Answer
$$\frac{5}{2}$$
Work Step by Step
Given
\begin{aligned} \int_0^2|2 x-1| d x\end{aligned}
Since
\begin{aligned} |2x-1|&= \begin{cases} -(2x-1) & 0\lt x\lt 1/2\\
(2x-1) & 1/2\lt x\lt 2 \end{cases}\end{aligned}
Then
\begin{aligned} \int_0^2|2 x-1| d x&=- \int_{0}^{1/2}(2x-1)dx+ \int_{1/2}^2(2x-1)dx\\
&= -(x^2-x)\bigg|_{0}^{1/2}+(x^2-x)\bigg|_{1/2}^{2}\\
&= -\left(\frac{1}{4}-\frac{1}{2}\right)+ \left(4-2-\frac{1}{4}+\frac{1}{2}\right)\\
&=\frac{5}{2}\end{aligned}
