IBP Help (1 Viewer)

[conics2008]

I(m)= S from 1-0 x^m*e^x (*=times)

show that I(m+1)= e-(m+1)I(m) ??

I got up to here

I(m) = { x^m*e^x} from 1 and 0 - S from 1 and 0 m*x^m-1*e^x

which leaves us with

I(m)= e-mI(m-1)

I tried subbin in m+1

I(m+1)= e - mI(m)

???

Help please Thanks xD

 

[conics2008]

sorry 3unitz I didn't understand what you were trying to say.

I got lost through out the working out..

I cant seem to understand computer handwriting..

sorry

 

[lolokay]

I_m+1 = S[1,0] x^m+1e^x (can you do that? as in assuming that is what I_m+1 is equal to)
= [1,0]x^m+1e^x - (m+1)*S[1,0]x^me^x
= e - (m+1)I_m

I(m)= e-mI(m-1)

I tried subbin in m+1

I(m+1)= e - mI(m)

???

Help please Thanks xD

you didn't sub in m+1 in mI(m-1) properly. It should be (m+1)I(m)

 

[conics2008]

^^

Thanks 3units and Lolkay..

I spoted my mistake..

when i let m=m+1 i have to sub in both I(m-1) and M lol xD

silly mistake thanks alot xD