The upper bound given in this problem is a good approximation for the integral:
Problem: Prove that
OK, I have come up with an incredibly contrived solution suggested by the answer. Clearly it is equivalent to
, so this suggests the inequality . If there's a better way...
Here we go. Start with and on . Therefore .
This rearranges to . The LHS is the derivative of , so integrate both sides from to , where .
Thus and so . Integrate this from to and pops up.
Guess what? , and we arrive at our destination. Where's my hat? Rabbit, anyone?