Question

$\underset{0}{\overset{1}{\int }}\left|2x-1\right|dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left|2x-1\right|=\left\{\begin{array}{l}-\left(2x-1\right),0\le x\le \frac{1}{2}\\ 2x-1,\frac{1}{2}\le x\le 1\end{array}\right. \\ \therefore {\int }_0^1\left|2x-1\right|dx \\ ={\int }_0^{\frac{1}{2}}-\left(2x-1\right)dx+{\int }_{\frac{1}{2}}^1\left(2x-1\right)dx \\ ={\left[-x^2+x\right]}_0^{\frac{1}{2}}+{\left[x^2-x\right]}_{\frac{1}{2}}^1 \\ =\frac{-1}{4}+\frac{1}{2}+1-1-\frac{1}{4}+\frac{1}{2} \\ =\frac{1}{2} \end{gathered}$$