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The problem asks us to evaluate the improper integral:

\[
\int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx
\]

### Solution

1. **Rewrite the integral as a limit:**
   Since the upper limit is infinity, we can handle this by taking the limit as \( t \to \infty \):

   \[
   \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{\sqrt{x}} \, dx
   \]

2. **Integrate \( \frac{1}{\sqrt{x}} \):**
   Rewrite \( \frac{1}{\sqrt{x}} \) as \( x^{-\frac{1}{2}} \) and integrate:

   \[
   \int x^{-\frac{1}{2}} \, dx = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2 \sqrt{x}
   \]

3. **Evaluate the integral from 1 to \( t \):**

   \[
   \int_{1}^{t} \frac{1}{\sqrt{x}} \, dx = \left[ 2 \sqrt{x} \right]_{1}^{t} = 2 \sqrt{t} - 2 \sqrt{1}
   \]

   Simplifying, we get:

   \[
   = 2 \sqrt{t} - 2
   \]

4. **Take the limit as \( t \to \infty \):**

   \[
   \lim_{t \to \infty} (2 \sqrt{t} - 2) = \infty
   \]

### Conclusion

The improper integral diverges, so:

\[
\int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx = \infty
\]

This integral does not converge to a finite value.

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Would you like more details, or have any specific questions on this solution?

### Related Questions

1. How would you handle an improper integral with a finite discontinuity?
2. What are the conditions for an improper integral to converge?
3. How does the exponent in the denominator affect convergence?
4. How can we test convergence of more complex improper integrals?
5. What are real-life applications of improper integrals?

### Tip

Always rewrite improper integrals with limits to make handling infinity more manageable.

Evaluate the improper integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx \) using the procedure covered in lecture.

Explore the evaluation of the improper integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx \) using limits. This solution demonstrates handling infinity in integration and provides step-by-step calculations. Suitable for college calculus students.

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