Question

Using the method of integration find area bounded by $|x|+|y|=1$

Answer 1

We know that $\left|x\right|=\{\begin{array}{c}x,x\ge 0\\ -x,x\le 0\end{array}$ and

$\left|y\right|=\{\begin{array}{c}y,y\ge 0\\ -y,y\le 0\end{array}$

so we can write $\left|x\right|+\left|y\right|=1$

as $\{\begin{array}{c}x+y=1forx>0,y>0\\ -x+y=1forx<0,y>0\\ x-y=1forx>0,y<0\\ -x-y=1forx<0,y<0\end{array}$

Since the curve is symmetrical about $x$ and $y$ axis

Required area$=4\times$Area$AOB$

Area $ABO={\int }_0^{1}y.dx$

where $x+y=1\Rightarrow y=1-x$

$\therefore$Area $ABO={\int }_0^1(1-x)dx$

$={[x-\frac{x^2}{2}]}_0^1$

$=1-\frac{1}{2}-{(0-\frac{0}{2})}^2$

$=1-\frac{1}{2}=\frac{1}{2}$

Hence,Required area$=4\times$Area$AOB$

$=4\times \frac{1}{2}=2$ sq.units.