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

1. The line meets the $x$-axis at $x = -\frac{2}{3}$; it lies below the axis on $\left(-1, -\frac{2}{3}\right)$ and above on $\left(-\frac{2}{3}, 1\right)$. 2. So area $= \left|\int_{-1}^{-2/3}(3x+2)\,dx\right| + \int_{-2/3}^{1}(3x+2)\,dx$ with antiderivative $\frac{3x^2}{2} + 2x$. 3. The first (below) piece gives $\frac{1}{6}$ in magnitude; the second (above) piece gives $\frac{25}{6}$. 4. Adding, $\frac{1}{6} + \frac{25}{6} = \frac{26}{6} = \frac{13}{3}$. 5. Forgetting the absolute value and subtracting would lose the sub-axis part, so $\frac{25}{6}$ alone is wrong. 6. Ignoring the split entirely (plain $\int_{-1}^{1}$) gives $4$, which under-counts the area below the axis. _Source: syllabus Mathematics Ch 8 "Application of Integrals", p.5_