Question

Evaluate: ${\int }_2^5(|x-2|+|x-3|+|x-5|)dx$.

Answer 1

Let $I={\int }_2^5(|x-2|+|x-3|+|x-5|)dx$

$=\frac{1}{2}(x-2)|x-2|+\frac{1}{2}(x-3)|x-3|+\frac{1}{2}(x-5)|x-5|$ (using indefinite integral form).

By the fundamental theorem of calculus ${\int }_a^{b}F\left(x\right)dx=f\left(b\right)-f\left(a\right)$,

so just evaluate integral endpoints and that's the answer.
$\left(\frac{1}{2}\right(x-2)|x-2|+\frac{1}{2}(x-3)|x-3|+\frac{1}{2}(x-5\left)|x-5|\right){|}_{(x=5)}=\frac{13}{2}$

$\left(\frac{1}{2}\right(x-2)|x-2|+\frac{1}{2}(x-3)|x-3|+\frac{1}{2}(x-5\left)|x-5|\right){|}_{(x=2)}=-5$

$I={\int }_2^5(|x-2|+|x-3|+|x-5|)dx$

$={\left(\frac{1}{2}\right(x-2)|x-2|+\frac{1}{2}(x-3)|x-3|+\frac{1}{2}(x-5\left)|x-5|\right)|}_{(x=5)}-{\left(\frac{1}{2}\right(x-2)|x-2|+\frac{1}{2}(x-3)|x-3|+\frac{1}{2}(x-5\left)|x-5|\right)|}_{(x=3)}$

$=\frac{23}{2}$

$I={\int }_2^5(|x-2|+|x-3|+|x-5|)dx=\frac{23}{2}=11.5$