Browse Questions

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

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