1. **State the problem:** Find the average value of the function $g(x) = -x \cdot e^{x^2}$ on the interval $[1, 3]$. 2. **Formula for average value:** The average value $A$ of a continuous function $f(x)$ on $[a,b]$ is given by $$A = \frac{1}{b-a} \int_a^b f(x) \, dx$$ 3. **Apply the formula:** Here, $a=1$, $b=3$, and $f(x) = g(x) = -x e^{x^2}$. So, $$A = \frac{1}{3-1} \int_1^3 -x e^{x^2} \, dx = \frac{1}{2} \int_1^3 -x e^{x^2} \, dx$$ 4. **Evaluate the integral:** Use substitution. Let $u = x^2$, then $du = 2x dx$ or $x dx = \frac{du}{2}$. Rewrite the integral: $$\int_1^3 -x e^{x^2} \, dx = -\int_1^3 x e^{x^2} \, dx = -\int_{u=1^2}^{3^2} e^u \frac{du}{2} = -\frac{1}{2} \int_1^9 e^u \, du$$ 5. **Integrate:** $$-\frac{1}{2} \int_1^9 e^u \, du = -\frac{1}{2} [e^u]_1^9 = -\frac{1}{2} (e^9 - e) = -\frac{e^9}{2} + \frac{e}{2}$$ 6. **Calculate average value:** $$A = \frac{1}{2} \times \left(-\frac{e^9}{2} + \frac{e}{2}\right) = \frac{-e^9 + e}{4}$$ **Final answer:** The average value is $$\boxed{\frac{-e^9 + e}{4}}$$ which corresponds to option (4).