Tackling Tricky Integration Homework

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SUMMARY

The discussion focuses on solving a complex integration problem involving the expression for E, defined as E = \frac{e^2}{4 \pi \epsilon_0 a_0^3 } \left( - b \int_0^b e^{- \frac{2r}{a_0}} dr + \int_0^b r e^{- \frac{2r}{a_0}} dr \right). The integrals used are \int_0^x e^{-u} du = 1 - e^{-x} and \int_0^x u e^{-u} du = 1 - e^{-x} - xe^{-x}. The solution attempt reveals a potential sign error in the third term of the expression, which should be negative. This correction is crucial for the accuracy of the final result.

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ronaldoshaky
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Homework Statement

[tex]E = \frac{e^2}{4 \pi \epsilon_0 a_0^3 } \left( - b \int_0^b e^{- \frac{2r}{a_0}} dr + \int_0^b r e^{- \frac{2r}{a_0}} dr \right)[/tex]

I have to use these integrals,

[tex]\int_0^x e^{-u} du = 1 - e^{-x}[/tex] and [tex]\int_0^x u e^{-u} du = 1 - e^{-x} - xe^{-x}[/tex]

I get:

The Attempt at a Solution



[tex]E = \frac{e^2}{4 \pi \epsilon_0 a_0^3 } \left( - \frac{a_0 b}{2} (1 - e^{- \frac{2b}{a_0}}) + \left(- \frac{- a_0 b}{2} e^{- \frac{2b}{a_0}} - \frac{a_0^2}{4} e^{- \frac{2b}{a_0}} + \frac{a_0^2}{4} \right) \right)[/tex]
 
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I think you blew a sign in the third term. It should be negative, not positive. Otherwise, it looks correct.
 

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