Question

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

Answer 1

$\begin{array}{c}y=\left|x+3\right|\\ firstdrawy=x+3,thenremovethepartthatisbelowx-axis\\ andmakeamironimageofthatpartabovethex-axis.\\ {\int }_{-6}^0\left|x+3\right|dx={\int }_{-6-3}^0-\left(x+3\right)dx+{\int }_{-6}^0\left(x+3\right)dx\\ =-{\left(\frac{x^2}{2}+3x\right)}_{-6}^{-3}+{\left(\frac{x^2}{2}+3x\right)}_{-3}^0\\ =-\left[18-18-\left(\frac{9}{2}-9\right)\right]+\left[0-\left(\frac{9}{2}-9\right)\right]\\ =\frac{9}{2}+\frac{9}{2}=0\end{array}$