Question

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

Answer 1

$I={\int }_0^4|x-1|dx$

We know that $|x-1|=\{\begin{array}{cc}1-x& 0<x<1\\ 0& x=1\\ x-1& 4\ge x>1\end{array}$

$I={\int }_0^{1}1-xdx+{\int }_1^{4}x-1dx$

$I={[x-\frac{x^2}{2}]}_0^1+{[\frac{x^2}{2}-x]}_1^4$

$I=[1-\frac{1}{2}]+[\frac{16}{2}-4-\frac{1}{2}+1]$

$I=\frac{1}{2}+[5-\frac{1}{2}]=\frac{1}{2}+\frac{9}{2}$

$[I=5]$

Hence value of given integral is $5$.