The region bounded by the line $3x - 2y + 6 = 0$, the $x$-axis and the line $x = 2$ is found with vertical strips. Between which limits is $y = \frac{3x + 6}{2}$ integrated?

Explanation

1. 01

   The line meets the $x$-axis where $3x + 6 = 0$, that is at $x = -2$.

2. 02

   The region runs from there to the line $x = 2$: $\int_{-2}^{2} \frac{3x + 6}{2}\,dx = 12$ square units.