${\int }_0^4 |\mathrm{x}-1|\mathrm{dx}$ is equal to

Solution:

Let ${\int }_0^4 |\mathrm{x}-1|\mathrm{dx}$

It can be seen that, $(\mathrm{x}-1)\le 0$ when $0\le \mathrm{x}\le 1$ and 

$(\mathrm{x}-1)\ge 0\text{ when }1\le \mathrm{x}\le 4$

$\begin{array}{l}\therefore \mathrm{I}={\int }_0^1 |\mathrm{x}-1|\mathrm{dx}+{\int }_1^4 |\mathrm{x}-1|\mathrm{dx}\\ \left.={\int }_0^1 (1-\mathrm{x})\mathrm{dx}+{\int }_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{\int }_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\right]\\ ={\left[\mathrm{x}-\frac{{\mathrm{x}}^2}{2}\right]}_0^1+{\left[\frac{{\mathrm{x}}^2}{2}-\mathrm{x}\right]}_1^4\\ =\left(1-\frac{1}{2}\right)-0+\left(\frac{4^2}{2}-4\right)-\left(\frac{1}{2}-1\right)=\frac{1}{2}+4+\frac{1}{2}=5\end{array}$