Question

Let $f\left(x\right)$ 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 $(\frac{a^2}{2}+\frac{a}{2}\sin a+\frac{\pi }{2}\cos a)$.sq.unit, then $f\left(\frac{\pi }{2}\right)$ is

• A. $\frac{1}{2}$
• B. $\frac{{\pi }^2}{8}+\frac{\pi }{4}$
• C. $\frac{\pi +1}{2}$
• D. none of these.

Answer 1

The correct option is A $\frac{1}{2}$

Since,
${\int }_0^{a}f\left(x\right)dx=\frac{a^2}{2}+\frac{a}{2}\sin a+\frac{\pi }{2}\cos a$ (given)
Differentiating both sides w.r.t $a$, we get
$\Rightarrow f\left(a\right)=a+\frac{a}{2}\cos a+\frac{1}{2}\sin a-\frac{\pi }{2}\sin a$
$\therefore f\left(\frac{\pi }{2}\right)=\frac{\pi }{2}+0+\frac{1}{2}.1-\frac{\pi }{2}.1=\frac{1}{2}$