To determine the convergence or divergence of the given improper integral

$I = \int_0^1 \frac{dx}{\sqrt{9x^5 + 4x^7}},$

we will use the comparison test. Here's a step-by-step breakdown to choose the correct reasoning:

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1. Behavior of the denominator:

The denominator of the integrand is $\sqrt{9x^5 + 4x^7}$. For small $x$ (near $0$), the term $9x^5$ dominates over $4x^7$, so $\sqrt{9x^5 + 4x^7} \approx \sqrt{9x^5} = 3x^{5/2}$. This suggests the integrand behaves like $\frac{1}{3x^{5/2}}$ near $x = 0$.

We compare this with the test function $\frac{1}{x^{5/2}}$, which leads to the integral:

$\int_0^1 \frac{dx}{x^{5/2}}.$

This integral is divergent because the power of $x$ in the denominator ($5/2 > 1$) makes the integral diverge at $x = 0$.

Thus, if the given integrand is greater than or comparable to $\frac{1}{3x^{5/2}}$, the integral $I$ will diverge.

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2. Comparison analysis:

We are given multiple reasoning options. Let us analyze them:

Option 1:

The integral is divergent since $9x^5 + 4x^7 \leq 13x^5$, hence $\frac{1}{\sqrt{9x^5 + 4x^7}} \geq \frac{1}{\sqrt{13x^5}} = \frac{1}{\sqrt{13} \cdot x^{5/2}}.$ If true, the integral diverges because $\int_0^1 \frac{dx}{x^{5/2}} = \infty.$

This reasoning is correct because $9x^5 + 4x^7 \leq 13x^5$ for $0 \leq x \leq 1$, and the comparison to $\frac{1}{x^{5/2}}$ shows divergence.

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Option 2:

The integral is convergent since $9x^5 + 4x^7 \geq 9x^5$, hence $\frac{1}{\sqrt{9x^5 + 4x^7}} \leq \frac{1}{\sqrt{9x^5}} = \frac{1}{3x^{5/2}}.$

This reasoning is incorrect. While $\frac{1}{\sqrt{9x^5 + 4x^7}} \leq \frac{1}{3x^{5/2}}$, the integral $\int_0^1 \frac{1}{x^{5/2}} dx$ still diverges, so the original integral also diverges.

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Option 3:

The integral is divergent since $9x^5 + 4x^7 \leq 13x^7$, hence $\frac{1}{\sqrt{9x^5 + 4x^7}} \geq \frac{1}{\sqrt{13x^7}} = \frac{1}{\sqrt{13} \cdot x^{7/2}}.$

This reasoning is incorrect. Near $x = 0$, the term $9x^5$ dominates, not $4x^7$, so the approximation $13x^7$ is not valid, and the divergence test using $x^{7/2}$ does not apply.

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Option 4:

The integral is convergent since $9x^5 + 4x^7 \geq 4x^7$, hence $\frac{1}{\sqrt{9x^5 + 4x^7}} \leq \frac{1}{\sqrt{4x^7}} = \frac{1}{2x^{7/2}}.$

This reasoning is incorrect. While $\frac{1}{2x^{7/2}}$ converges near $x = 0$, the actual behavior of the integrand is determined by $x^{5/2}$, which causes divergence. The comparison to $x^{7/2}$ is not appropriate here.

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3. Correct Answer:

The correct reasoning is:

Option 1: The integral is divergent since $9x^5 + 4x^7 \leq 13x^5$, hence $\frac{1}{\sqrt{9x^5 + 4x^7}} \geq \frac{1}{\sqrt{13} \cdot x^{5/2}}$, and the integral diverges.

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Would you like a detailed breakdown of any step or concept?

Here are 5 related questions to consider:

1. What is the general method for determining if an improper integral converges or diverges?
2. How does the behavior of the denominator influence the convergence of an improper integral?
3. What is the significance of comparing to a simpler function using the comparison test?
4. Can $\sqrt{9x^5 + 4x^7}$ ever lead to convergence for large exponents?
5. What happens if we attempt a limit comparison test instead of direct comparison here?

Tip: For improper integrals, always consider the dominant term in the denominator or numerator when $x \to 0$ or $x \to \infty$.