Question: User: $\int_{0}^{2}$ 4x $e^{2x}$ dx
1. **State the problem:** We need to evaluate the definite integral $$\int_0^2 4x e^{2x} \, dx$$.
2. **Formula and method:** This integral involves a product of a polynomial and an exponential function, so we use integration by parts. Recall the formula:
$$\int u \, dv = uv - \int v \, du$$
3. **Choose parts:** Let
$$u = 4x \implies du = 4 \, dx$$
$$dv = e^{2x} \, dx \implies v = \frac{e^{2x}}{2}$$
4. **Apply integration by parts:**
$$\int_0^2 4x e^{2x} \, dx = \left.4x \cdot \frac{e^{2x}}{2}\right|_0^2 - \int_0^2 \frac{e^{2x}}{2} \cdot 4 \, dx$$
Simplify the integral:
$$= \left.2x e^{2x}\right|_0^2 - 2 \int_0^2 e^{2x} \, dx$$
5. **Evaluate the remaining integral:**
$$\int_0^2 e^{2x} \, dx = \left. \frac{e^{2x}}{2} \right|_0^2 = \frac{e^{4} - 1}{2}$$
6. **Substitute back:**
$$= 2 \cdot 2 e^{4} - 2 \cdot \frac{e^{4} - 1}{2} - 0$$
$$= 4 e^{4} - (e^{4} - 1)$$
7. **Simplify:**
$$= 4 e^{4} - e^{4} + 1 = 3 e^{4} + 1$$
**Final answer:**
$$\int_0^2 4x e^{2x} \, dx = 3 e^{4} + 1$$