1. **State the problem:** Calculate the definite integral $$\int_0^1 \frac{-2x}{e^{2x}} \, dx$$. 2. **Rewrite the integral:** Note that $$\frac{1}{e^{2x}} = e^{-2x}$$, so the integral becomes $$\int_0^1 -2x e^{-2x} \, dx$$. 3. **Use integration by parts:** Let $$u = -2x$$ and $$dv = e^{-2x} dx$$. 4. **Compute derivatives and integrals:** - $$du = -2 dx$$ - $$v = \int e^{-2x} dx = \frac{e^{-2x}}{-2} = -\frac{1}{2} e^{-2x}$$ 5. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ So, $$\int_0^1 -2x e^{-2x} dx = \left[-2x \cdot \left(-\frac{1}{2} e^{-2x}\right)\right]_0^1 - \int_0^1 \left(-\frac{1}{2} e^{-2x}\right)(-2) dx$$ 6. **Simplify the expression:** $$= \left[x e^{-2x}\right]_0^1 - \int_0^1 e^{-2x} dx$$ 7. **Evaluate the remaining integral:** $$\int_0^1 e^{-2x} dx = \left[ -\frac{1}{2} e^{-2x} \right]_0^1 = -\frac{1}{2} e^{-2} + \frac{1}{2}$$ 8. **Substitute back:** $$= (1 \cdot e^{-2} - 0) - \left(-\frac{1}{2} e^{-2} + \frac{1}{2}\right) = e^{-2} + \frac{1}{2} e^{-2} - \frac{1}{2} = \frac{3}{2} e^{-2} - \frac{1}{2}$$ 9. **Final answer:** $$\boxed{\frac{3}{2} e^{-2} - \frac{1}{2}}$$