The area of the region bounded by the line y = 3x + 2, the X-axis and the ordinates x = - 1 and x = 1 is

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

$$\begin{gathered} Required area=\left|{\int }_{-1}^{\frac{-2}{3}}(3x+2) dx\right|+\left|{\int }_{\frac{-2}{3}}^1(3x+2) dx\right| \\ =\left|{(3\frac{x^2}{2}+2x)}_{-1}^{\frac{-2}{3}}\right|+\left|{(3\frac{x^2}{2}+2x)}_{\frac{-2}{3}}^1\right| \\ =\left|\frac{-2}{3}+\frac{1}{2}\right|+\left|\frac{7}{2}+\frac{2}{3}\right| \\ =\left|\frac{-1}{6}\right|+\left|\frac{25}{6}\right| \\ =\frac{26}{6} \\ =\frac{13}{3} \end{gathered}$$