1. **State the problem:** Find the area between the curves $y = x e^{x^2}$ and $y = x^2$ for $x \geq 0$ from $x=0$ to $x=2$. 2. **Formula used:** The area between two curves $y=f(x)$ and $y=g(x)$ from $a$ to $b$ is given by $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ 3. **Determine which function is on top:** For $x \geq 0$, compare $y = x e^{x^2}$ and $y = x^2$ at $x=2$: $$x e^{x^2} = 2 e^{4} \approx 2 \times 54.598 = 109.196$$ $$x^2 = 4$$ Since $x e^{x^2} > x^2$ for $x=2$, $y = x e^{x^2}$ is the upper curve. 4. **Set up the integral:** $$\text{Area} = \int_0^2 \left(x e^{x^2} - x^2\right) dx$$ 5. **Evaluate the integral:** Split the integral: $$\int_0^2 x e^{x^2} dx - \int_0^2 x^2 dx$$ 6. **First integral:** Use substitution $u = x^2$, so $du = 2x dx$ or $x dx = \frac{du}{2}$. Change limits: when $x=0$, $u=0$; when $x=2$, $u=4$. $$\int_0^2 x e^{x^2} dx = \int_0^4 e^u \frac{du}{2} = \frac{1}{2} \int_0^4 e^u du = \frac{1}{2} \left[e^u\right]_0^4 = \frac{1}{2} (e^4 - 1)$$ 7. **Second integral:** $$\int_0^2 x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3}$$ 8. **Combine results:** $$\text{Area} = \frac{1}{2} (e^4 - 1) - \frac{8}{3}$$ 9. **Final answer:** $$\boxed{\text{Area} = \frac{e^4 - 1}{2} - \frac{8}{3}}$$