Answer:
Given equation of curve is \[|x|+|y|=1\] \[\Rightarrow \] \[\pm \,x\pm y=1\] \[\left[ \because \,\,|x|=\left\{ \begin{matrix} x,\,\,\text{if}\,\,x\ge 0 \\ -x,\,\,\text{if}\,\,x<0 \\ \end{matrix} \right. \right]\] The above equation represents these four lines, \[x+y=1\] ? (i) \[x-y=1\] ? (ii) \[-x+y=1\] ? (iii) and \[-x-y=1\] ? (iv) The graphical representation of these lines is given below Since, the required area is symmetrical in all the four quadrants. Required area = 4 (Area of region OABO) \[=4\int_{0}^{1}{ydx=4\int_{0}^{1}{(1-x)}}dx\] \[[\because \,\,x+y=1]\] \[=4\left[ x-\frac{{{x}^{2}}}{2} \right]_{0}^{1}=4\left[ \left( 1-\frac{{{1}^{2}}}{2} \right)-\left( 0-\frac{{{0}^{2}}}{2} \right) \right]\] \[=4\left( 1-\frac{1}{2} \right)=4\left( \frac{1}{2} \right)=2\,\text{sq}.\] units
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