Question

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

Answer 1

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