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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.

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    $\begingroup$ I had to look up "King property of integration"... $\endgroup$ – lhf Jul 4 '18 at 17:42
  • $\begingroup$ Binomial theorem? $\endgroup$ – copper.hat Jul 4 '18 at 17:42
  • $\begingroup$ try with $n=2$ to reach that result $\endgroup$ – Dr. Sonnhard Graubner Jul 4 '18 at 17:47
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    $\begingroup$ There are a couple of things to notice. The right-hand side is negative, so $n$ must be odd, and we only have to try $n=1,3,5.$ Also, the integrand decreases in absolute value as $n$ increases, so that if we check $n=3,$ if that doesn't work, we'll know which one to check next. $\endgroup$ – saulspatz Jul 4 '18 at 17:49
  • $\begingroup$ @Dr.SonnhardGraubner Are you sure? $16-6e<0$ $\endgroup$ – saulspatz Jul 4 '18 at 17:51
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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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$$I_n = \int_0^1 e^x(x-1)^n \,\text{d}x $$

For $n\geq 1$

$$ I_n = \int_0^1 e^x(x-1)^n \,\text{d}x = (-1)^{n+1} -n\int_0^1 e^x(x-1)^{n-1}\,\text{d}x = (-1)^{n+1}-nI_{n-1} $$ and $I_0 = e-1$. $I_n$ is of the form $a_ne+b_n$ where $a_n$ and $b_n$ are integers. We can easily see that $a_n=(-1)^nn!$ so that $n$ must equal $3$.

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Here's a hint. I think what may be bothering you is that you think you have to compute many integrals to do the problem, but there's an easier way. Integrating by parts, $$\int{x^ne^x\mathrm{dx}}=x^ne^x-n\int{x^{n-1}e^x\mathrm{dx}},$$ and since we know $\int{e^x\mathrm{dx}}=e^x+C,$ we can deduce $\int{xe^x\mathrm{dx}}=xe^x-e^x+C,$ and so on.

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Characteristic function approach: \begin{eqnarray} \int_0^1 e^x e^{t(x-1)} dx &=& {1 \over 1+t} (e-e^{-t}) \end{eqnarray}

Differentiate with respect to $t$ $k$ times and set $t=0$.

Or eyeball the first few terms: $(1-t+t^2-t^3+\cdots)(e - 1 +t -{1 \over 2!} t^2 +{1 \over 3!} t^3 -\cdots)$.

Since the multiplier of $e$ is $6 = 3!$, we can guess that $n=3$ and quickly verify correctness.

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$\begin{array}\\ I_n &=\int_0^1 e^x(x-1)^ndx\\ &=\int_{-1}^0 e^{x+1}x^ndx\\ &=e\int_{-1}^0 e^{x}x^ndx\\ &=e\int_0^1 e^{-x}(-x)^ndx\\ &=e(-1)^n\int_0^1 e^{-x}x^ndx\\ &=e(-1)^n\left(-e^{-x}\sum_{k=0}^n x^k\dfrac{n!}{k!}\right)\big|_0^1 \qquad\text{(Incomplete Gamma function for integer parameter)}\\ &=e(-1)^{n+1}\left(e^{-x}\sum_{k=0}^n x^k\dfrac{n!}{k!}\right)\big|_0^1\\ &=e(-1)^{n+1}\left(e^{-1}\sum_{k=0}^n \dfrac{n!}{k!}-n!\right)\\ &=(-1)^{n+1}\left(\sum_{k=0}^n \dfrac{n!}{k!}-n!e\right)\\ &=(-1)^{n+1}\sum_{k=0}^n \dfrac{n!}{k!}+(-1)^nn!e\\ \end{array} $

For your case, $(-1)^nn! =-6$ so $n=3$.

Note that

$\begin{array}\\ I_n &=(-1)^{n+1}n!\sum_{k=0}^n \dfrac{1}{k!}+(-1)^nn!e\\ &=(-1)^{n+1}n!(e-\dfrac1{(n+1)!}+O(\dfrac1{(n+2)!}))+(-1)^nn!e\\ &=(-1)^{n+1}n!e-(-1)^{n+1}\dfrac1{n+1}+O((-1)^{n+1}\dfrac1{(n+1)(n+2)}))+(-1)^nn!e\\ &=(-1)^{n}\dfrac1{n+1}+O((-1)^{n+1}\dfrac1{(n+1)(n+2)})\\ \end{array} $

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