Integral by part

**dwsmith** · January 5th 2011, 01:45 PM

1. Isn't too bad have you done anything for this one?

**dwsmith** · January 5th 2011, 01:47 PM

2. $\displaystyle I_{n+1}=\int_0^1x^{n+1}e^{x-1}dx$

3. $\displaystyle I_{3}=\int_0^1x^{3}e^{x-1}dx$

$\displaystyle uv-\int vdu$

**pickslides** · January 5th 2011, 01:51 PM

You are correct to say this is an integral by parts.

First step is to find $I_0$ and $I_1$

Recall that $e^{1-x} = \frac{e^1}{e^x} = e\times e^{-x}$

**Random Variable** · January 5th 2011, 02:22 PM

$\displaystyle I(n+1) = \int^{1}_{0} x^{n+1} e^{1-x} \ dx = - x^{n+1} e^{1-x} \Big|^{1}_{0} + \ (n+1) \int^{1}_{0} x^{n} e^{1-x} \ dx$ ( let $u = x^{n+1}$ and $dv = e^{1-x} \ dx )$

$\displaystyle = (-1-0) + (n+1) I(n) = (n+1) I(n) - 1$

**Random Variable** · January 5th 2011, 02:31 PM

$I(0) = e-1$

$I(1) = 1(e-1) - 1 = e-2$

$I(2) = 2(e-2) - 1 = 2e-5$

$I(3) = 3(2e-5) - 1 = 6e - 16$

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