Question

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 $–x–y=11$]

Answer 1

The correct answer is:$=2units$
The area bounded by the curve,$|x|+|y|=1$, is represented by the shaded region 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 $y-axis.$
$∴Area\,\, ADCB=4\times Area\,\, OBAO$
$=∫^1_0(1-x)dx$
$=4\bigg(x-\frac{x^2}{2}\bigg)^1_0$
=$=4[1-\frac{1}{2}]$
$=4(\frac{1}{2})$
$=2units$