Question

Solve:
${\int }_1^4\left[x-1|+|x-2|+|x-3\right]dx$

Answer 1

$I={\int }_1^4|x-1|+|x-2|+|x-3|dx$

$\Rightarrow x\in [1,2]$

$|x-1|=x-1$

$|x-2|=2-x$

$|x-3|=3-x$

$\Rightarrow x\in [2,3]$

$|x-1|=x-1$

$|x-2|=x-2$

$|x-3|=3-x$

$\Rightarrow x\in [3,4]$

$|x-1|=x-1$

$|x-2|=x-2$

$|x-3|=x-3$

$I={\int }_1^2|x-1|+|x-2|+|x-3|dx+{\int }_2^3|x-1|+|x-2|+|x-3|dx+{\int }_2^4|x-1|+x-2|+|x-3|dx$

$={\int }_1^2(x-1)+(2-x)+(3-x)dx+{\int }_2^3(x-1)+(x-2)+(3-x)dx+$

${\int }_1^2(4-x)dx+{\int }_2^{3}xdx+{\int }_3^{4}3(x-2)dx$

$={4x-\frac{x^2}{2}]}_1^2+{\frac{x^2}{2}]}_2^3+{\frac{3x^2}{2}]}_3^4-{2x]}_3^4$

$=4(2-1)-\left[\frac{4-1}{2}\right]+\left[\frac{9-4}{2}\right]+\frac{3}{2}[16-9]+2[4-3]$

$=4-\frac{3}{2}+\frac{5}{2}+\frac{21}{2}-2$

$=2+\frac{23}{2}=\frac{27}{2}$

$\therefore {\int }_1^4[|x-1|+|x-2|+|x-3|]=\frac{27}{2}$.