## Calculus: Early Transcendentals 8th Edition

$\frac{25}{4}$
On the interval $-4 \leq x \leq 3$, the graph of $\vert \frac{1}{2}x \vert$ forms two triangles above the x-axis. We can find the area of these two triangles: $A = \frac{1}{2}(4)(2)+\frac{1}{2}(3)(\frac{3}{2})$ $A = 4+\frac{9}{4}$ $A = \frac{25}{4}$ Therefore: $\int_{-4}^{3} \vert \frac{1}{2}x \vert~dx = \frac{25}{4}$