TITLE: Proving if an integral is positive, negative, or zero
QUESTION [8 upvotes]: Find $$\text{sgn}\left[\int_0^{2\pi} e^{-2014x^2}\sin(x) \, dx\right]$$
Source: TCU Calculus Bee 2014 (Held 8 months ago)
By logic because the $\displaystyle\int_0^\pi \sin(x)\,dx=-\int_\pi^{2\pi} \sin(x)\,dx$ and the exponential is smaller in the interval $[\pi,2\pi]$, the area of the 2nd part is minimized, making it positive.
Can anyone provide a more rigorous proof of the answer?

REPLY [2 votes]: Well, the following is more rigorous, but is essentially the same answer:
We have
$\begin{align}
I&=\int_0^{2\pi} e^{-2014x^2}\sin(x) \, dx
\\&=\int_0^{\pi} e^{-2014x^2}\sin(x)\,dx+\int_\pi^{2\pi} e^{-2014x^2}\sin(x)\,dx
\\&=\int_0^{\pi} e^{-2014x^2}\sin(x)\,dx-\int_\pi^{0} e^{-2014(2\pi-x)^2}\sin(2\pi-x)\,dx
\\&=\int_0^{\pi} e^{-2014x^2}\sin(x)\,dx+\int_0^{\pi} e^{-2014(2\pi-x)^2}\sin(2\pi-x)\,dx
\\&=\int_0^{\pi} e^{-2014x^2}\sin(x)\,dx-\int_0^{\pi} e^{-2014(2\pi-x)^2}\sin(x)\,dx
\\&=\int_0^{\pi} \left(e^{-2014x^2}-e^{-2014(2\pi-x)^2}\right)\sin(x)\,dx
\end{align}$
and since $2\pi-x> x$ on the interval $[0,\pi)$, we have $2014x^2<2014(2\pi-x)^2$  and so $e^{-2014x^2}> e^{-2014(2\pi-x)^2}$  for this interval, and finally since the value of sine is positive for this interval, we have 
$$\int_0^{2\pi} e^{-2014x^2}\sin(x) \, dx=\int_0^{\pi} \left(e^{-2014x^2}-e^{-2014(2\pi-x)^2}\right)\sin(x)\,dx>0$$