Let \(f(x)\) be a continuous function, whose first and second derivatives are continuous on \([0,2\pi]\) and \(f"(x) \ge 0 \:\: \forall \:x \in [0,2\pi]\). Show that \[\int _0^{2\pi} f(x) \cos x dx \ge 0\]

- anonymous

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- anonymous

@myininaya @amistre64 How would you prove it?

- anonymous

@Mr.Math

- Mr.Math

\[\int_0^{2\pi}f(x)\cos(x)dx=f(x)\sin(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)\sin(x)dx=-\int_0^{2\pi}f'(x)\sin(x)dx\]
\[=f'(x)\cos(x)|_0^{2\pi} +\int_0^{2\pi}f''(x)\cos(x)dx=f'(2\pi)-f'(0)+\int_0^{2\pi}f''(x)\cos(x)dx.\]
We can see here that \(f'(2\pi)-f'(0)\ge 0\), since \(f'(x)\) is an increasing function. So we're left to show that \(\int_0^{2\pi}f''(x)\cos(x)dx\ge 0\) knowing that \(f''(x)\ge 0\).

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## More answers

- Mr.Math

Let \(g(x)=\int f''(x)\cos(x)dx\). The question here is how can we show that \(g(2\pi)\ge g(0)\)?
We know that \(g(x)\) is increasing on \((0,\frac{\pi}{2})∪(\frac{3\pi}{2},2\pi)\), and decreasing on \((\frac{\pi}{2}, \frac{3\pi}{2}).\)

- anonymous

f"(x) must be constant, right?

- Mr.Math

Why?

- Mr.Math

It doesn't have to be constant.

- myininaya

mr.math do you mean g is positive on those intervals...and g is negative on that interval...

- Mr.Math

I meant \(g'\) is positive on those interval and negative on that interval.

- anonymous

I know, I just wanted it to be constant I mean not any obvious reason, just like that.

- Mr.Math

Note that \(g'(x)=f''(x)\cos(x)\), where \(f''(x)\ge 0\). So the sign of \(g'(x)\) (that determines whether g is increasing or decreasing) can be determined by \(\cos(x)\).

- Mr.Math

I didn't defined it as a constant.

- Mr.Math

define*

- Mr.Math

I don't understand why you want \(f''(x)\) to be a constant. If it was then the solution would be obvious since g(2pi)=g(0), and the integral will be 0.

- Mr.Math

Am I making sense Ishaan?

- anonymous

Of course, you're... it's me, who didn't

- Mr.Math

This problem is not difficult, but I'm missing something. What am I missing?!

- anonymous

Do you think I should ping Zarkon?

- Mr.Math

I think I have it. One minute.

- Mr.Math

You can ping him though, but he should not answer until I finish :P

- anonymous

If I'm not mistaken,\[- \int_0^{2\pi} f'(x) \sin x dx = f'(x) \cos x|_0^{2\pi} - \int _0^{2\pi} f''(x) \cos xdx\]

- Mr.Math

Correct! I just noticed that :)

- Mr.Math

So we have to show that \(f'(2\pi)+g(0)\ge g(2\pi)+f'(0).\)

- Mr.Math

I'm getting tired. @Zarkon should come from his planet now.

- anonymous

Zarkon's offline :(... @satellite73 @eseidl

- anonymous

i am thinking parts

- anonymous

oh i should pay attention. looks like mr.math did parts right?

- Mr.Math

Yes.

- anonymous

yeah

- Mr.Math

I think I have it this time. By the squeeze theorem and property of inequality in integrals we have:
\[-\int_0^{2\pi}f''(x) dx\le \int_0^{2\pi}f''(x)\cos(x)dx\le \int_0^{2\pi}f''(x)dx.\] Thus \(g(2\pi)-g(0)\le f'(2\pi)-f'(0)\) as required.

- anonymous

that was nice.
i was trying to come up with a counter example, and example of a positive function where this integral would be negative. can't seem to do it.

- anonymous

still not sure why i cannot though. why can't it be postive, but larger on the interval
\[[\frac{\pi}{2},\frac{3\pi}{2}]\] then on the rest? wouldn't that make the integral negative?

- Mr.Math

After this proof, I am sure you can't find such an example :P
@Ishaan, I will go for few minutes and be back to summarize my solution in one post, so it can be read and understood easily.

- anonymous

i believe you, and understand the first part. but it looks like the second part is hinging on the fact that if
\[f''>0\] then
\[\int f''(x)\cos(x)dx\geq 0\] and i am not sure exactly why that has to be

- anonymous

It is a given in the problem

- Mr.Math

Well, what I proved is that a continuous function \(f(x)\) that has a positive second derivative has the property:
\[\int_0^{2\pi} f(x)\cos(x)dx\ge 0.\]

- anonymous

ooooooooh ok i see i misinterpreted the last step

- Mr.Math

@satellite I first said that, but I don't need that statement anymore (I don't know if it's true or not). Read the last two posts.

- anonymous

yes, sorry i see what you wrote, i simply read it wrong.

- Mr.Math

We can use integration by parts and write
\[\int_0^{2\pi}f(x)\cos(x)dx=f(x)\sin(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)\sin(x)dx=-\int_0^{2\pi}f'(x)\sin(x)dx\]
\[=f'(x)\cos(x)|_0^{2\pi} +\int_0^{2\pi}f''(x)\cos(x)dx=f'(2\pi)-f'(0)-\int_0^{2\pi}f''(x)\cos(x)dx.\]
From here it's sufficient to show that \(f'(2\pi)-f'(0)\ge\int_0^{2\pi}f''(x)\cos(x)dx.\)
We know that \(\cos(x)\le 1 \implies f''(x)\cos(x)\le f''(x)\). So
\[\int_0^{2\pi}f''(x)\cos(x)dx \le \int_0^{2\pi}f''(x)dx=f'(2\pi)-f'(0). ■\]

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