Alan

a) Presumably the customer paid $120

b) If it was discounted by 20% then this is 80% (or 0.8) of its original price, so 0.8*original price = $120

Divide both sides by 0.8 to get original price = $150

Yes!  Both are correct.

The volume of the full bale is V = w*pi*d²/4  where w is width and d is diameter.

If damage depth is s then the undamaged inner cylinder has volume Vu = (w-2s)*pi*(d-2s)²/4 because the width reduces by s at both left and right, and the diameter reduces by s on "both" sides.

The damaged volume is therefore  the difference between these. Here, we have w = 4', d = 5' and s = (2/3)' (because 8" is 2/3 of a foot). So:

$${\mathtt{V}} = {\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}}}{{\mathtt{4}}}} \Rightarrow {\mathtt{V}} = {\mathtt{78.539\: \!816\: \!339\: \!744\: \!831}}$$ cubic ft.

$${\mathtt{Vu}} = {\frac{\left({\mathtt{4}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}{{\mathtt{4}}}} \Rightarrow {\mathtt{Vu}} = {\mathtt{28.157\: \!978\: \!598\: \!841\: \!850\: \!5}}$$ cubic ft.

Damaged volume Vd ≈ 78.54 - 28.16 = 50.38 cubic ft.

As a percentage Vd/V ≈ 100*50.38/78.54 = 64.15%

Doesnt n+1 happen in taking the integration...

No.  You are thinking of

 $$\int t^ndt=\frac{t^{n+1}}{n+1}$$ 

But 

$$\int e^{kt}dt = \frac{e^{kt}}{k}$$

Depends on the cross-sectional shape of the ring.  Is it rectangular/circular/elliptical...?

Either this is

1.  x(t) = Ce^3t where C is a constant and you are expected to find C from the fact that x at time 0 is 5.  If so, then x(0) = C*e⁰ = C, so C = 5

or

2. this is meant to be x`(t) = Ce<sup>3t</sup> (note the `) or dx(t)/dt = Ce^3t , where C is a constant, with initial condition x at time 0 being 5, and you are expected to integrate it to find x as a function of t. If so, then

$$$$\int dx = \int Ce^{3t}dt$$ 

$$$$x(t)=\frac{C}{3}e^{3t}+k$$

where k is another constant, found from x(0) = 5

x(0) = (C/3) + k = 5 so k = 5 - (C/3) and

$$x(t)=5+\frac{C}{3}(e^{3t}-1)$$

Angle EOB is just 180° - angle EOD, so yes, subtract!

asin is the same as sin^-1

acos is the same as cos^-1

atan is the same as tan^-1

$$\frac{d}{dx}\sin x = \cos x\\\\
\frac{d}{dx}\cos x = -\sin x$$

These should be committed to memory!