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.

We have, $y = 1 + |x + 1|x = - 3,x = 3$ and $y = 0$.
Now $|x + 1| = \left\{ {\begin{array}{llllllllllllllllllll}{ - x - 1,}&{x < - 1}\\{x + 1,}&{x \ge - 1}\end{array}} \right.$
$\therefore$ $y = 1 + |x + 1| = \left\{ {\begin{array}{cccccccccccccccccccc}{ - x,}&{x < - 1}\\{x + 2,}&{x \ge - 1}\end{array}} \right.$
Graph of the above function with $x = - 3,x = 3$ as shown in the following figure

From the figure, Area of shaded region,
$A = \int_{ - 1}^{ - 1} - xdx + \int_{ - 1}^3 {(x + 2)} dr$
$= - \left[ {\frac{{{x^2}}}{2}} \right]_{ - 3}^{ - 1} + \left[ {\frac{{{x^2}}}{2} + 2x} \right]_{ - 1}^3$
$= - \left[ {\frac{1}{2} - \frac{9}{2}} \right] + \left[ {\frac{9}{2} + 6 - \frac{1}{2} + 2} \right] = 4 + 12 = 16$ sq units.