Question

Sketch the graph of $y=|x+4|$. Using integration, find the area of the region bounded by the curve $y=|x+4|$ and $x=-6$ and $x=0$.

Answer 1

The area of the region bounded by the curve $y=|x+4|,x=-6,x=0$ is shown below.

Required area $=$ Area of region $ABC+$Area of region $AOD$

$={\int }_{-6}^{-4}|x+4|dx+{\int }_{-4}^0|x+4|dx$

$={\int }_{-6}^{-4}(-x-4)dx+{\int }_{-4}^0(x+4)dx$

$={[\frac{-x^2}{2}-4x]}_{-6}^{-4}+{[\frac{x^2}{2}+4x]}_{-4}^0$

$=\left[\right(-8+16)-(-18+24\left)\right]+\left[\right(0+0)-(8-16\left)\right]$

$=[8-6]+[0+8]$

$=2+8$

$=10$

Hence Required area is $10sq.units$