# 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.

-
Are you sure that you have copied the problem correctly? –  mrf Feb 7 '13 at 11:31
@mrf - Yes I am pretty sure I copied it correctly. –  Derek 朕會功夫 Feb 7 '13 at 14:37

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 $f(2) = 2 \sin (\frac{\pi e^c}{2}) + 1$

-
So what do I do with the $c$? And when I plug in $2$ or $0$ into $c$, both does not seem to give me a correct answer. –  Derek 朕會功夫 Feb 7 '13 at 8:57
I used the Laplace Transformation to solve this, but cannot get a better answer than you gave here. In fact, we need to solve $\int\frac{\sin k}{k}dk$ for some $k$. –  Babak S. Feb 7 '13 at 9:31
$c$ lives in the soul of the interval. Therefore, $c$ cannot be $0$ or $2$. –  Marvin Gaye Feb 7 '13 at 9:34
@BabakSorouh: From Citizen solution, one can get $f(2)\in [-3,3].$ –  Mhenni Benghorbal Feb 7 '13 at 9:34
I added a graph for the OE with initial condition. I hope you don't mind. If you don't like it, remove it. ;-) –  Babak S. Feb 7 '13 at 9:51

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.

-
This is very helpful, thanks! –  Derek 朕會功夫 Feb 7 '13 at 16:30
@Derek: You are welcome. –  Mhenni Benghorbal Feb 8 '13 at 3:26

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].

-