Question

Sketch the graph y = | x + 3 |. Evaluate $\underset{-6}{\overset{0}{\int }}\left|x+3\right|dx$. What does this integral represent on the graph?

Answer 1

We have,
y = | x + 3 | intersect x = 0 and x = −6 at (0, 3) and (−6, 3)
Now,
$$\begin{gathered} y=\left|x+3\right| \\ =\left\{\begin{array}{l}\left(x+3\right)\mathrm{For}\mathrm{all}x>-3\\ -\left(x+3\right)\mathrm{For}\mathrm{all}x<-3\end{array}\right. \end{gathered}$$

Integral represents the area enclosed between x = −6 and x = 0
$$\begin{gathered} \therefore A={\int }_{-6}^0\left|y\right|dx \\ ={\int }_{-6}^{-3}\left|y\right|dx+{\int }_{-3}^0\left|y\right|dx \\ ={\int }_{-6}^{-3}-\left(x+3\right)dx+{\int }_{-3}^0\left(x+3\right)dx \\ =-{\left[\frac{x^2}{2}+3x\right]}_{-6}^{-3}+{\left[\frac{x^2}{2}+3x\right]}_{-3}^0 \\ =-\left[\frac{9}{2}-9-\frac{36}{2}+18\right]+\left[0+0-\frac{9}{2}+9\right] \\ =-\frac{9}{2}+9+\frac{36}{2}-18-\frac{9}{2}+9 \\ =9\mathrm{sq}.\mathrm{units} \end{gathered}$$