Question

The area bounded by the curve $y^2=1-x$ and the lines $y=\frac{|x|}{x},x=-1$ 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,
$A=\underset{-1}{\overset{0}{\int }}\left(-1-\left(-\sqrt{1-x}\right)dx+\underset{0}{\overset{\frac{1}{2}}{\int }}\left(1-\sqrt{1-x}\right)dx\right)$
$={\left[-x-\frac{{\left(1-x\right)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{-1}^0+{\left[x+\frac{{\left(1-x\right)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{\frac{1}{2}}$
$=\left[-\frac{2}{3}-1-\frac{2\times 2^{\frac{3}{2}}}{3}\right]+\left[\frac{1}{2}+\frac{2}{3\times 2^{\frac{3}{2}}}-\frac{2}{3}\right]$
$=\frac{2}{3\times 2^{\frac{3}{2}}}+\frac{2\times 2^{\frac{3}{2}}}{3}-\frac{4}{3}-\frac{1}{2}$
$=\frac{3}{\sqrt{2}}-\frac{4}{3}-\frac{1}{2}$
$=\frac{3}{\sqrt{2}}-\frac{11}{6}sq.$ unit