Question

The area bounded by $y={\sin }^{-1}x,x=\frac{1}{\sqrt{2}}$ and x-axis is

• A. $\left[\frac{1}{\sqrt{2}}+1\right]$
• B. $\left[1-\frac{1}{\sqrt{2}}\right]$
• C. $\frac{\mathrm{\pi }}{4\sqrt{2}}$
• D. $\left(\frac{\mathrm{\pi }}{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1\right)$

Answer 1

The correct option is D

$\left(\frac{\mathrm{\pi }}{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1\right)$

Explanation for correct option

Finding the area of the region bounded by the curves

Given, $y={\sin }^{-1}x,x=\frac{1}{\sqrt{2}}$

Visualize the two equations on a graph.

We have to calculate the area of the shaded region $OAB$

The required area will be,$\begin{array}{rcl}& =& Area\left(OAB\right)\\ & =& areaofrec\tan gleOABC-areaofcurveOBCO\\ & =& \left(\frac{1}{\sqrt{2}}\times \frac{\mathrm{\pi }}{4}\right)-{\int }_0^{\mathrm{\pi }/4}\sin ydy\\ & =& \frac{\mathrm{\pi }}{4\sqrt{2}}+{\left[\cos y\right]}_0^{\mathrm{\pi }/4}\\ & =& \left[\frac{\mathrm{\pi }}{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1\right]\end{array}$

Hence, the correct answer is Option (D).