Reduction integration** (1 Viewer)

norelle · Jun 7, 2009

Given that I(2n+1) = integral (0->1) x^(2n+1) e^(x^2) dx

Show that,
I (2n+1) = 1/2 e - n I(2n+1)

Thanks a lot.

And can anyone share their method when dealing with reduction integration?
I always struggling with this type of questions..
How do you normally approach it?
like.. I can never know which method should I use, I have to try each method..
sometimes u need to use by parts, sometimes split I(n) into I(n-2) x I(2) ..
sometimes u need to + I(n-1) - I (n+1) into the question..
urggg this is so annoying........

Drongoski · Jun 7, 2009

Let u = x^2n; v=e^x^2; dv = 2xe^x^2dx

Then integrate by parts

$\100dpi I_{2n+1}\ \ =\ \ \int_{0}^{1}x^{2n+1}\ e^{x^{2}}\ dx\ =\ \frac{1}{2}\int_{0}^{1}x^{2n}\cdot 2xe^{x^{2}}\ dx\\ \\ =\ \ \frac{1}{2}\left [x^{2n}e^{x^{2}} \right ]^{1}_{0}\ -\ \frac{1}{2}\int_{0}^{1} e^{x^{2}}\cdot 2n\ x^{2n-1}dx\ \ =\ \ \frac{1}{2}\left [ e-0 \right ]\ -\ nI_{2n-1}\\ \\ \ =\ \ \frac{1}{2}e\ -\ nI_{2n-1}$

norelle · Jun 7, 2009

Drongoski said:
Let u = x^2n; v=e^x^2; dv = 2xe^x^2dx

Then integrate by parts

$\100dpi I_{2n+1}\ \ =\ \ \int_{0}^{1}x^{2n+1}\ e^{x^{2}}\ dx\ =\ \frac{1}{2}\int_{0}^{1}x^{2n}\cdot 2xe^{x^{2}}\ dx\\ \\ =\ \ \frac{1}{2}\left [x^{2n}e^{x^{2}} \right ]^{1}_{0}\ -\ \frac{1}{2}\int_{0}^{1} e^{x^{2}}\cdot 2n\ x^{2n-1}dx\ \ =\ \ \frac{1}{2}\left [ e-0 \right ]\ -\ nI_{2n-1}\\ \\ \ =\ \ \frac{1}{2}e\ -\ nI_{2n-1}$

I dont get the second step~~ where is the 2x ?

norelle · Jun 7, 2009

oh ye i get it now thankss

Reduction integration** (1 Viewer)

norelle

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Drongoski

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