Combining functions

Let $g(x)=\displaystyle \int_0^{\pi/2} |x-t| sin(t) dt$ .

Setting $s=x-t$ , we get the integral

$\displaystyle \int_{x-\pi/2}^{x} |s| sin(x-s) ds$

which can be evaluated by naturally considering the cases $x<0$ , $0 \leq x \leq \pi/2$ , and $x>\pi/2$ to simplify the $|s|$ function. This yields the piecewise function

$\left\{ \begin{array}{ll} -\int_{x-\pi/2}^x s \cdot sin(x-s) ds & x < 0 \\ \int_{0}^x s \cdot sin(x-s) ds - \int_{x-\pi/2}^0 s \cdot sin(x-s) ds & 0\leq x\leq \pi/2 \\ \int_{x-\pi/2}^x s \cdot sin(x-s) ds & x > \pi/2 \end{array} \right.$

which simplifies to

\left\{ \begin{array}{11} 1-x & x < 0 \\ x+1-2sin(x) & 0\leq x\leq \pi/2 \\ x-1 & x > \pi/2 \end{array} \right. .

Now since we know $f(x)=a\cdot g(x) +bx+c$ , we end up needing to solve the equations $a+b=0$ , $c-a=1$ , $a+c=0$ , $b-a=1$ , and $-2a=1$ , which has the solution $a=-1/2$ , $b=1/2$ , and $c=1/2$ , yielding an answer of $\boxed{8}$ .