Question

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

Answer 1

Solution :- given

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

$\therefore$ Required integration can be written as

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

$={\int }_1^2(x-1)+(-x+2)+(-x+4)dx+{\int }_2^4(x-1)+(x-2)+(-x+4)dx$

$\therefore$ Evaluating 1st part:

$\Rightarrow {\int }_1^2(x-1-x+2-x+4)dx$

$={\int }_1^2(5-x)dx{=}_1^2[5x-\frac{x^2}{2}]=\left[\right(10-\frac{4}{2})-(5-\frac{1}{2}\left)\right]=8-\frac{9}{2}$

$=\frac{7}{2}$

$\therefore$ Evaluating 2nd part

$\Rightarrow {\int }_2^4(x-1+x-2-x+4)dx={\int }_2^4(x+1)$

${=}_2^4[\frac{x^2}{2}+x]=\left[\right(\frac{4^2}{2}+4)-(\frac{2^2}{2}+2\left)\right]=12-4=8$

$\therefore {\int }_1^4[|x-1|+|x-2|+|x-4|]dx=\frac{7}{2}+8=\frac{23}{2}$ Ans