Spoiler:

Is this correct?

Cause the texbook gives.

$\displaystyle (b^{b}a^{-a})^{\frac{1}{b-a}}e^{-1}$

The original question is at the bottom on the picture i posted.

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- May 19th 2009, 07:34 PMjustchillinTough Integral.
__Spoiler__:

Is this correct?

Cause the texbook gives.

$\displaystyle (b^{b}a^{-a})^{\frac{1}{b-a}}e^{-1}$

The original question is at the bottom on the picture i posted. - May 19th 2009, 10:21 PMIsomorphism
Yes thats correct. You can simplify your answer to the one given in the textbook. Remember that $\displaystyle e^{\ln x} = x$ $\displaystyle \, \, $ !

- May 19th 2009, 10:23 PMmatheagle