Question

The integral ${\int }_0^2\left|\left|x-1\right|-x\right|dx$ is equal to :

Answer 1

Evaluate the integral

$$\begin{gathered} {\int }_0^2\left|\left|x-1\right|-x\right|dx={\int }_0^1\left|1-x-x\right|dx+{\int }_1^2\left|x-1-x\right|dx \\ ={\int }_0^1\left|2x-1\right|dx+{\int }_1^{2}1dx \\ ={\int }_0^{\frac{1}{2}}(1-2x)dx+{\int }_{\frac{1}{2}}^1(2x-1)dx+{\int }_1^{2}1dx \\ =\left[\left(\frac{1}{2}\right)-0-\left(\frac{1}{4}\right)-0\right]+\left(1-\left(\frac{1}{4}\right)\right)-\left(1-\left(\frac{1}{2}\right)\right)+1 \\ =\left(\frac{1}{2}\right)-\left(\frac{1}{4}\right)+\left(\frac{3}{4}\right)-\left(\frac{1}{2}\right)+1 \\ =\frac{3}{2} \end{gathered}$$

Hence, the required integral is $\frac{3}{2}.$