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To evaluate the integral 

\[
\int_{1}^{9} \frac{1}{2} \sqrt{u} \, du,
\]

we can follow these steps:

1. **Factor out the constant**: Since \(\frac{1}{2}\) is a constant, we can factor it out of the integral:

\[
\frac{1}{2} \int_{1}^{9} \sqrt{u} \, du.
\]

2. **Express the integrand**: The square root of \(u\) can be rewritten as a power of \(u\):

\[
\sqrt{u} = u^{1/2}.
\]

So the integral becomes:

\[
\frac{1}{2} \int_{1}^{9} u^{1/2} \, du.
\]

3. **Integrate**: Use the power rule for integration, which states:

\[
\int u^n \, du = \frac{u^{n+1}}{n+1} + C,
\]

where \(n \neq -1\). In our case, \(n = \frac{1}{2}\), so:

\[
\int u^{1/2} \, du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3} u^{3/2}.
\]

4. **Apply the limits**: Now, substitute this result back into the integral and apply the limits from 1 to 9:

\[
\frac{1}{2} \cdot \frac{2}{3} \left[u^{3/2}\right]_{1}^{9}.
\]

5. **Simplify and evaluate**: 

\[
\frac{1}{3} \left[ \left(9^{3/2}\right) - \left(1^{3/2}\right) \right].
\]

- \(9^{3/2} = (9^1)^{3/2} = 27\),
- \(1^{3/2} = 1\).

So the integral becomes:

\[
\frac{1}{3} (27 - 1) = \frac{1}{3} \times 26 = \frac{26}{3}.
\]

Thus, the evaluated integral is:

\[
\frac{26}{3}.
\]

Would you like more details on any of these steps, or do you have any questions?

Here are five related questions to explore:

1. How do you apply the power rule for integration to different exponents?
2. What are the steps to evaluate definite integrals with variable limits?
3. How can constants be factored out of integrals, and why is this useful?
4. What is the significance of the limits of integration in a definite integral?
5. How does the power rule differ when integrating functions involving roots?

**Tip**: Always check if the function within the integral can be simplified before integrating, as this can make the process easier.

Evaluate the integral from 1 to 9 of (1/2) * sqrt(u) du

Learn how to evaluate the definite integral of (1/2) * sqrt(u) from 1 to 9 with step-by-step instructions, utilizing the power rule and applying limits of integration. Suitable for college-level calculus students.

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