Question

Evaluate the definite integral ${\int }_1^4[|x-1|+|x-2|+|x-3|]dx$

Answer 1

Let $I={\int }_1^4[|x-1|+|x-2|+|x-3|]dx$\Rightarrow I=\int_1^4|x-1|dx+\int_1^4|x-2|dx+\int_1^4|x-3|dxI=I_1+I_2+I_3$...............\left(1\right)where,$I_1=\int_1^4|x-1|dx, I_2=\int_1^4|x-2|dx$,and$I_3=\int_1^4|x-3|dxI_1=\int_1^4|x-1|dx(x-1)\geq 0$for$1\leq x\leq 4\therefore I_1=\int_1^4(x-1)dx\Rightarrow I_1=\left [\dfrac {x^2}{2}-x\right ]_1^4\Rightarrow I_1=\left [8-4-\dfrac {1}{2}+1\right ]=\dfrac {9}{2}$...............\left(2\right)$I_2=\int_1^4|x-2|dxx-2\leq 0$for$2\leq x\leq 4$and$x-2\leq 0$for$1\leq x\leq 2\therefore I_2=\int_1^2(2-x)dx+\int_2^4(x-2)dx\Rightarrow I_2=\left [2x-\dfrac {x^2}{2}\right ]_1^2+\left [\dfrac {x^2}{2}-2x\right ]_2^4\Rightarrow I_2=[4-2-2+\dfrac {1}{2}]+[8-8-2+4]\Rightarrow I_2=\dfrac {1}{2}+2=\dfrac {5}{2}$..............\left(3\right)$I_3=\int_1^4|x-3|dxx-3\geq 0$for$3\leq x\leq 4$and$x-3\leq 0$for$1\leq x\leq 3\therefore I_3=\int_1^3(3-x)dx+\int_3^4(x-3)dx\Rightarrow I_3=\left [3x-\dfrac {x^2}{2}\right ]_1^3+\left [\dfrac {x^2}{2}-3x\right ]_3^4\Rightarrow I_3=\left [9-\dfrac {9}{2}-3+\dfrac {1}{2}\right ]+\left [8-12-\dfrac {9}{2}+9\right ]\Rightarrow I_3=[6-4]+\left [\dfrac {1}{2}\right ]=\dfrac {5}{2}$.............\left(4\right)Fromequations\left(1\right),\left(2\right),\left(3\right),and\left(4\right),weobtain$I=\displaystyle \frac {9}{2}+\frac {5}{2}+\frac {5}{2}=\frac {19}{2}$