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We are given the integral:

\[
\int_0^7 \frac{1}{\sqrt{7 - x}} \, dx
\]

We are tasked with determining whether this integral is convergent or divergent, and if convergent, we need to evaluate it.

### Step 1: Check the behavior at the limits

- **Lower limit:** At \( x = 0 \), \( 7 - 0 = 7 \), and the integrand behaves well, so there's no issue at the lower limit.
- **Upper limit:** At \( x = 7 \), \( 7 - 7 = 0 \), so the integrand becomes \( \frac{1}{\sqrt{0}} \), which means there is a singularity (the integrand blows up). This suggests the integral might be improper at the upper limit, and we need to check if it converges.

### Step 2: Set up the improper integral

We deal with the singularity by taking a limit. We rewrite the integral as:

\[
\lim_{\epsilon \to 0^+} \int_0^{7 - \epsilon} \frac{1}{\sqrt{7 - x}} \, dx
\]

### Step 3: Solve the integral

To evaluate this integral, we use the substitution \( u = 7 - x \), which implies \( du = -dx \). The limits of integration change accordingly:
- When \( x = 0 \), \( u = 7 \).
- When \( x = 7 - \epsilon \), \( u = \epsilon \).

Thus, the integral becomes:

\[
\lim_{\epsilon \to 0^+} \int_7^\epsilon \frac{-1}{\sqrt{u}} \, du
\]

The negative sign can be removed by reversing the limits of integration:

\[
\lim_{\epsilon \to 0^+} \int_\epsilon^7 \frac{1}{\sqrt{u}} \, du
\]

Now we evaluate this integral:

\[
\int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u}
\]

Thus, we get:

\[
\lim_{\epsilon \to 0^+} \left[ 2\sqrt{u} \right]_\epsilon^7 = \lim_{\epsilon \to 0^+} \left( 2\sqrt{7} - 2\sqrt{\epsilon} \right)
\]

As \( \epsilon \to 0 \), \( \sqrt{\epsilon} \to 0 \), so we are left with:

\[
2\sqrt{7}
\]

### Step 4: Conclusion

The integral converges, and its value is:

\[
2\sqrt{7}
\]

Would you like further details or clarification on any step?

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Here are 5 related questions for further exploration:

1. How do you determine whether an integral is improper at a given limit?
2. What are the conditions for an improper integral to converge?
3. How does substitution help in solving integrals involving square roots?
4. What methods can be used to evaluate improper integrals with singularities at the limits?
5. Can you explain how to compute the integral of other functions with similar singularities?

**Tip:** Always check the behavior of the integrand near singularities before evaluating an improper integral.

Determine whether the integral is convergent or divergent. If it is convergent, evaluate it: ∫(0 to 7) 1/(3√(7-x)) dx.

This problem involves determining the convergence of an improper integral ∫(0 to 7) 1/√(7 - x) dx and evaluating it. The integral has a singularity at the upper limit, requiring substitution and careful evaluation. Suitable for college-level calculus students.

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