find f if f''(x) = 2e^x + 3sin(x), f(0) = -4, f(pi) = -5

Hi, I'm new here and it seems like a good site to get help. Well, I have not been able to figure out one of my calculus problems.

Problem: find f if f''(x) = 2e^x + 3sin(x), f(0) = -4, f(pi) = -5.

I integrated f''(x) and attained f'(x) = 2e^x - 3cos(x) + C.
So, now how do I find what C is when I don't have any known value as you can see above I only have f(0) and f(pi), but no f'(x) value. Please help.

\(\displaystyle \L f''(x) = 2e^x + 3\sin{x}\)

\(\displaystyle \L f'(x) = 2e^x - 3\cos{x} + C_1\)

\(\displaystyle \L f(x) = 2e^x - 3\sin{x} + C_1 x + C_2\)

now ... use your two initial conditions for f(x) to solve for the two constants.

wow!!! thanks that was a lot of help. so the answer is f(x) = 2e^x - 3sin(x) -14.41351x - 6.

This was the only problem left and it was taking me forever. Thanks again.