Question

Evaluate: ${\int }_1^4\{|x-1|+|x-2|+|x-4|\}dx$.

Answer 1

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

We know that ${\int }_a^{b+c}f\left(x\right)dx={\int }_a^{b}f\left(x\right)dx+{\int }_a^{c}f\left(x\right)dx$

${\int }_1^{4}f\left(x\right)dx={\int }_1^{2}f\left(x\right)dx+{\int }_2^{4}f\left(x\right)dx$

${\int }_1^{2}f\left(x\right)dx={\int }_1^2\left\{\right(x-1)-(x-2)-(x-4\left)\right\}dx$

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

$={\int }_1^2(5-x)dx={[5x-\frac{x^2}{2}]}_1^2$

$=(10-2)-(5-\frac{1}{2})=\frac{7}{2}$

${\int }_2^{4}f\left(x\right)dx={\int }_2^4\left\{\right(x-1)+(x-2)+(x-3\left)\right\}dx$

$={\int }_2^4\{3x-6\}$

$={(\frac{3x^2}{2}-6x)}_2^4$

$=\left(\right(24-24)-(6-12\left)\right)=6$

${\int }_1^{4}f\left(x\right)dx={\int }_1^{2}f\left(x\right)dx+{\int }_2^{4}f\left(x\right)dx$

$=\frac{7}{2}+6=\frac{19}{2}$.