Question

The area bounded by the curve $y^{2}=1-x$ and the lines $y=\frac{\left|x\right|}{x}, \, x=-1$ and $x=\frac{1}{2}$ is

• A. $\left(\frac{3}{\sqrt{2}} - \frac{11}{6}\right)sq.$ unit
• B. $\left(3 \sqrt{2} - \frac{11}{4}\right)sq.$ unit
• C. $\left(\frac{6}{\sqrt{2}} - \frac{11}{5}\right)sq.$ unit
• D. None of these

Answer 1

Answer: (A)

From the diagram, the area of the shaded region,
$A=\int\limits_{- 1}^{0}\left(- 1 - \left(- \sqrt{1 - x}\right)dx+\int\limits _{0}^{\frac{1}{2}}\left(1 - \sqrt{1 - x}\right)dx \right)$
$=\left[- x - \frac{\left(1 - x\right)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{- 1}^{0}+\left[x + \frac{\left(1 - x\right)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{\frac{1}{2}}$
$=\left[- \frac{2}{3} - 1 - \frac{2 \times 2^{\frac{3}{2}}}{3}\right]+\left[\frac{1}{2} + \frac{2}{3 \times 2^{\frac{3}{2}}} - \frac{2}{3}\right]$
$=\frac{2}{3 \times 2^{\frac{3}{2}}}+\frac{2 \times 2^{\frac{3}{2}}}{3}-\frac{4}{3}-\frac{1}{2}$
$=\frac{3}{\sqrt{2}}-\frac{4}{3}-\frac{1}{2}$
$=\frac{3}{\sqrt{2}}-\frac{11}{6}sq.$ unit