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The given integral is:

\[
\int_0^{\infty} \frac{x^2}{\sqrt{2 + x^3}} \, dx
\]

We are asked to determine whether this integral is convergent or divergent. If it converges, we are to evaluate it.

### Step 1: Check for convergence
We need to check the behavior of the integrand at both bounds: \( x = 0 \) and \( x = \infty \).

#### Behavior at \( x = 0 \):
For small values of \( x \), let's approximate the function \( \frac{x^2}{\sqrt{2 + x^3}} \). As \( x \to 0 \):
- \( x^3 \) is very small compared to 2, so \( \sqrt{2 + x^3} \approx \sqrt{2} \).
- Thus, near \( x = 0 \), the integrand behaves like \( \frac{x^2}{\sqrt{2}} \).

This means the integrand behaves like \( x^2 \) as \( x \to 0 \). The integral of \( x^2 \) near 0 converges because \( \int_0^\epsilon x^2 \, dx = \frac{x^3}{3} \Big|_0^\epsilon = \frac{\epsilon^3}{3} \), which tends to 0 as \( \epsilon \to 0 \).

So, the integrand does not cause divergence at \( x = 0 \).

#### Behavior as \( x \to \infty \):
Now, for large values of \( x \), we analyze the asymptotic behavior of the integrand. As \( x \to \infty \):
- \( 2 \) becomes negligible compared to \( x^3 \), so \( \sqrt{2 + x^3} \approx \sqrt{x^3} = x^{3/2} \).
- Thus, for large \( x \), the integrand behaves like \( \frac{x^2}{x^{3/2}} = x^{1/2} \).

The integral of \( x^{1/2} \) as \( x \to \infty \) is divergent, since \( \int_1^\infty x^{1/2} \, dx = \frac{2}{3} x^{3/2} \Big|_1^\infty = \infty \).

### Step 2: Conclusion
Since the integrand behaves like \( x^{1/2} \) as \( x \to \infty \), the integral diverges at the upper bound.

Thus, the given integral **diverges**.

Let me know if you would like more details on the analysis, or have any further questions!

Here are some related questions to explore:
1. What is the general criterion for testing the convergence of improper integrals?
2. How can we use comparison tests to determine the convergence of integrals?
3. What are the differences between improper integrals of the first and second kind?
4. Can the integral be evaluated numerically even if it diverges analytically?
5. What are some typical integrands that lead to divergent integrals?

**Tip:** When analyzing improper integrals, focus on the behavior of the integrand at the boundaries (limits) of integration, especially at infinity or near singularities.

Determine whether the integral \( \int_0^{\infty} \frac{x^2}{\sqrt{2 + x^3}} \, dx \) is convergent or divergent. If it is convergent, evaluate it. (If the quantity diverges, enter DIVERGES.)

This page explores the convergence or divergence of the improper integral \( \int_0^{\infty} \frac{x^2}{\sqrt{2 + x^3}} \, dx \). Using asymptotic analysis and convergence tests, we determine whether the integral diverges due to its behavior at infinity.

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