Question

For these problems find y.
1. y'= 3x-5

so basically I'm suppose to go backwards this time instead of finding the derivative I'm finding the original y?

Answer 1

correct. Use the power rule
Since d/dx x^n = n x^(n-1)
That is, when taking the derivative, you divide by the power and subtract one.
To go backwards, reverse the operations -- add one to the power and then divide by it.
If y' = x^n, then y = 1/(n+1) x^(n+1)

y = 3/2 x^2 - 5x + c
where c is an arbitrary constant (whose derivative is just 0)

Answer 2

Ok! I was able to do a few problems like that^

What about y'= 1/x^2+1/x ?

Answer 3

come on, man. Don't forget your Algabra I now that you're taking calculus!
y' = 1/x^2 + 1/x = x^-1 + x^-1
The x^-2 is easy -- y = 1/-1 x^-1 = -1/x
The y' = 1/x is a bit different. It should look familiar, though. Recall that
d/dx lnx = 1/x
so, if y' = 1/x, y = lnx

So this one winds up as -1/x + lnx + c