Draw a rough sketch of the given curve $y = 1 + |x + 1|$, $x = -3$, $x = 3$, $y = 0$ and find the area of the region bounded by them, using integration.

$16 \text{ sq. units}$

The correct answer is Option (2) → $16 \text{ sq. units}$

We have, $y = 1 + |x + 1|$, $x = -3$, $x = 3$ and $y = 0$

$∵\quad y = \begin{cases} -x, & \text{if } x < -1 \\ x + 2, & \text{if } x \geq -1 \end{cases}$

Put $y = -x$ for $-3$ to $-1$ and $y = x + 2$ for $-1$ to $3$.

$∴\quad \text{Area of shaded region} = \int_{-3}^{-1} -x \, dx + \int_{-1}^{3} (x + 2) \, dx$

$= -\left[ \frac{x^2}{2} \right]_{-3}^{-1} + \left[ \frac{x^2}{2} + 2x \right]_{-1}^{3}$

$= -\left[ \frac{1}{2} - \frac{9}{2} \right] + \left[ \frac{9}{2} + 6 - \frac{1}{2} + 2 \right]$

$= -[-4] + [8 + 4]$

$= 4 + 12 = 16 \text{ sq. units}$