
C. 4pi
D. 8pi
E. infinity

think about the graph of $y = \sqrt{4-x^2}$ ... what geometric figure is it?

2. (the big S thing in calculus) -> S 2 to -2 ( 2 - radical(4-x^2) ) dx =
A. 12.566
B. 14.2832
C. 7.1416
D. 1.717
E. 0.8585

same clue as before ... sketch the graph of $y = 2 - \sqrt{4-x^2}$

A particle's acceleration along a straight line is given by 3 + 2t at any time t. At t=4, the velocity is 4. Which of these gives the correct function for the velocity of this particle at any time t?
A. v(t) = 3t^2 + t^2- 24
B. v(t) = 3
C. v(t) = 3t + t^2 + 4
D. none of these
E. not enuff info. to decide

the antiderivative of acceleration is velocity ... you are also given an initial condition for v(t) so you may determine the value of the constant of integration ... integrate the acceleration and determine the velocity function.

A paperweight is thrown up with initial vel. of 60 m/s and intial height of 6 meters. The accel(due to gravity) of the object is a(t) = 9.8 m/s^2, where t is in secs.
4. The equation that describes the paperweight's velocity as a function of time t is :
A. v(t) = -9.8t + C
B. v(t) = -9.8t - 60
C. v(t) = -4.9t^2 + 60
D. v(t) = -9.8t + 60
E. v(t) = -4.9t^2 - 60

$y(t) = 6 + 60t - 4.9t^2$ ... velocity is the derivative of position.

5. The maximum height of the paperweight is:
A. 6.122 m
B. 189.673 m
C. -4.9t^2 + 60t + 6 m
D. 3.06 m
E. 144.692 m

when the paperweight achieves its max height v = 0

.
• Mar 17th 2009, 12:27 PM
azn989
ok, i was able to solve the original problems with the help of the hints given, could I get some more for these other 4 problems?
• Mar 17th 2009, 03:26 PM
skeeter