Question

The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ $\frac{\mathrm{\pi }}{2}$ is
(a) 2 $\left(\sqrt{2}-1\right)$

(b) $\sqrt{2}-1$

(c) $\sqrt{2}+1$

(d) $\sqrt{2}$

Answer 1

(b) $\sqrt{2}-1$



$$\begin{gathered} \mathrm{Points}\mathrm{of}\mathrm{intersection}\mathrm{is}\mathrm{obtained}\mathrm{by}\mathrm{solving} \\ y=\sin x\mathrm{and}y=\cos x \\ \therefore \sin x=\cos x \\ \Rightarrow x=\frac{\mathrm{\pi }}{4} \\ \mathrm{Thus}\mathrm{the}\mathrm{two}\mathrm{functions}\mathrm{intesect}\mathrm{at}x=\frac{\mathrm{\pi }}{4} \\ \Rightarrow y=\sin \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}} \\ \mathrm{Hence}A\left(\frac{\mathrm{\pi }}{4},\frac{1}{\sqrt{2}}\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}. \\ \therefore \mathrm{Area}\mathrm{bound}\mathrm{by}\mathrm{the}\mathrm{curves}\mathrm{and}\mathrm{the}y\mathit{-}\mathrm{axis}when0\le x\le \frac{\mathrm{\pi }}{2}, \\ A={\int }_0^{\frac{1}{\sqrt{2}}}\left|x_1\right|dy+{\int }_{\frac{1}{\sqrt{2}}}^1\left|x_2\right|dy \\ ={\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}{x}_{1}dy+{\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}^1{x}_{2}dy \\ ={\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}{\sin }^{-1}ydy+{\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}^1{\cos }^{-1}ydy \\ ={\left[y{\sin }^{-1}y+\sqrt{1-y^2}\right]}_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}+{\left[y\cos {}^{-1}y-\sqrt{1-y^2}\right]}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}^1 \\ =\left[\frac{1}{\sqrt{2}}{\sin }^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}-1\right]+\left[1\times \cos {}^{-1}1-0-\frac{1}{\sqrt{2}}\cos {}^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}\right] \\ =\left[\frac{1}{\sqrt{2}}\times \frac{\mathrm{\pi }}{4}+\frac{1}{\sqrt{2}}-1\right]+\left[0-\frac{1}{\sqrt{2}}\times \frac{\mathrm{\pi }}{4}+\frac{1}{\sqrt{2}}\right] \\ =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1 \\ =\frac{2}{\sqrt{2}}-1 \\ =\left(\sqrt{2}-1\right)\mathrm{sq}\mathrm{units} \end{gathered}$$