1. We are asked to evaluate the definite integral $$\int_0^1 x^2 e^{2x} \, dx$$. 2. To solve this integral, we use integration by parts. Let: $$u = x^2 \implies du = 2x \, dx$$ $$dv = e^{2x} \, dx \implies v = \frac{e^{2x}}{2}$$ 3. Applying integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ $$= x^2 \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} \cdot 2x \, dx$$ $$= \frac{x^2 e^{2x}}{2} - \int x e^{2x} \, dx$$ 4. We now need to evaluate $$\int x e^{2x} \, dx$$. Use integration by parts again: Let: $$u = x \implies du = dx$$ $$dv = e^{2x} \, dx \implies v = \frac{e^{2x}}{2}$$ 5. Applying integration by parts again: $$\int x e^{2x} \, dx = x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} \, dx$$ $$= \frac{x e^{2x}}{2} - \frac{1}{2} \int e^{2x} \, dx$$ 6. Evaluate $$\int e^{2x} \, dx$$: $$\int e^{2x} \, dx = \frac{e^{2x}}{2} + C$$ 7. Substitute back: $$\int x e^{2x} \, dx = \frac{x e^{2x}}{2} - \frac{1}{2} \cdot \frac{e^{2x}}{2} + C = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C$$ 8. Substitute this result back into step 3: $$\int_0^1 x^2 e^{2x} \, dx = \left[ \frac{x^2 e^{2x}}{2} - \left( \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} \right) \right]_0^1$$ $$= \left[ \frac{x^2 e^{2x}}{2} - \frac{x e^{2x}}{2} + \frac{e^{2x}}{4} \right]_0^1$$ 9. Evaluate at the bounds: At $$x=1$$: $$\frac{1^2 e^{2}}{2} - \frac{1 e^{2}}{2} + \frac{e^{2}}{4} = \frac{e^{2}}{2} - \frac{e^{2}}{2} + \frac{e^{2}}{4} = \frac{e^{2}}{4}$$ At $$x=0$$: $$\frac{0^2 e^{0}}{2} - \frac{0 e^{0}}{2} + \frac{e^{0}}{4} = 0 - 0 + \frac{1}{4} = \frac{1}{4}$$ 10. Subtract lower bound from upper bound: $$\frac{e^{2}}{4} - \frac{1}{4} = \frac{e^{2} - 1}{4}$$ **Final answer:** $$\int_0^1 x^2 e^{2x} \, dx = \frac{e^{2} - 1}{4}$$