Reduction integration** (1 Viewer)

[norelle]

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]

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

Then integrate by parts

 

[norelle]

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

Then integrate by parts

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

 

[norelle]

oh ye i get it now thankss