Finding the second antiderivative given f(x) values 
Find $f$.
$f''(t)=7e^t + 3\sin(t)$, $f(0) = 0$ and $f(\pi)=0$

I found that:
$f'(t)= 7e^t + 3\cos(t) + C$
$f(t) = 7e^t - 3\sin(t) + Ct + D$
How do I know what equation to plug the $x$-values into?
 A: You made a mistake  in taken derivative:
$$ f^{''}(t)=7e^{t}+3\sin(t) $$
$$ f^{'}(t)=7e^{t}-3\cos(t)+C$$
$$ f(t)=7e^{t}-3\sin(t)+Ct+D$$
$$ f(0)=0 : 7-0+0+D=0 : D=-7$$
$$ f(\pi)=0 : 7e^{\pi}-0+C\pi+-7=0 : C=\frac{7-7e^{\pi}}{\pi}$$
A: If you have $$ f(t) = 7e^t-3\sin(t)+Ct+D$$ (although I would check the signs on the trig functions) and you know $f(0)=0$ and $f(\pi)=0$.  The above is what you plug into!
$$
f(0) = 7e^0-3\sin(0)+C(0)+D = 0
$$
and
$$
f(\pi) = 7e^\pi-3\sin(\pi)+C(\pi)+D = 0
$$
This is a system of two equations with two unknowns ($C$ and $D$), which you can solve.
A: $$
f''(t)=7e^t + 3\sin{t}\Longleftrightarrow \int\left(7e^t + 3\sin{t}\right)\,dt=f'(t)\implies\\
f'(t)=\int\left(7e^t + 3\sin{t}\right)\,dt=7e^t - 3\cos{t}+C_1
$$
Use the same logic to find the original function itself (in fact, it's going to be a family of functions because of the constants that appear as a result of the integration process):
$$
f'(t)=7e^t - 3\cos{t}+C_1\Longleftrightarrow \int\left(7e^t - 3\cos{t}+C_1\right)\,dt=f(t)\implies\\
f(t)=\int\left(7e^t - 3\cos{t}+C_1\right)\,dt=7e^t - 3\sin{t}+C_1t+C_2
$$
Now, use the information you're given as part of the problem statement to find out what these constants are equal to:
$$
f(0)=0\implies 7e^0 - 3\sin{0}+C_1\cdot 0+C_2=0\implies C_2=-7\\
f(\pi)=0\implies 7e^\pi - 3\sin{\pi}+C_1\pi-7=0\implies C_1=\frac{7(1-e^\pi)}{\pi}
$$
So, the original function $f(t)$ is then:
$$
f(t)=7e^t - 3\sin{t}+\frac{7(1-e^\pi)}{\pi}-7.
$$
