Let's start by solving problem 37 from the first image, which involves the improper integral:

$I = \int_{2}^{\infty} \frac{1}{\sqrt[3]{x(x-1)}} \, dx$

Step 1: Analyze the Integrand

The integrand is:

$\frac{1}{\sqrt[3]{x(x-1)}}$

This can be written as:

$\frac{1}{(x(x-1))^{1/3}} = \frac{1}{(x^{4/3} - x^{1/3})}$

We have an improper integral due to the infinite limit at the upper bound. To determine whether the integral converges or diverges, we'll test the behavior of the integrand as $x \to \infty$.

Step 2: Behavior as $x \to \infty$

As $x \to \infty$, the term $x - 1$ behaves like $x$. So:

[ \sqrt[3]{x(x-1)} \sim \sqrt[3]{x^2} ]

This implies that, for large $x$:

$\frac{1}{\sqrt[3]{x(x-1)}} \sim \frac{1}{x^{2/3}}$

Now we check the integral of $\frac{1}{x^{2/3}}$ from 2 to $\infty$:

$\int_{2}^{\infty} x^{-2/3} \, dx$

The integral of $x^p$ is $\frac{x^{p+1}}{p+1}$, so:

$\int_{2}^{\infty} x^{-2/3} \, dx = \left[ \frac{x^{1/3}}{1/3} \right]_{2}^{\infty}$

As $x \to \infty$, $x^{1/3} \to \infty$, so the integral diverges.

Conclusion

Since the integral of the approximated form diverges, the original integral $\int_{2}^{\infty} \frac{1}{\sqrt[3]{x(x-1)}} \, dx$ also diverges.

Would you like a detailed explanation for any step?

Related Questions:

1. How do you determine the convergence or divergence of an improper integral?
2. What are the general tests for improper integrals with infinite limits?
3. Can the Comparison Test be applied to this type of integral? If so, how?
4. How would you approach an integral with a singularity at the lower bound?
5. What happens if the power of $x$ in the integrand is slightly changed (e.g., $x^{-5/3}$)?

Tip:

When dealing with improper integrals, analyzing the behavior of the integrand near problematic points (infinities or singularities) is crucial to determining convergence.