Why is the integration $\int_{\frac6\pi}^{\frac2{3\pi}}\frac{\cos\left(\frac1x\right)}{x^2}dx$ non-negative?

Question

Why does the following inequality hold?

$$\int_{\frac6\pi}^{\frac2{3\pi}}\frac{\cos\left(\frac1x\right)}{x^2}dx\ge0$$

I am new to integration so I may be blind to a careless mistake or need my handheld in the answer.

I have gone through the calculation and always end up with 1.5, but I am aware that the upper bound is smaller than the lower bound, and I am supposed to take the negative of my result.

Also from the graph from Wolfram Alpha, I do not understand why they even show a red negative section. I thought we were concerned with just the blue section.

Thank you to all who try to help.

Answer 1

As you've noted, $\frac6\pi>\frac{2}{3\pi}$, thus note that: $$\int_{\beta}^{\alpha}{f(x)dx}=-\int_{\alpha}^{\beta}{f(x)dx}$$ You most likely did $$\int_{\frac{2}{3\pi}}^{\frac6\pi}{\frac{\cos(\frac 1x)}{x^2}dx}$$so just take the negative of that.

Answer 2

Substitute $y=1/x$ so $dx/x^2=-dy$ and $$\int_{6/\pi}^{2/(3\pi)}\cos\frac{1}{x}\frac{dx}{x^2}=\int_{\pi/6}^{3\pi/2}-\cos y dy=[-\sin y]_{\pi/6}^{3\pi/2}=\sin\frac{\pi}{6}-\sin\frac{3\pi}{2}=\frac{1}{2}-(-1)=1.5.$$Note that $\cos y<0$ for $y\in(\frac{\pi}{2},\,\frac{3\pi}{2})$, so the $y\ge\frac{\pi}{2}$ part of the integral over $y$ is actually positive. Equivalently, the original integral includes a positive contribution from $x\in(\frac{2}{3\pi},\,\frac{2}{\pi})$ so that $\frac{1}{x}\in(\frac{\pi}{2},\,\frac{3\pi}{2})$. In both cases the integral also has a negative part, not quite sufficient to make the total negative.