Question

Using integration, find the area bounded by the curves = |x - 1| and y = 3 - |x|

Answer 1

We have, $y=|x-1|.........\left(1\right)$

$y=3-\left|x\right|.......\left(2\right)$

The curve intersect each other at 2 point obtained by solving

From equation (1),we get

$|x-1|\to x-1$ $ifx\ge 1$

$1-x,ifx<1$

And from (2),we get

$3-\left|x\right|\to 3-x$ $ifx\ge 0$

$x-3,ifx<0$

we have $3-x=x-1$

if $x\ge 1\to x=2$

if $0\le x\le 1\to 3-x=2$ which is not possible

Also when $x<0$ then $3+x=1-x$

Thus we get two points $x=2$ and $x=-1$

The required area is

$S={\int }_{-1}^{2}3-\left|x\right|-|x-1|dx$

$\Rightarrow {\int }_{-1}^{0}3-x-(1-x)dx+$ ${\int }_0^1(3-x)-(1-x)dx+$${\int }_1^2(3-x)-(x-1)dx$

$\Rightarrow {\int }_{-1}^0(2+2x)dx+$ ${\int }_0^{1}2dx-$${\int }_1^2(4-2x)dx$

$\Rightarrow |2x+2\frac{x^2}{2}{|}_{-1}^0$+$|2x{|}_0^1$+$|4x-2\frac{x^2}{2}{|}_1^2$

$\Rightarrow [0-(-2+1)+2+[(8-4)-(4-1)]=1+2+1=4$ $sq.units$