Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y=f(x), x$-axis…
Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y=f(x), x$-axis and the ordinates $x=\frac{\pi}{4}$ and $x=\beta>\frac{\pi}{4}$ is $\left(\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta\right)$. Then $f\left(\frac{\pi}{2}\right)$ is
$\left(\frac{\pi}{4}+\sqrt{2}-1\right)$
$\left(\frac{\pi}{4}-\sqrt{2}+1\right)$
$\left(1-\frac{\pi}{4}-\sqrt{2}\right)$
$\left(1-\frac{\pi}{4}+\sqrt{2}\right)$
Solution
Given that $\int_{\pi / 4}^\beta f(x) d x=\beta \sin \beta+\frac{\pi}{4} \cos \beta+2 \beta$
Differentiating w. r. t $\beta$
$f(\beta)=\beta \cos \beta+\sin \beta-\frac{\pi}{4} \sin \beta+\sqrt{2}$
$f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}\right) \sin \frac{\pi}{2}+\sqrt{2}=1-\frac{\pi}{2}+\sqrt{2}$.