Given `int_(0)^(1)e^(x)(x-1)^(n)dx=16-6e`
where `n epsilonN` and `nge5`.
To find the value f `n`, let
`I_(n)=int_(0)^(1)e^(x)(x-1)^(n)dx`
`=[(x-1)^(n)e^(x)]_(0)^(1)-int_(0)^(1)n(x-1)^(n-1)e^(x)dx`
`=-(-1)^(n)-int_(0)^(1)n(x-1)^(n-1)e^(x)dx`
or `I_(n)=(-1)^(n+1)-nI_(n-1)`.............1
Also `I_(1)=int_(0)^(1)e^(x)(x-1)dx`
`=[e^(x)(x-1)]_(0)^(1)-int_(0)^(1)e^(x)dx`
`=-(-1)-(e^(x))_(0)^(1)`
`=1-(e-1)=2-e`
Using equation 1 we get
`I_(2)=(-1)^(3)-2I_(1)=-1-2(2-e)=2e-5`
Similarly`, I_(3)=(-1)^(4)-3I_(2)=1-3(2e-5)`
Thus, `n=3`
