Question

The value of integral ${\int }_0^4|x-1|dx$ is

• A. 4
• B. 5
• C. 7
• D. 9

Answer 1

Answer: (B)

${\int }_0^4|x-1|dx={\int }_0^1(1-x)dx+{\int }_1^4(x-1)dx$
$={\left[x-\frac{x^2}{2}\right]}_0^1+{\left[\frac{x^2}{2}-x\right]}_1^4$
$=1-\frac{1}{2}+\left[8-4-\left(\frac{1}{2}-1\right)\right]$
$=\frac{1}{2}+4+\frac{1}{2}=5$