Question: User: a \int_1^4 \sqrt{x} \, dx
1. **State the problem:** We need to evaluate the definite integral $$\int_1^4 \sqrt{x} \, dx$$.
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
$$\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} + C = \frac{x^{3/2}}{3/2} + C = \frac{2}{3} x^{3/2} + C$$.
4. **Evaluate the definite integral:**
$$\int_1^4 \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_1^4 = \frac{2}{3} (4^{3/2} - 1^{3/2})$$.
5. **Calculate powers:**
$$4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$$ and $$1^{3/2} = 1$$.
6. **Substitute values:**
$$\frac{2}{3} (8 - 1) = \frac{2}{3} \times 7 = \frac{14}{3}$$.
7. **Final answer:**
$$\int_1^4 \sqrt{x} \, dx = \frac{14}{3}$$.