1. **State the problem:** Evaluate the definite integral $$\int_1^4 \frac{2}{\sqrt{x}} \, dx$$. 2. **Recall the formula and rules:** The integral of $$x^n$$ with respect to $$x$$ is $$\frac{x^{n+1}}{n+1} + C$$ for $$n \neq -1$$. 3. **Rewrite the integrand:** $$\frac{2}{\sqrt{x}} = 2x^{-\frac{1}{2}}$$. 4. **Set up the integral:** $$\int_1^4 2x^{-\frac{1}{2}} \, dx$$. 5. **Integrate:** $$\int 2x^{-\frac{1}{2}} \, dx = 2 \int x^{-\frac{1}{2}} \, dx = 2 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = 2 \cdot 2x^{\frac{1}{2}} + C = 4\sqrt{x} + C$$. 6. **Evaluate the definite integral:** $$\int_1^4 \frac{2}{\sqrt{x}} \, dx = \left[4\sqrt{x}\right]_1^4 = 4\sqrt{4} - 4\sqrt{1} = 4 \cdot 2 - 4 \cdot 1 = 8 - 4 = 4$$. **Final answer:** $$4$$.