Two calculus questions: Euler's Rule and integration?

if you hate math go here http://ashleyrecordsvirgy.wordpress.com/

First, you'd divide by dx to move it to the right side, and divide by y squared to move it to the left side. Now, you should have:

dy/(y^2) = x/dx

Plug that into your 89 and integrate. Plug in 0 for your x, and solve for your constant, C, so that y(0) = 1. Once you've solved for C, solve for y(2).

The integral of (3x-2)^-1 from 1 to 0 is going to need U substitution.

Let u = 3x-2, and derive so that dx = du/3.

Since this integral is a definite integral, you are going to have to change you limits with respect to x, to limits with respect to u.

When x =1, u = 3(1) -2 = 1

When x = 0, u = 3(0) - 2 = -2

Your new limits are now from -2 to 1

Now, you should have:

Integral of u^-1 (du/3) from -2 to 1

The integral of anything to the -1 power is going to be Ln(anything). Don't forget to pull out the constant 1/3.

So, this is what you should get:

(1/3)Ln(u) from -2 to 1

Plug in u

(1/3)Ln(3x-2) from -2 to 1

And solve :)