1. The problem is to evaluate the definite integral $$\int_e^1 \frac{1}{x^3} \, dx$$. 2. Rewrite the integrand as $$x^{-3}$$ to make integration easier. 3. Find the antiderivative of $$x^{-3}$$: $$\int x^{-3} \, dx = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$$. 4. Apply the limits from $$e$$ to $$1$$: $$\left[-\frac{1}{2x^2}\right]_e^1 = -\frac{1}{2(1)^2} - \left(-\frac{1}{2e^2}\right) = -\frac{1}{2} + \frac{1}{2e^2}$$. 5. Simplify the result: $$\frac{1}{2e^2} - \frac{1}{2} = \frac{1}{2}\left(\frac{1}{e^2} - 1\right)$$. Final answer: $$\int_e^1 \frac{1}{x^3} \, dx = \frac{1}{2}\left(\frac{1}{e^2} - 1\right)$$.