Question

Let $fx$ be a continuous function such that the area bounded by the curve $y=f\left(x\right)$, the $x$-axis and the two ordinates $x=0$ and $x=a$ is $\frac{a^2}{2}+\frac{a}{2}\sin a+\frac{\pi }{2}\cos a$.Then $f\left(\frac{\pi }{2}\right)$ is

Answer 1

$\stackrel{a}{\underset{0}{\int }}f\left(x\right)=\frac{a^2}{2}+\frac{a^2}{2}\sin a+\frac{\pi }{2}\cos a$

By Leibniz rule $\to d=\stackrel{g\left(x\right)}{\underset{h\left(x\right)}{\int }}\left(x\right)dx$

$\frac{d.d}{dx}=f\left(g\right(x\left)\right)g^{\prime }\left(x\right)-fh\left(x\right)\left)h^{\prime }\right(x)$

$\therefore f\left(a\right)-0=a+\frac{a^2}{2}\cos a+a\sin A-\frac{\pi }{2}\sin a$

$\therefore f\left(a\right)=a+(a-\frac{\pi }{2})\sin A+\frac{a^2}{2}\cos a$

$\therefore f\left(x\right)=x+\frac{x^2}{2}\cos x+(x-\frac{\pi }{2})\sin x$