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int(1)^(2013)[(x-1)(x-2)...(x-2013)]dx...

`int_(1)^(2013)[(x-1)(x-2)...(x-2013)]dx`

A

`(2013)^(2)`

B

`(2012)(2013)(2014)`

C

`2013!`

D

`0`

Text Solution

Verified by Experts

The correct Answer is:
D
`I=int_(1)^(2013)(x-1)(x-2)(x-3)...(x-2013)dx`
Using `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`
`int_(1)^(2013)(2013-x)(2012-x)...(1-x)=-I`
`rArr" "2I=0rArr I=0`
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