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Q.
The area bounded by the curve $|x+y|+|x-y|=1$ is
Application of Integrals
Solution:
We have, $\left|x+y\right|+\left|x-y\right|=1$
$\left(i\right) $ If $x, y \,\ge\,0$
$\left(a\right)$ If $x\, <\,y$
$x + y + x - y = 1$
$\Rightarrow x=\frac{1}{2}$
$\left(b\right)$ If $x\, >\,y$
$x + y + y - x = 1$
$\Rightarrow y=\frac{1}{2}$
$\left(ii\right)$ If $x, y \,<\,0$
$\left(a\right)$ If $x\, <\,y$
$-\left(x+y\right)-\left(x-y\right)=1$
$\Rightarrow x=-\frac{1}{2}$
$\left(b\right)$ If $x\, >\,y$
$-\left(x+y\right)-\left(y-x\right)=1$
$\Rightarrow y=-\frac{1}{2}$
Required area $=4\int\limits_{0}^{1 /2}\frac{1}{2}dx=2\left[x\right]_{0}^{1 /2}$
$= 1$ sq. unit
