If $f'(x) = \sin{\frac{\pi e^x}{2}}$ and $f(0)= 1$, what is $f(2)$? If $f'(x) = \sin{\dfrac{\pi e^x}{2}}$ and $f(0)= 1$, then what will be $f(2)$?
This is what I tried to find the antiderivative of $f'(x)$ with u-substitution, 
$$
\begin{align}
u &=\frac{\pi e^x}{2} \\
\frac{2}{\pi}du &=e^x dx
\end{align}
$$
I don't know what to do next.
 A: Some thoughts: Apply mean value theorem to $f$ on the interval $[0,2]$ to obtain:
$$ \frac{f(2) - f(0)}{2} = \sin (\frac{\pi e^c}{2}) $$
for some $c$ in the interval. Then, we have that $f(2) = 2 \sin (\frac{\pi e^c}{2}) + 1$

A: Note that
$$ {f'(x) = \sin{\frac{\pi e^x}{2}}}\implies \int_{0}^{x}f'(t)dt = \int_{0}^{x}\sin\left(\frac{\pi e^t}{2}\right) $$
$$\implies f(x)=f(0)+\int_{0}^{x}\sin\left(\frac{\pi e^t}{2} \right)dt $$
$$ \implies f(2)=1+\int_{0}^{2}\sin\left(\frac{\pi e^t}{2} \right)dt \longrightarrow (*)$$
$$=1-{\it Si} \left( \frac{\pi}{2}  \right) +{\it Si} \left( \frac{1}{2}\,{{\rm e}^{2}
}\pi  \right)  \sim 1.157117528,$$
where $Si$ is the sine integral.
A: another option.. you could go with  a taylor expansion of sin(x) take the integral, and get  f(2) [in form of an infinite series].  
A: To get result from Mean Value Theorem we have to make sure that the sine function in the last step is $\ge 0$. The argument of sine function lies between $-\pi/2$ to $\pi/2$. From there we can get $x$ lies between $-\infty$ to $0$. If $x$ lies between $-\infty$ to $0$ then min value of sine function is $0$ and max value is $1$. In this way we can make sure that sine function is greater than $0$. I am not sure though. 
