Let's go through the questions on applying the Comparison Theorem to determine the behavior of the integral:

$\int_2^{\infty} \frac{1}{\sqrt[3]{x - 1}} \, dx.$

Question 1

We are asked to compare this integral with a p-integral of the form

$\int_2^{\infty} \frac{1}{x^p} \, dx.$

We need to determine the value of $p$ for this comparison.

1. To find an appropriate value of $p$, let's analyze the behavior of the integrand $\frac{1}{\sqrt[3]{x - 1}}$ as $x$ becomes large. For large $x$, $x - 1$ behaves similarly to $x$, so:

   $\frac{1}{\sqrt[3]{x - 1}} \approx \frac{1}{x^{1/3}}.$

   Therefore, we can compare our integral with the p-integral where $p = \frac{1}{3}$.

Answer for Question 1:

$p = \frac{1}{3}.$

Question 2

Now, we need to determine if the p-integral $\int_2^{\infty} \frac{1}{x^{1/3}} \, dx$ converges or diverges.

• For a p-integral $\int_a^{\infty} \frac{1}{x^p} \, dx$, it converges if $p > 1$ and diverges if $p \leq 1$.
• Since $p = \frac{1}{3}$ here, which is less than 1, the p-integral diverges.

Answer for Question 2:

The p-integral diverges.

Question 3

We need to determine the relationship between the two functions

$\frac{1}{\sqrt[3]{x - 1}} \quad \text{and} \quad \frac{1}{x^{1/3}}$

for large values of $x$. Since $x - 1 \approx x$ when $x$ is large, we have:

$\frac{1}{\sqrt[3]{x - 1}} \approx \frac{1}{x^{1/3}}.$

This suggests that:

\frac{1}{\sqrt[3]{x - 1}} \leq \frac{1}{x^{1/3}} \quad \text{when x is sufficiently large.}

Answer for Question 3:

The correct inequality is:

\frac{1}{\sqrt[3]{x - 1}} \leq \frac{1}{x^{1/3}} \quad \text{when x is sufficiently large.}

Question 4

Based on the Comparison Theorem and the results from previous questions, we conclude:

• Since $\frac{1}{\sqrt[3]{x - 1}} \leq \frac{1}{x^{1/3}}$ and the integral $\int_2^{\infty} \frac{1}{x^{1/3}} \, dx$ diverges, we can use the Comparison Theorem to state that the original integral $\int_2^{\infty} \frac{1}{\sqrt[3]{x - 1}} \, dx$ also diverges.

Answer for Question 4:

The correct choice is:

• The comparison theorem tells us that $\int_2^{\infty} \frac{1}{\sqrt[3]{x - 1}} \, dx$ diverges.

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

Related Questions:

1. What is the Comparison Theorem, and how does it help in determining convergence or divergence?
2. Can we apply a different comparison function for this integral? If so, which one?
3. How would the answer change if $p > 1$ in the p-integral?
4. What are other common integrals that use the Comparison Theorem for analysis?
5. Why is it important to consider the behavior of the function as $x \to \infty$ when applying the Comparison Theorem?

Tip:

When using the Comparison Theorem, always ensure that the inequality relationship holds for all large values of $x$ to correctly apply the theorem's results.