Let f(x) be a non-negative continuous function such that the area bounded by the curve y=f…

Question

Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y=f(x)$, the $x$-axis, and the ordinates $x=\dfrac{\pi}{4}$ and $x=\beta>\dfrac{\pi}{4}$ is $\beta\sin\beta+\dfrac{\pi}{4}\cos\beta+\sqrt2\beta$. Then $f\left(\dfrac{\pi}{2}\right)$ is:

Accepted Answer

Given that

$$\int\limits_{\pi /4}^\beta  {f\left( x \right)} dx = \beta \sin \beta  + {\pi  \over 4}\cos \,\beta  + \sqrt 2 \beta $$

Differentiating $w.r.t$ $$\beta $$

$$f\left( \beta  \right) = \beta \cos \beta  + \sin \beta  - {\pi  \over 4}\sin \beta  + \sqrt 2 $$

$$f\left( {{\pi  \over 2}} \right) = \left( {1 - {\pi  \over 4}} \right)\sin {\pi  \over 2} + \sqrt 2 $$

$$ = 1 - {\pi  \over 4} + \sqrt 2 $$

Solutions · 1