Evaluate `int_(0)^(1)x(1-x)^3dx`

Answer

A, B, D

Explanation

`f(2-x)=f(2+x),f(4-x)=f(4+x)`
or `f(4+x)=f(4-x)=f(2+2-x)=f(2-(2-x))=f(x)`
Thus, the period of `f(x)` is 4.
`int_(0)^(50)f(x)dx=int_(0)^(48)f(x)dx+int_(48)^(50)f(x)dx`
`=12 int_(0)^(4)f(x)dx+int_(0)^(2)f(x)dx`
[In second integral, replacing `x` by `x+48` and then using `f(x)=f(x+48)`]
`=12(int_(0)^(2)f(x)dx+int_(0)^(2)f(4-x)dx)+5`
`=12(int_(0)^(2)f(x)dx+int_(0)^(2)f(4+x)dx)+5`
`=24int_(0)^(2)f(x)dx+5=125`
`int_(-4)^(46)f(x)dx=int_(-4)^(-2)f(x)dx+int_(-2)^(-2+48)f(x)dx`
`=int_(0)^(2)f(x+4)dx+12int_(0)^(4)f(x)dx`
`=int_(0)^(2)f(x)dx+24int_(0)^(2)f(x)dx`
`=125`
also `int_(2)^(52)f(x)dx=int_(2)^(4)f(x)dx+int_(4)^(4+48)f(x)dx`
`=int_(0)^(2)f(4-x)dx+12int_(0)^(4)f(x)dx`
`=int_(0)^(2)f(4+x)dx+24int_(0)^(2)f(x)dx`
`=int_(0)^(2)f(x)dx+24int_(0)^(2)f(x)dx`
`=125`
`int_(1)^(51)f(x)dx=int_(1)^(3)f(x)dx+int_(3)^(3+48)f(x)dx`
`=int_(1)^(3)f(x)dx+12int_(0)^(4)f(x)dx`
`=int_(0)^(2)f(x+1)dx+24int_(0)^(2)f(x)dx`
`!=125`