Evaluate : `(i) int_(0)^(1)x(1-x)^(n)dx` `(ii) int_(0)^(1)x(1-x)^(3//2)dx`

Explanation

`(i)` We have `int_(0)^(1)x(1-x)^(n)dx=int_(0)^(1)(1-x)[1-(1-x)]^(n)dx` `[:'int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`=int_(0)^(1)(a-x)x^(n)dx=int_(0)^(1)x^(n)dx=int_(0)^(1)x^(n+1)dx`
`=[(x^(n+1))/((n+1))]_(0)^(1)-[(x^(n+2))/(n+2)]_(0)^(1)=((1)/(n+1)-(1)/(n+2))`
`=(1)/((n+1)(n+2))`
`(ii)` We have `int_(0)^(1)x(1-x)^(3//2)dx=int_(0)^(1)(1-x)[1-(1-x)]^(3//2)dx` `[:'int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`=int_(0)^(1)(1-x)x^(3//2)dx=int_(0)^(1)x^(3//2)dx-int_(0)^(1)x^(5//2)dx`
`=[(2)/(5)x^(5//2)]_(0)^(1)-[(2)/(7)x^(7//2)]_(0)^(1)=((2)/(5)-(2)/(7))=(4)/(35)`.