The upper bound given in this problem is a good approximation for the integral:

Problem: Prove that

Printable View

- Jul 6th 2009, 04:28 AMNonCommAlgAn inequality
The upper bound given in this problem is a good approximation for the integral:

__Problem__: Prove that - Jul 6th 2009, 11:53 AMSampras
By the Rectangle Rule, . Or we know that has a bounded derivative over . Now by the MVT, for some . Also, . Then

- Jul 6th 2009, 12:00 PMhalbard
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?