Evaluate: ∫₋₆³ |x + 3| dx

$|x + 3| = -(x+3)$ for $x < -3$ and $x + 3$ for $x \ge -3$, so split at $x = -3$. $\int_{-6}^{3}|x+3|dx = \int_{-6}^{-3}-(x+3)dx + \int_{-3}^{3}(x+3)dx$ $= \left[-\frac{(x+3)^2}{2}\right]_{-6}^{-3} + \left[\frac{(x+3)^2}{2}\right]_{-3}^{3}$ $= \left(0 + \frac{9}{2}\right) + \left(18 - 0\right)$ Hence the value is $\frac{45}{2}$.