Question

$\underset{-a}{\overset{a}{\int }}(|x|+|x-2|)dx=22,a>2,$ then the value of $2\underset{0}{\overset{a}{\int }}|x-3|dx$ is

Answer 1

Given: $\underset{-a}{\overset{a}{\int }}(|x|+|x-2|)dx=22$
$\Rightarrow \underset{-a}{\overset{0}{\int }}(2-2x)dx+\underset{0}{\overset{2}{\int }}2dx+\underset{2}{\overset{a}{\int }}(2x-2)dx=22$
$\Rightarrow [2x-x^2{]}_{-a}^0+[2x{]}_0^2+[x^2-2x{]}_2^a=22$
$\Rightarrow 2a+a^2+4+a^2-2a-0=22$
$\Rightarrow a^2=9\Rightarrow a=3$
$\therefore 2\underset{0}{\overset{3}{\int }}|x-3|dx$
$=2\underset{0}{\overset{3}{\int }}(3-x)dx$
$=2{(3x-\frac{x^2}{2})}_0^3$
$=9$