Find the area of the region bounded by the line $y = 3x + 2$, the $x$-axis and the ordinates $x = -1$ and $x = 1$.

$\frac{13}{3}$

The correct answer is Option (1) → $\frac{13}{3}$

As shown in Figure, the line $y = 3x + 2$ meets the $x$-axis at $x = \frac{-2}{3}$ and its graph lies below the $x$-axis for $x \in \left( -1, \frac{-2}{3} \right)$ and above the $x$-axis for $x \in \left( \frac{-2}{3}, 1 \right)$.

The required area $=$ Area of the region ACBA $+$ Area of the region ADEA

$= \left| \int\limits_{-1}^{-2/3} (3x + 2) \, dx \right| + \int\limits_{-2/3}^{1} (3x + 2) \, dx$

$= \left| \left[ \frac{3x^2}{2} + 2x \right]_{-1}^{-2/3} \right| + \left[ \frac{3x^2}{2} + 2x \right]_{-2/3}^{1} = \frac{1}{6} + \frac{25}{6} = \frac{13}{3}$