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
\[y = 1 + \left| x + 1 \right|\]
\[ = \begin{cases}1 - \left( x + 1 \right) &x < - 1\\1 + \left( x + 1 \right)& x \geq 1\end{cases}\]
\[ = \begin{cases} - x x& < - 1\\x + 2& x \geq 1\end{cases}\]
Let the required area be A. Since limits on x are given, we use horizontal strips to find the area:
\[A = \int_{- 2}^3 \left| y \right| d x\]
\[ = \int_{- 2}^{- 1} \left| y \right| d x + \int_{- 1}^3 \left| y \right| d x\]
\[ = \int_{- 2}^{- 1} - x d x + \int_{- 1}^3 \left( x + 2 \right) d x\]
\[ = - \left[ \frac{x^2}{2} \right]_{- 2}^{- 1} + \left[ \frac{x^2}{2} + 2x \right]_{- 1}^3 \]
\[ = - \left[ \frac{1}{2} - \frac{4}{2} \right] + \left[ \frac{9}{2} + 6 - \frac{1}{2} + 2 \right]\]
\[ = \frac{3}{2} + \left[ 8 + \frac{8}{2} \right]\]
\[ = \frac{3}{2} + \left[ 8 + 4 \right]\]
\[ = \frac{27}{2}\text{ sq . units }\]