Using the method of integration find the area bounded by the curve $|x|+|y|=1$ [Hint: the required region is bounded by lines $x+y=1, x-y=1,-x+y=1$ and $-x$ $-y=11]$

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Using the method of integration find the area bounded by the curve $|x|+|y|=1$ [Hint: the required region is bounded by lines $x+y=1, x-y=1,-x+y=1$ and $-x$ $-y=11]$

Using the method of integration find the area bounded by the curve $|x|+|y|=1$ [Hint: the required region is bounded by lines $x+y=1, x-y=1,-x+y=1$ and $-x$ $-y=11]$

Solution:
Area bounded by the curve, $|x|+|y|=1$ , is represented by the shaded region $\mathrm{ADCB}$ as

The curve intersects the axes at points $A(0,1), B(1,0), C(0,-1)$ , and $D(-1,0)$ . It can be observed that the given curve is symmetrical about $x$ -axis and $y$ -axis.
$\therefore$ Area ADCB $=4 \times$ Area OBAO
$\begin{array}{l} =4 \int_{n}(1-x) d x \\ =4\left(x-\frac{x^{2}}{2}\right)_{0}^{1} \\ =4\left[1-\frac{1}{2}\right] \\ =4\left(\frac{1}{2}\right) \\ =2 \text {sq. units } \end{array}$