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
