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Let's analyze each of these improper integrals using the comparison test to determine if they converge or diverge.

1. \(\int_{1}^{\infty} \frac{e^{-x}}{x^2} \, dx\)

   - As \(x \to \infty\), \(e^{-x}\) approaches zero very quickly compared to \(x^2\). Thus, \(\frac{e^{-x}}{x^2}\) converges faster than \(\frac{1}{x^2}\), which is known to converge.
   - By comparison with \(\int_{1}^{\infty} \frac{1}{x^2} \, dx\), which converges, we conclude that this integral converges.
   - **Answer:** \( \text{A} \) (Convergent) and \( \text{F} \) (Comparison with \(\int_{1}^{\infty} \frac{1}{x^2} \, dx\)).

2. \(\int_{1}^{\infty} \frac{\sqrt{x}}{9 + \sin(x)} \, dx\)

   - For large \(x\), the function \(9 + \sin(x)\) is bounded and does not affect the asymptotic behavior of \(\frac{\sqrt{x}}{9 + \sin(x)}\).
   - The integral \(\int_{1}^{\infty} \sqrt{x} \, dx\) diverges because \(\sqrt{x} = x^{1/2}\) and \(\int_{1}^{\infty} x^{1/2} \, dx\) diverges for \(p = 1/2 < 1\).
   - By comparison with \(\int_{1}^{\infty} \sqrt{x} \, dx\), we conclude that this integral diverges.
   - **Answer:** \( \text{B} \) (Divergent) and \( \text{D} \) (Comparison with \(\int_{1}^{\infty} \sqrt{x} \, dx\)).

3. \(\int_{0}^{\pi/2} \frac{\sqrt{x}}{\cos^2(x)} \, dx\)

   - As \(x \to \frac{\pi}{2}\), \(\cos(x) \to 0\), which makes \(\cos^2(x)\) approach zero, causing the integrand to grow large.
   - We can compare this to \(\int_{0}^{\pi/2} \frac{1}{\cos^2(x)} \, dx\), which diverges.
   - Since \(\int_{0}^{\pi/2} \frac{1}{\cos^2(x)} \, dx\) diverges, this integral diverges as well.
   - **Answer:** \( \text{B} \) (Divergent) and \( \text{C} \) (Comparison with \(\int_{0}^{\pi/2} \frac{1}{\cos^2(x)} \, dx\)).

4. \(\int_{1}^{\infty} \frac{x^2 + 3}{x^3 + 3} \, dx\)

   - For large \(x\), the \(x^3\) term in the denominator and the \(x^2\) term in the numerator dominate, so this behaves asymptotically like \(\frac{x^2}{x^3} = \frac{1}{x}\).
   - The integral \(\int_{1}^{\infty} \frac{1}{x} \, dx\) diverges (harmonic series), so this integral diverges as well.
   - **Answer:** \( \text{B} \) (Divergent) and \( \text{E} \) (Comparison with \(\int_{1}^{\infty} \frac{1}{x} \, dx\)).

### Summary of Answers:

1. **AF**
2. **BD**
3. **BC**
4. **BE**

Would you like further details on any step, or do you have questions about these solutions?

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1. What conditions make a function suitable for the comparison test in improper integrals?
2. How can you determine whether a function's integral converges based on its asymptotic behavior?
3. What are some common divergent integrals used as comparison benchmarks?
4. Why does \(\int_{1}^{\infty} \frac{1}{x} \, dx\) diverge, and what does this tell us about other integrals with similar behavior?
5. How does the behavior of \(e^{-x}\) affect integrals compared to polynomial terms?

**Tip:** When applying the comparison test, it's often useful to simplify the behavior of functions by focusing on the dominant terms for large values of \(x\) (or near points of discontinuity) to judge convergence or divergence.

For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to G that best applies, and if the comparison test does not apply, enter only G. Determine the convergence or divergence of each integral and identify the appropriate comparison function if applicable.

Explore how to determine the convergence or divergence of improper integrals using the comparison test. This solution provides a step-by-step approach to analyzing integrals and applying appropriate comparisons for each case. Ideal for undergraduate calculus students.

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