Is okay to distribute the e^(-x) in this case or is that just more work than expected?

Question

#int_0^ln2 2pi(ln2-x)e^-xdx#

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Answer 1 · 2018-03-04T17:33:39

It certainly is OK. Try it to see whether you have a preference for which way to do it. (I have an opinion, but yours may be different.)

You may do either:

#int_0^ln2 2pi(ln2-x)e^-xdx#

or

#int_0^ln2 2pi ln2 e^(-x) dx - 2 pi int_0^ln2 xe^-xdx#