[Math] Find $n$ such that $\int_0^1 e^x(x-1)^n \,dx = 16-6e$.

calculusdefinite integralsintegration

Find the value of $n$ (where $n$ is a positive integer less than or equal to $5$) such that $$\int_0^1 e^x(x-1)^n \,dx = 16-6e.$$

I have tried this question by two methods.

  1. Integration by parts
  2. Using king property

Neither of these lead me anywhere conclusive. The integral calculator has used a gamma function which is not in my high school syllabus.

Best Answer

Let $I(n) = \int_{0}^{1} e^{x}(x-1)^n dx$. Integration by parts results in $$I(n)=\int_{0}^{1} e^{x}(x-1)^n dx = (-1)^{n+1} -n \int_{0}^{1} e^x(x-1)^{n-1} dx=(-1)^{n+1}-nI(n-1).$$ Now use the fact that $I(0)=e-1$ to get $I(1) = 1-I(0)=2-e$, $I(2)=-1-2I(1)=-5+2e$, and $I(3)=1-3I(2) = 16-6e$.

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