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# How would demonstrate equality int_1^n (|x-1|+-----+|x-n|)dx =1^2+2^2+------+(n-1)^2 if...

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How would demonstrate equality int_1^n (|x-1|+-----+|x-n|)dx =1^2+2^2+------+(n-1)^2 if n>=2, n natural number?

Posted by ruals on September 24, 2013 at 10:11 AM via web and tagged with calculus, equality, integral, math, sum

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`I=int_1^n(|x-1|+|x-2|+|x-3|+.......|x-n|)dx`

`=1^2+2^2+3^2+.....+(n-1)^2`

`Let`

`f(x)=|x-1|+|x-2|+|x-3|+.....+|x-n|`

`int_1^nf(x)dx=int_1^2f(x)dx+int_2^3f(x)dx+....+int_{n-1}^nf(x)dx`

`` Consider

`I_1=int_1^2f(x)dx=int_1^2(x-1-x+2-x+3..........-x+n)dx`

`=int_1^2(2(x-1)-nx+(n(n+1))/2) dx`

`I_1=2int_1^2(x-1)dx-int_1^2(nx-(n(n+1))/2)dx`

`similarly`

`I_2=2int_2^3(x-1)dx+2int_2^3(x-2)dx-int_2^3(nx-(n(n+1))/2)dx`

`......`

`.......`

Adding

`I_1+I_2+.........+I_(n-1)=2int_1^n(x-1)dx+2int_2^n(x-2)dx+....-int_1^n(nx-(n(n+1))/2)dx`

`I_1=2int_1^n(x-1)dx=(n-1)^2`

`I_2=2int_2^n(x-2)dx=(n-2)^2`

`..............`

`.................`

`I_(n-1)=1`

Thus

`I=I_1+I_2+....+I_(n-1)`

Hence proved.

``

Posted by aruv on September 24, 2013 at 11:40 AM (Answer #1)

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