Evaluate:
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Since and x is close to 0, for the interval we're integrating in. At this point we can use the Maclaurin series for : I'll leave the rest to you.
The answer given in my text is simply . Does that mean ? EDIT: Yes! I forgot the Maclaurin Series.
Yes, you can see that when using the Maclaurin series since
We can also do it like this : Differentiating numerator and denominator,
Hi, Originally Posted by fardeen_gen Evaluate: A slightly different solution : Let be an antiderivative of . We have hence
Last edited by flyingsquirrel; June 10th 2009 at 07:03 AM. Reason: + one step
Leibnitz Rule: If then = = =
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