a: Given an integral $\int_{0}^{\pi/2}|\sin{x}-c|dx$. I found the result as $-1 +2 c\arcsin(c)+ 2\cos(\arcsin(c)) - \frac{c\pi}{2}$. Which is right but the second part if that :
b: The median of $\sin{x}$ over the interval $[0, \pi/2]$ is the value of $c\in \Bbb{R}$ for which the integral from a is as small as possible. Calculate the median. Can someone give a hint how can i calculate that median?i have no idea where to start :(
Just derive relative to $c$. The median is the value of $c$ for which one has
$$2\arcsin{c}-{\pi\over 2}=0$$
The solution is
$$c={\sqrt{2}\over 2}$$
Suppose $f(x)$ is continuous on $[a, b]$. Partition $[a, b]$ into $n$ equal subintervals and call the midpoints of these subintervals $x_1, x_2,\cdots, x_n $. Let $\operatorname {med}_f(n)$ be the median value of the n function values $f (x_1), f (x_2),\cdots,f (x_n)$. Then the median value of $f$ on $[a, b]$ is $$f_{\text {med}}= \lim_{n\to \infty} \operatorname {med}_f (n)$$
Also see here. Hope it helps.