Question

Evaluate the definite integral: $∫^4_1[|x-1|+|x-2|+|x-3|]dx$

Answer 1

$Let I=∫^4_1[|x-1|+|x-2|+|x-3|]dx$

$⇒I=∫^4_1|x-1|dx+∫^4_1|x-2|dx+∫^4_1|x-3|dx$

$I=I_1+I_2+I_3...(1)$

$Where,I_1=∫_1^4|x-1|dx,I_2=∫_1^4|x-2|dx,and\, I_3=∫_1^4|x-3|dx$

$I_1=∫_1^4|x-1|dx$

$(x-1)≥0 for 1≤x≤4$

$∴I_1=∫_1^4(x-1)dx$

$⇒I_1=[\frac{x^2}{x}-x]_1^4$

$⇒I_1=[8-4-\frac{1}{2}+1]=\frac{9}{2}...(2)$

$I_2=∫_1^4|x-2|dx$

$x-2≥0\, for\, 2≤x≤4\, and\, x-2≤0\, for\, 1≤x≤2$

$∴I_3=∫_1^2(2-x)dx+∫_2^4(x-2)dx$

$⇒I_2=[4-2-2+\frac{1}{2}]+[8-8-2+4]$

$⇒I_2=\frac{1}{2}+2=\frac{5}{2}...(3)$

$I_3=∫^4_1|x-3|dx$
$x-3\geq0\,for\, 3\leq x \leq4\,and\,x-3\leq0\,for\,1\leq x\leq3$
$\therefore I_3=\int^3_1(3-x)dx+\int^4_3(x-3)dx$

$⇒I_3=[3x-\frac{x^2}{2}]_1^3+[(\frac{x^2}{2}-3x)]^4_3$

$⇒I_3=[9-\frac{9}{2}-3+\frac{1}{2}]+[8-12-\frac{9}{2}+9]$

$⇒I_3=[6-4]+[\frac{1}{2}]=\frac{5}{2}...(4)$

$From equation(1),(2),(3),and(4),we obtain$

$I=\frac{9}{2}+\frac{5}{2}+\frac{5}{2}=\frac{19}{2}$