The value of $\int_{-6}^{0} |x + 3|\,dx$ is

1. $|x + 3|$ has a corner at $x = -3$; split there: $|x+3| = -(x+3)$ on $[-6, -3]$ and $(x+3)$ on $[-3, 0]$. 2. $\int_{-6}^{-3}-(x+3)\,dx = \left[-\frac{x^2}{2} - 3x\right]_{-6}^{-3} = \frac{9}{2}$. 3. $\int_{-3}^{0}(x+3)\,dx = \left[\frac{x^2}{2} + 3x\right]_{-3}^{0} = \frac{9}{2}$. 4. Adding, the value $= \frac{9}{2} + \frac{9}{2} = 9$. 5. Each piece is a right triangle of legs $3$ and $3$ (area $\frac{9}{2}$), confirming $9$; treating it as one triangle gives the wrong $\frac{9}{2}$. _Source: syllabus Mathematics Ch 8 "Application of Integrals", p.6_