You are welcome.

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Evaluate this integral:

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Yes that is almost correct!

]]>Hi;

Evaluate this integral:

Prove the inequality:

1/2[x - log(sinx + cosx)]

]]>Hi;

Another integral this one is also easy:

(e^2x)/2 - (e^-2x)/2

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Another integral this one is also easy:

]]>The inequality was the easy one. Now how about the integral it is even easier. If you start substituting you will go mad.

]]>Jane wrote:

The inequality holds for

as well, in which case the method will not be trivial for .

But the problem states x > 0 and mathsy is right using the expansion of e^x makes it trivial.

Jane wrote:

And note that there is an apostrophe in

Thats.

That is true Jane, poor typing and high speeds, a lethal combination. Thanks for the correction.

Jane wrote:

Ever seen this theorem?

Yes Jane, I have heard of that theorem but I forgot it.

]]>Are we allowed to express the LHS of the inequality as a Taylor series? If so, the problem becomes trivial.

bobbym wrote:

Thats how I would do it.

The inequality holds for

as well, in which case the method will not be trivial for .And note that there is an apostrophe in **Thats**.

Let be convex and differentiable. Then

[align=center]

[/align]So you just put

and .(What Im saying is that no point on a convex and differentiable curve lies below the corresponding point on a straight line that is tangent to any part of the curve.)

PS: I just looked up Wikipedia. This theorem is stated there without proof.

]]>Thats how I would do it. But in this Putnam problem book he doesn't use that method. Why, I can't say.

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