Define a function f : R `to` R by
`f(x)=max {|x|,|x-1|, . . .|x-n|}`
where n is a fixed natural number. Then `underset(0)overset(2n)(f)f(x)dx` is -

Options

A. n
B. `n^(2)`
C. 3n
D. `3n^(2)`

Answer

D

Explanation

`f(x)=max {|x|,|x-1|, . . .|x-n|}`

`int_(0)^(2n)f(x)dx=int_(0)^(n)f(x)dx+int_(n)^(2n)f(x)dx`
`=int_(0)^(n)|x-2n|dx+int_(n)^(2n)|x|dx`
`=int_(0)^(n)(2n-x)dx+int_(n)^(2n)x.dx`
`=[2nx-(x^(2))/(2)]_(0)^(n)+[(x^(2))/(2)]_(n)^(2n)`
`(2n^(2)-(n^(2))/(2))+((4n^(2))/(2)-(n^(2))/(2))`
`=(3n^(2))/(2)+(3n^(2))/(2)=3n^(2)``underset(0)overset(2n)(f)f(x)dx=underset(0)overset(n)(f)f(x)dx+underset(0)overset(2n)(f)f(x)dx`
`=underset(0)overset(n)(f)|x-2n|dx+underset(0)overset(2n)(f)|x|dx`
`=underset(0)overset(n)(f)(2n-x)dx+underset(0)overset(2n)(f)x.dx`