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- Jan 5th 2011, 01:41 PMBAHADEENIntegral by part
- Jan 5th 2011, 01:45 PMdwsmith
1. Isn't too bad have you done anything for this one?

- Jan 5th 2011, 01:47 PMdwsmith
2. $\displaystyle \displaystyle I_{n+1}=\int_0^1x^{n+1}e^{x-1}dx$

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

$\displaystyle \displaystyle uv-\int vdu$ - Jan 5th 2011, 01:51 PMpickslides
You are correct to say this is an integral by parts.

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

Recall that $\displaystyle e^{1-x} = \frac{e^1}{e^x} = e\times e^{-x}$ - Jan 5th 2011, 02:22 PMRandom Variable
$\displaystyle \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 $\displaystyle u = x^{n+1} $ and $\displaystyle dv = e^{1-x} \ dx )$

$\displaystyle \displaystyle = (-1-0) + (n+1) I(n) = (n+1) I(n) - 1$ - Jan 5th 2011, 02:31 PMRandom Variable
$\displaystyle I(0) = e-1 $

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

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

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