How can the sign of this integral-based function at a particular point be determined? The function $g(x)$ is defined thusly:
$$g(x) = \int_1^x (t^2 - 3t + 2) e^{-t^2} dt$$
Is it the case that $g(3) > 0$?
This question is part of a sample quiz provided by a high school student I'm tutoring.  The theme of the quiz is the Fundamental Theorem of Calculus, with lots of functions defined in terms of an integral, as above.  Every other question has been fairly effortless for me, but this one has me stumped.  The integral appears too intractable for direct evaluation of $g(3)$.  The integrand is negative on $(1, 2)$ and positive on $(2, 3)$, but I haven't been able to compare the magnitude of the areas.  My intuition tells me that the rapidly decreasing $e^{-t^2}$ factor dominates the polynomial factor and leaves the overall integral negative at $x = 3$, but I'd prefer a more definitive explanation for my student.
I've almost concluded that the quiz is misprinted and should be asking whether $g'(3) > 0$, which is trivial to answer, but on the chance that it isn't, I thought I'd ask here.
 A: 
As a preface, I agree with you that the question is wrong. Aside from some sneaky way to get around the problem, I see no other way to answer the question other than evaluating the integral which isn't the nicest.

You can split the integral into three parts
$$
\int_1^xt^2\exp(-t^2)\mathrm{d}t-3\int_1^xt\exp(-t^2)\mathrm{d}t+2\int_1^x\exp(-t^2)\mathrm{d}t
$$
By knowing that
$$
\frac{\mathrm{d}}{\mathrm{d}t}\exp(-t^2)=-2t\exp(-t^2)
$$
we can make use of this with the chain rule to evaluate the entire integral.
Considering the first integral you have
$$
\begin{aligned}
\int_1^xt^2\exp(-t^2)\mathrm{d}t&=\int_1^xt\cdot t\exp(-t^2)\mathrm{d}t\\
&=-\frac{1}{2}\int_1^xt\frac{\mathrm{d}}{\mathrm{d}t}\exp(-t^2)\mathrm{d}t\\
&=-\frac{1}{2}\left[t\exp(-t^2)\right]_1^x+\frac{1}{2}\int_1^x\exp(-t^2)\mathrm{d}t\\
&=\frac{1}{2}(e^{-1}-x\exp(-x^2))+\frac{1}{2}\int_1^x\exp(-t^2)\mathrm{d}t\\
\end{aligned}
$$
Considering the second integral we have
$$
\begin{aligned}
\int_1^xt\exp(-t^2)\mathrm{d}t&=-\frac{1}{2}\int_{-1}^{-x^2}e^u\mathrm{d}u&\{u=-t^2\}\\
&=-\frac{1}{2}\left[e^u\right]_{-1}^{-x^2}\\
&=\frac{1}{2}\left(e^{-1}-\exp(-x^2)\right)
\end{aligned}
$$
The third integral (linked with the result from the first) can be evaluated in terms of the well defined error function
$$
\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x\exp(-t^2)\mathrm{d}t
$$
We need deal with the different bounds however, so we do some nice integral subtraction
$$
\begin{aligned}
\mathrm{erf}(x)-\mathrm{erf}(1)&=\frac{2}{\sqrt{\pi}}\int_0^x\exp(-t^2)\mathrm{d}t-\frac{2}{\sqrt{\pi}}\int_0^1\exp(-t^2)\mathrm{d}t\\
&=\frac{2}{\sqrt{\pi}}\int_1^x\exp(-t^2)\mathrm{d}t
\end{aligned}
$$
meaning that
$$
\int_1^x\exp(-t^2)\mathrm{d}t=\frac{\sqrt{\pi}}{2}(\mathrm{erf}(x)-\mathrm{erf}(1))
$$
Knowing this, we can now put the separate integrals together again to end up with
$$
\left[\frac{1}{2}(e^{-1}-x\exp(-x^2))+\frac{1}{2}\int_1^x\exp(-t^2)\mathrm{d}t\right]-3\left[\frac{1}{2}\left(e^{-1}-\exp(-x^2)\right)\right]+2\left[\int_1^x\exp(-t^2)\mathrm{d}t\right]
$$
which after some simplification, along with substitution of integrals for the error function, turns into
$$
\frac{1}{4}\left[5\sqrt{\pi}\left(\mathrm{erf}(x)-\mathrm{erf}(1)\right)-2\exp\left(-x^2\right)(x-3)-4e^{-1}\right]
$$
which by no means is a simple integral. $g(3)\approx-0.0194$ according to WolframAlpha.
I don't think there's a way to explain this without having to calculate (at least in part) the integral. Typically if the question wants you to compare two areas, the areas will differ by a large enough amount that the answer is fairly evident; but in this case the areas under the integrand on (1,2) and (2,3) differ by less than 0.02.
The only way I could think of getting the answer without evaluating the integral would be numerical integration which I think unlikely due to the nature of the question.
Hope this helps in some way
