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Let $f(x)$ be a non-negative continous function such that the area bounded by the curve $y=f(x), x$-axis and the ordinates $x=\frac{\pi}{4}, \quad x=\beta\gt\frac{\pi}{4}$ is $\left(\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta\right)$. Then $f\left(\frac{\pi}{2}\right)$ is
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Verified Answer
The correct answer is:
$\left(1-\frac{\pi}{4}+\sqrt{2}\right)$
Given that, $\int_{\pi / 4}^\beta f(x) d x=\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta$
Differentiating w.r.t. $\beta$, we get
$\therefore f(\beta)=\sin \beta+\beta \cos \beta-\frac{\pi}{4} \sin \beta+\sqrt{2}$
Hence,
$f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}+\sqrt{2}\right)$
Differentiating w.r.t. $\beta$, we get
$\therefore f(\beta)=\sin \beta+\beta \cos \beta-\frac{\pi}{4} \sin \beta+\sqrt{2}$
Hence,
$f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}+\sqrt{2}\right)$
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