Answer
$\frac{25}{4}$
Work Step by Step
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:
$A = \frac{1}{2}(0-(-4))(\frac{1}{2}(4))+\frac{1}{2}(3-0)(\frac{1}{2}(3))$
$A = 4+\frac{9}{4}$
$A = \frac{25}{4}$
Therefore: $\int_{-4}^{3} \vert \frac{1}{2}x \vert~dx = \frac{25}{4}$
