Proving if an integral is positive, negative, or zero 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?
 A: We divide the integral in the parts where $\sin x\geq 0$ and $\sin x\leq 0$,
$$
\int_0^{2\pi}e^{-2014x^2}\sin x\, dx
=
\int_0^{\pi}e^{-2014x^2}\sin x\, dx
+
\int_\pi^{2\pi}e^{-2014x^2}\sin x\, dx.
$$
In the second integral we do the change of variables $t=x-\pi$, to get
$$
\int_\pi^{2\pi}e^{-2014x^2}\sin x\, dx =
\int_0^{\pi}e^{-2014(t+\pi)^2}\sin (t+\pi)\, dt
$$
Since $\sin(t+\pi)=-\sin(t)$, we can write the integral as (we go back to the variable $x$ again)
$$
\int_0^\pi(e^{-2014x^2}-e^{-2014(x+\pi)^2})\sin x\, dx.
$$
Can you get the sign from here?
A: 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 
$$\int_0^{2\pi} e^{-2014x^2}\sin(x) \, dx=\int_0^{\pi} \left(e^{-2014x^2}-e^{-2014(2\pi-x)^2}\right)\sin(x)\,dx>0$$
