I'm trying to prove that the integral: $$ \int_0^{2\pi} f(x)\cos(x)\, dx $$ is positive. It has continuous first and second derivatives and such that $f''(x)>0$ for $0<x<2π$. I know I need to use the integration by parts and the fundamental theorem of calculus. But I'm not sure how such a thing can be proved.
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By integration by parts, $$ \int_0^{2\pi}f(x)\cos xdx=f(x)\sin(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)\sin xdx=-\int_0^{2\pi}f'(x)\sin xdx. $$ By integration by parts once more, $$ -\int_0^{2\pi}f'(x)\sin xdx=f'(x)\cos(x)|_0^{2\pi}-\int_0^{2\pi}f''(x)\cos(x)dx\\ =f'(2\pi)-f'(0)-\int_0^{2\pi}f''(x)\cos(x)dx=\int_0^{2\pi}f''(x)(1-\cos(x))dx\ge0 $$ where the last equality uses the fundamental theorem of calculus.
Pauly B
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Kim Jong Un
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3I prefer this proof because of the last equality. – marty cohen Sep 05 '14 at 00:46
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We use integration by parts:
$$\begin{align*}\int_0^{2 \pi} f(x) \cos x dx&=f(x) \sin x |_0^{2 \pi} - \int_{0}^{2 \pi} f'(x) \sin x dx\\&=-\int_0^{2 \pi} f'(x) \sin x dx\\&=f'(x) \cos x|_0^{2 \pi}+ \int_0^{2 \pi} f''(x) \cos x dx\\&=f'(2 \pi)-f'(0)-\int_0^{2 \pi} f''(x) \cos x\end{align*}$$
So,now,it remains to show that:
$$f'(2 \pi)-f'(0) \geq \int_0^{2 \pi} f''(x) \cos x dx$$
We know that $\cos x \leq 1 \Rightarrow f''(x) \cos x \leq f''(x)$
So,
$$\int_0^{2 \pi} f''(x) \cos x dx \leq \int_0^{2 \pi} f''(x) dx=f'(2\pi)-f'(0)$$
