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Mark44
#6
Dec1-10, 05:24 PM
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Quote Quote by unknownuser9 View Post
Sorry, i really cant use Latex. I did it in word and this is the best i can do:
It's not that hard to use LaTeX, and it's much more readable that what you have below. For me the limits of integration show up as a rectangular pattern of dots, like this: ▒.
Bring the constants out of the integral.
[tex]\frac{4}{a_0^3}\int_0^{a_0}r^2e^{-2r/a_0}dr[/tex]

To see what I did in LaTeX, click the expression above.

After you make the substitution, use integration by parts twice. After you integrate, undo your substitution. Check your final result by differentiating - you should get back to the original integrand.
Quote Quote by unknownuser9 View Post

P= ∫▒4/(a_0^3 ) r^2 e^((-2r)/a_0   ) dr
P= ∫▒4/(a_0^3 ) 〖a_0〗^2 r^2 e^(-2x ) a_0 dx
P= ∫▒〖4x〗^2  e^(-2x) dx
P=4∫▒x^2  e^(-2x) dx
P= -1/2 x^2 e^(-2x)+∫▒〖xe^(-2x) 〗 dx
P= -1/2 x^2 e^(-2x)+xe^(-2x) ∫▒1/2  e^(-2x) dx
P= -1/2 x^2 e^(-2x)+xe^(-2x)-1/4  e^(-2x)
So now if i use r = 0 and r = a_0 as my limits i get:

P= |-1/2 x^2 e^(-2x)+xe^(-2x)-1/4  e^(-2x) |  a_00
P=(-1/2 a_0^2 e^(-2a_0 )+a_0 e^(-2a_0 )-1/4 e^(-2a_0 ) )- (0+0-1/4 e^0)
P=-1/2 a_0^2 e^(-2a_0 )+a_0 e^(-2a_0 )-1/4 e^(-2a_0 )+1/4
Am i close? Thanks.