Let us work through the problem step by step. The problem asks us to determine whether the improper integrals provided are convergent or divergent by using the comparison test. If applicable, you should also identify which comparison function (from options C–F) was used. If the comparison test does not apply, choose G.

Step-by-step analysis:

1. Integral 1:

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

• Analysis:

  • For $x \geq 1$, note that $\cos^2(x) \leq 1$, so $x^3 + 4 \cos^2(x) \geq x^3$.
  • This gives $\frac{1}{x^3 + 4 \cos^2(x)} \leq \frac{1}{x^3}$.
  • The comparison function is $\frac{1}{x^3}$, whose integral from $1$ to $\infty$ is convergent (as the exponent $3 > 1$).
  • By the comparison test, this integral is convergent.

  Answer for Integral 1: AF

2. Integral 2:

$\int_{1}^{\infty} \frac{1}{x^2 + 4 + 9 \sin(x)} \, dx$

• Analysis:

  • For $x \geq 1$, $-1 \leq \sin(x) \leq 1$, so $x^2 + 4 + 9 \sin(x) \geq x^2 - 5$.
  • Thus, $\frac{1}{x^2 + 4 + 9 \sin(x)} \leq \frac{1}{x^2 - 5}$, but this comparison is not suitable since $x^2 - 5$ changes sign.
  • A better comparison is $\frac{1}{x^2}$, as $x^2 + 4 + 9 \sin(x) \geq x^2$.
  • The comparison function $\frac{1}{x^2}$ has a convergent integral on $[1, \infty)$ (since $2 > 1$).

  Answer for Integral 2: AF

3. Integral 3:

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

• Analysis:

  • The integrand is $\frac{1}{x^{1/2}}$, and the exponent $1/2 < 1$.
  • The improper integral $\int_{1}^{\infty} x^{-1/2} dx$ is divergent.
  • No comparison function is needed, as we can directly determine divergence.

  Answer for Integral 3: B

4. Integral 4:

$\int_{1}^{\infty} \frac{e^{-x}}{x^2} \, dx$

• Analysis:

  • For large $x$, $e^{-x}$ decays exponentially, so $\frac{e^{-x}}{x^2} \leq \frac{1}{x^2}$.
  • The comparison function is $\frac{1}{x^2}$, which is convergent (since $2 > 1$).
  • Since $e^{-x}$ further suppresses the integrand, the integral is convergent.

  Answer for Integral 4: AF

5. Integral 5:

$\int_{1}^{\infty} \frac{1}{x^2} \, dx$

• Analysis:

  • The integrand is $\frac{1}{x^2}$, and its exponent $2 > 1$.
  • The integral is known to be convergent.

  Answer for Integral 5: A

Final Answers:

1. AF
2. AF
3. B
4. AF
5. A

Would you like me to explain any specific comparison in more detail? Let me know!

Expanding Questions:

1. How does the comparison test determine convergence or divergence?
2. What are other common tests for improper integrals?
3. Can you describe why the function $e^{-x}$ makes an integral more likely to converge?
4. Why does the integral $\int_{1}^{\infty} \frac{1}{\sqrt{x}} dx$ diverge despite the decreasing function?
5. How does the exponent of $x$ in $\frac{1}{x^p}$ affect convergence?

Tip:

Always check the behavior of the integrand near any singularity points and at infinity when deciding on convergence.