1. **State the problem:** Calculate the definite integral $$\int_0^5 \sqrt{x} \, dx$$. 2. **Recall the formula:** The integral of $$\sqrt{x}$$, which is $$x^{1/2}$$, is given by $$\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C$$ for $$n \neq -1$$. 3. **Apply the formula:** Here, $$n = \frac{1}{2}$$, so $$\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3} x^{3/2} + C$$. 4. **Evaluate the definite integral:** $$\int_0^5 \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^5 = \frac{2}{3} (5)^{3/2} - \frac{2}{3} (0)^{3/2} = \frac{2}{3} (5^{3/2})$$. 5. **Simplify the expression:** $$5^{3/2} = (\sqrt{5})^3 = \sqrt{5} \times \sqrt{5} \times \sqrt{5} = 5 \sqrt{5}$$. 6. **Final answer:** $$\int_0^5 \sqrt{x} \, dx = \frac{2}{3} \times 5 \sqrt{5} = \frac{10}{3} \sqrt{5}$$. This is the exact value of the integral.