Question

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

Answer 1

Area under the curve:

It is given, ${\int }_{\frac{\mathrm{\pi }}{4}}^{\beta }f\left(x\right)dx=(\beta \sin \beta +(\frac{\pi }{4})\cos \beta +\sqrt{2}\beta )$

Now, differentiating both sides with respect to $\beta$, we get:

$$\begin{gathered} \Rightarrow f\left(\beta \right)=\sin \beta +\beta \cos \beta –\left(\frac{\pi }{4}\right)\sin \beta +\sqrt{2} \\ \Rightarrow f\left(\frac{\mathrm{\pi }}{2}\right)=\sin \left(\frac{\mathrm{\pi }}{2}\right)+\frac{\mathrm{\pi }}{2}\cos \left(\frac{\mathrm{\pi }}{2}\right)-\left(\frac{\pi }{4}\right)\sin \left(\frac{\mathrm{\pi }}{2}\right)+\sqrt{2}\mathbf{[}\mathbf{P}\mathbf{u}\mathbf{t}\mathbf{}\mathbf{\beta }\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{]} \\ \Rightarrow f\left(\frac{\mathrm{\pi }}{2}\right)=1+0-\frac{\mathrm{\pi }}{4}+\sqrt{2} \\ \Rightarrow f\left(\frac{\mathrm{\pi }}{2}\right)=1-\frac{\mathrm{\pi }}{4}+\sqrt{2} \end{gathered}$$

Hence, $\mathit{f}\left(\frac{\pi }{2}\right)$ is equal to $\mathbf{1}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\sqrt{\mathbf{2}}$.