Whenever a function is represented by y=|x| there arises two cases.

(i)y=x if $x\geq 0$

(ii)y=-x if x<0.

Where $x\geq 0$ represents the portion that lies to the right side of the curve and x<0 represents the portion that lies to the left side of the curve.

The function |x|+|y|=1 has 4 cases.

case(i) x+y=1

case(ii) x+y=1

case(iii) x-y=1

case(iv) -x-y=1

By sketching a graph we can find that the area bounded by the curve |x|+|y|=1 is represented by the shaded region as shown in the fig:

The curve intersects the points (0,1),(1,0),(0,-1),(-1,0).

It can also be observed that the curve is symmetrical about x-axis and also y-axis.

We can conclude that the required area can be taken as 4$\times $area of OBAO.

$A=4\times \int_0^1(1-x)dx$

on integrating we get,

$\;\;\;=4\big(x-\frac{x^2}{2}\big)_0^1$

on applying the limits we get,

$\;\;\;=4\begin{bmatrix}1-\frac{1}{2}\end{bmatrix}$

$\;\;\;=4\times \frac{1}{2}$

$\;\;\;=2sq.units.$

Hence the required area is 2sq.units