Answer
$\frac{1}{2}$
Work Step by Step
On the interval $0 \leq x \leq 1$, the graph of $\vert 2x-1 \vert$ forms two triangles above the x-axis. We can find the area of these two triangles:
$A = \frac{1}{2}(\frac{1}{2})(1)+ \frac{1}{2}(\frac{1}{2})(1)$
$A = \frac{1}{4}+\frac{1}{4}$
$A = \frac{1}{2}$
Therefore: $\int_{0}^{1} \vert 2x-1 \vert~dx = \frac{1}{2}$
