Question

Sketch the graph y = | x + 1 |. Evaluate $\underset{-4}{\overset{2}{\int }}\left|x+1\right|dx$. What does the value of this integral represent on the graph?

Answer 1

We have,
y = | x + 1 | intersect x = −4 and x = 2 at (−4, 3) and (2, 3) respectively.
Now,
$$\begin{gathered} y=\left|x+1\right| \\ =\left\{\begin{array}{l}\left(x+1\right)\mathrm{For}\mathrm{all}x>-1\\ -\left(x+1\right)\mathrm{For}\mathrm{all}x<-1\end{array}\right. \end{gathered}$$

Integral represents the area enclosed between x = −4 and x = 2
$$\begin{gathered} A={\int }_{-4}^2\left|y\right|dx \\ ={\int }_{-4}^{-1}\left|y\right|dx+{\int }_{-1}^2\left|y\right|dx \\ ={\int }_{-4}^{-1}-\left(x+1\right)dx+{\int }_{-1}^2\left(x+1\right)dx \\ =-{\left[\frac{x^2}{2}+x\right]}_{-4}^{-1}+{\left[\frac{x^2}{2}+x\right]}_{-1}^2 \\ =-\left[\frac{1}{2}-1-\frac{16}{2}+4\right]+\left[\frac{4}{2}+2-\frac{1}{2}+1\right] \\ =-\left[3-\frac{15}{2}\right]+\left[5-\frac{1}{2}\right] \\ =-3+\frac{15}{2}+5-\frac{1}{2} \\ =9\mathrm{sq}.\mathrm{units} \end{gathered}$$