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Question

# Evaluate the definite integral ∫41[|x−1|+|x−2|+|x−3|]dx

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Solution

## Let I=∫41[|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...............(1)where,I_1=\int_1^4|x-1|dx, I_2=\int_1^4|x-2|dx,andI_3=\int_1^4|x-3|dxI_1=\int_1^4|x-1|dx(x-1)\geq 0for1\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}...............(2)I_2=\int_1^4|x-2|dxx-2\leq 0for2\leq x\leq 4andx-2\leq 0for1\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}..............(3)I_3=\int_1^4|x-3|dxx-3\geq 0for3\leq x\leq 4andx-3\leq 0for1\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}.............(4)Fromequations(1),(2),(3),and(4),weobtainI=\displaystyle \frac {9}{2}+\frac {5}{2}+\frac {5}{2}=\frac {19}{2}\$

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