Question

${\int }_0^2(|x|+|x-1|)dx=$

• A. 1
• B. -1
• C. 2
• D. 3

Answer 1

The correct option is D 3

${\int }_0^2|x|+|x-1|dx$
$|x|+|x-1|$
$=-2x+1x<0$
$=10⩽x<1$
$=2x-1x\ge 1$
${\int }_0^{1}1dx+{\int }_1^2(2x-1)dx$
$=x{|}_0^1+\frac{x^2}{2}-x{|}_1^2$
$1+[4-2]+[1-1]$
$=3$