Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^4|x-1|dx$

Answer 1

$I={\int }_0^4|x-1|dx$
It can be seen that, $(x-1)\le 0$ when $0\le x\le 1$ and $(x-1)\ge 0$ when $1\le x\le 4$
$I={\int }_0^1|x-1|dx+{\int }_1^4|x-1|dx,(\because {\int }_a^{b}f(x)={\int }_a^{c}f(x)+{\int }_c^{b}f(x\left)\right)$
$={\int }_0^1-(x-1)dx+{\int }_0^4(x-1)dx$
$={[x-\frac{x^2}{2}]}_0^1+{[\frac{x^2}{2}-x]}_1^4$
$=1-\frac{1}{2}+\frac{(4{)}^2}{2}-4-\frac{1}{2}+1$
$=1-\frac{1}{2}+8-4-\frac{1}{2}+1=5$