Question

If $f\left(x\right)$ be a continuous function such that the area bounded by the curve $y=f\left(x\right)$, the $x$-axis, and the lines $x=0$ and $x=a$ is $1+\frac{a^2}{2}\sin a$, then

• A. $f\left(\frac{\pi }{2}\right)=1+\frac{{\pi }^2}{8}$
• B. $f\left(a\right)=1+\frac{a^2}{2}\sin a$
• C. $f\left(a\right)=a\sin a+\frac{1}{2}{a}^2\cos a$
• D. none of these

Answer 1

The correct option is C $f\left(a\right)=a\sin a+\frac{1}{2}{a}^2\cos a$

It is given that area of the function $f\left(x\right)$ from $x=0$ to $x=a$ is $1+\frac{a^2}{2}\sin a$

Hence, ${\int }_0^{a}f\left(x\right)dx=1+\frac{a^2}{2}\sin a$

$⟹{\int }_0^{x}f\left(x\right)dx=1+\frac{x^2}{2}\sin x$

Now differentiating both sides with respect to $x$:-

$⟹f\left(x\right)=\frac{d}{dx}(1+\frac{x^2}{2}\sin x)$

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

$⟹f\left(a\right)=a\sin a+\frac{1}{2}{a}^2\cos a$

Hence, answer is option-(C).