Question

Using integration, find the area of the region bounded by the following curves, after making a rough sketch: y = 1 + | x + 1 |, x = −2, x = 3, y = 0.

Answer 1

We have,
y = 1 + | x + 1 | intersect x = − 2 and at ( −2, 2) and x = 3 at (3, 5).
And y = 0 is the x-axis.
The shaded region is our required region whose area has to be found

$$\begin{gathered} y=1+\left|x+1\right| \\ =\left\{\begin{array}{l}1-\left(x+1\right)x<-1\\ 1+\left(x+1\right)x\ge 1\end{array}\right. \\ =\left\{\begin{array}{l}-xx<-1\\ x+2x\ge 1\end{array}\right. \end{gathered}$$
Let the required area be A. Since limits on x are given, we use horizontal strips to find the area:
$$\begin{gathered} A={\int }_{-2}^3\left|\mathit{y}\right|\mathit{d}x \\ ={\int }_{-2}^{-1}\left|y\right|\mathit{d}x+{\int }_{-1}^3\left|y\right|\mathit{d}x \\ ={\int }_{-2}^{-1}-x\mathit{d}x+{\int }_{-1}^3\left(x+2\right)\mathit{d}x \\ =-{\left[\frac{x^2}{2}\right]}_{-2}^{-1}+{\left[\frac{x^2}{2}+2x\right]}_{-1}^3 \\ =-\left[\frac{1}{2}-\frac{4}{2}\right]+\left[\frac{9}{2}+6-\frac{1}{2}+2\right] \\ =\frac{3}{2}+\left[8+\frac{8}{2}\right] \\ =\frac{3}{2}+\left[8+4\right] \\ =\frac{27}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$