Question

Sketch the graph of $y=|x+3|$ and evaluate ${\int }_{-6}^0|x+3|dx$

Answer 1

The given equation is $y=|x+3|$
Graph is plotted in the diagram.
It is known that, $(x+3)\le 0$ for $-6\le x\le -3$ and $(x+3)\ge 0$ for $-3\le x\le 0$
$\therefore {\int }_{-6}^0\left|\right(x+3\left)\right|dx=-{\int }_{-6}^{-3}(x+3)dx+{\int }_{-3}^0(x+3)dx$
$=-{[\frac{x^2}{2}+3x]}_{-6}^{-3}+{[\frac{x^2}{2}+3x]}_{-3}^0$
$=-[(\frac{(-3{)}^2}{2}+3(-3\left)\right)-(\frac{(-6{)}^2}{2}+3(-6\left)\right)]+[0-(\frac{(-3{)}^2}{2}+3(-3\left)\right)]$
$=-[-\frac{9}{2}]-[-\frac{9}{2}]=9$