1. **State the problem:** Evaluate the double integral $$\int_0^1 \int_x^1 y^2 e^{x^2} \, dy \, dx$$ by integrating first with respect to $y$ and then with respect to $x$. 2. **Integrate with respect to $y$ first:** Treat $x$ as constant. The integral inside is $$\int_x^1 y^2 e^{x^2} \, dy = e^{x^2} \int_x^1 y^2 \, dy$$ because $e^{x^2}$ does not depend on $y$. 3. Compute $$\int_x^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_x^1 = \frac{1^3}{3} - \frac{x^3}{3} = \frac{1}{3} - \frac{x^3}{3} = \frac{1 - x^3}{3}.$$ 4. Substitute back: $$\int_x^1 y^2 e^{x^2} \, dy = e^{x^2} \times \frac{1 - x^3}{3} = \frac{e^{x^2}(1 - x^3)}{3}.$$ 5. **Now integrate with respect to $x$:** $$\int_0^1 \frac{e^{x^2} (1 - x^3)}{3} \, dx = \frac{1}{3} \int_0^1 (1 - x^3) e^{x^2} \, dx.$$ 6. Split the integral: $$\frac{1}{3} \left( \int_0^1 e^{x^2} \, dx - \int_0^1 x^3 e^{x^2} \, dx \right).$$ 7. **Evaluate** $$\int_0^1 x^3 e^{x^2} \, dx$$ using substitution: let $u = x^2$, then $du = 2x dx$ so $x dx = \frac{du}{2}$. Rewrite $x^3 dx = x^2 (x dx) = u \times \frac{du}{2} = \frac{u du}{2}$. When $x=0$, $u=0$, and when $x=1$, $u=1$. So, $$\int_0^1 x^3 e^{x^2} \, dx = \int_0^1 e^{u} \frac{u}{2} \, du = \frac{1}{2} \int_0^1 u e^{u} \, du.$$ 8. **Use integration by parts on** $$\int_0^1 u e^{u} \, du$$ with $a=u$, $db=e^{u} du$: $$da = du, \quad b = e^{u}.$$ Thus, $$\int u e^{u} du = u e^{u} - \int e^{u} du = u e^{u} - e^{u} + C = e^{u}(u - 1) + C.$$ Evaluate from 0 to 1: $$\left[ e^{u}(u - 1) \right]_0^1 = e^{1} (1 - 1) - e^{0} (0 - 1) = 0 - (-1) = 1.$$ 9. Therefore, $$\int_0^1 x^3 e^{x^2} dx = \frac{1}{2} \times 1 = \frac{1}{2}.$$ 10. The problem reduces to $$\frac{1}{3} \left( \int_0^1 e^{x^2} dx - \frac{1}{2} \right).$$ 11. **Note:** The integral $$\int_0^1 e^{x^2} dx$$ does not have an elementary antiderivative but is understood as a non-elementary function (often represented using the error function or approximated numerically). **Final answer:** $$\boxed{\frac{1}{3} \left( \int_0^1 e^{x^2} dx - \frac{1}{2} \right)}.$$