Here's a SS of the questions. #6 and #7.

If you can write it out with accuracy I can work on selling some CS skins for FG.
http://i.imgur.com/Xo39qUA.png

Could not edit post... Here are the problems.

#6

First, realize they want the total area. We need to know what this graph looks like. You hopefully should know it's a cubic function. Since they want actual area, we need to bust this puppy up into two integrals. Graph it on your calculator if you want a picture of the graph (no way to draw it on here).

Hopefully, you can realize that the zero of the graph lies at x = 2

So we need to integrate from -1 to 2, then 2 to 4

u = x/2 - 1

du = x/2 hence 2du = dx

so our integral is 2u^3 du

which becomes .5u^4 and throwing back in, .5(x/2 - 1)^4

From here it's just algebra, plug in your bounds -1 to 2, that will give you one part of the area (DO NOT FORGET TO TAKE THE NEGATIVE OF THIS, since the curve is under the x-axis in this region), then plug in the bounds from 2 to 4. Add the two together and you're good to go. Again, it's just algebra/arithmetic at this point so I'm not going to bother finishing.

#7 is a classic "separation of variables" problem

Put the dx on the other side to get dy = (4x/9y)dx

Put the 9y on the other side to get 9ydy = 4xdx

Slap an integral sign on both sides, integrate, then plug in what you are given (3,5) to find C.

Again, it'd take too much typing to finish this here. If you have any difficulties, let me know though. These are relatively easy problems.

#8

No way in hell I'm typing all of that out!

Point is to treat one variable as just a constant.

For example, if I wanted partial x, I would treat y as a constant. Then just take the derivative with respect to x. Boom done.

Ditto for partial y

#8

f(x,y) = ln(3x-27) + x^3y - xy^2 + 6x - 8y +14.

partial of f(x,y) = (3/(3x-27) dx + 3x^2y dx + x^3 dy - y^2 dx - 2xydy + 6dx - 8dy.

need to break this into two parts so

The partial with respect to x = 1/(x-9) + 3x^2y - y^2 +6 and to y is x^3 -2xy - 8