The area bounded by the curve $y = x^3$, the $x$-axis and the ordinates $x = -2$ and $x = 1$ is

1. $y = x^3$ is negative on $[-2, 0]$ and positive on $[0, 1]$, so split at $0$ and use magnitudes. 2. $\left|\int_{-2}^{0}x^3\,dx\right| = \left|\left[\frac{x^4}{4}\right]_{-2}^{0}\right| = \left|0 - 4\right| = 4$. 3. $\int_{0}^{1}x^3\,dx = \left[\frac{x^4}{4}\right]_0^1 = \frac{1}{4}$. 4. Area $= 4 + \frac{1}{4} = \frac{17}{4}$. 5. Skipping the split, $\int_{-2}^{1}x^3\,dx = \frac{1}{4} - 4 = \frac{-15}{4}$, which is the signed value, not the area. _Source: syllabus Mathematics Ch 8 "Application of Integrals", p.6_