Some help with AP Calc. problems, please help.

[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.

 

[Deleted member 4993]

There are 3 problems I need help with, if you don't have time you can just help me out with one or two or whatever. Thank you very much, REALLY APPRECIATE any help given. I tried or at least thought about each problem but I am very unsure about all three because some of answers are not even in the available choices.
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
i honestly tried this prob. but hav no idea how to do it... so its says right hand end pts, so im plugging in -.1, -.2, -.3, into the equation, and etc. into it and not getting any answer large enough to even be pi...
2. (the big S thing in calculus) -> S 2 to -2 ( 2 - radical(4-x^2) ) dx =

\(\displaystyle \int_{-2}^{2} \sqrt{4-x^2}dx\)

substitute

\(\displaystyle x \, = \, 2\cdot \sin\theta\)

\(\displaystyle dx \, = \, 2\cdot \cos\theta \, d\theta\)

\(\displaystyle \int_{-2}^{2} \sqrt{4-x^2}dx\)

\(\displaystyle = \, 4\cdot \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos^2\theta \, d\theta\)

Now continue.....

A. 12.566
B. 14.2832
C. 7.1416
D. 1.717
E. 0.8585
this is a prob. that i never am able to do. I get the eq. to 2x - (1/2) radical(4-x^2) (2/3)<<< This is incorrect as shown above but then if i plug in 2 and -2 that would cancel out the last half of the equation and it would be just 4 - - 4 ?

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 + t^2- 24
B. v(t) = 3
C. v(t) = 3t + t^2 + 4
D. none of these
E. not enuff info. to decide
im thinkin its a since if the vel. is 4 and t is 4 and you plug in 4 for t, A fits it seems like it anyway.

Your thinking is correct

 

[BigGlenntheHeavy]

3rd problem:

a(t) = 2t+3

v(t) = t^2+3t+C, v(4) = 4 = 16+12+C, C = -24,

Hence v(t) = t^2+3t-24.

 

[azn989]

I solved the original 3 problems with the hints given, but need some help now with these other problems given out today.