Question

The area bounded by the region by the curves $\left|\begin{array}{c}x\end{array}\right|=1-y^2$ and $\left|\begin{array}{c}x\end{array}\right|+\left|\begin{array}{c}y\end{array}\right|=1$ is

• A. $\frac{1}{3}$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{1}{3}$

$y=\sqrt{1-|x|}$.
And
$y=\pm (1-|x|)$.
Considering $y=1-x$ ...(first quadrant).
Then
$\sqrt{1-x}=1-x$
Or
$x=0,x=1$.
Hence the required area will be
${\int }_{-1}^0\sqrt{1-x}-(1-x.)dx-{\int }_0^1\sqrt{1-x}-(1-x).dx$
Now the graph $\sqrt{1-x}-(1-x)$ is symmetric to y axis.
Hence
$I={\int }_{-1}^0\sqrt{1-x}-(1-x.)dx-{\int }_0^1\sqrt{1-x}-(1-x).dx$
$=2{\int }_0^1\sqrt{1-x}-(1-x).dx$
$=2{\int }_0^1\sqrt{1-x}+x-1.dx$
$=2[\frac{x^2}{2}-x-\frac{2(1-x{)}^{\frac{3}{2}}}{3}{]}_0^1$
$=2[\frac{1}{2}-1-(-\frac{2}{3}\left)\right]$
$=2[\frac{2}{3}-\frac{1}{2}]$
$=\frac{1}{3}$ sq units.