The integral you provided is:

$\int_0^{\infty} x \cdot 8 \, dx$

Let's solve it step by step.

Step 1: Simplify the expression

Since $8$ is a constant, it can be factored out of the integral:

$8 \int_0^{\infty} x \, dx$

Step 2: Evaluate the remaining integral

Now, we need to evaluate the integral of $x$ from 0 to infinity:

$\int_0^{\infty} x \, dx$

This is a standard improper integral. Let's compute it:

$\int_0^{\infty} x \, dx = \lim_{b \to \infty} \int_0^b x \, dx$

The integral of $x$ is:

$\int x \, dx = \frac{x^2}{2}$

So,

$\int_0^b x \, dx = \left[ \frac{x^2}{2} \right]_0^b = \frac{b^2}{2} - 0 = \frac{b^2}{2}$

As $b \to \infty$, this expression grows without bound (it approaches infinity). Therefore:

$\int_0^{\infty} x \, dx = \infty$

Step 3: Conclusion

Multiplying by the constant 8:

$8 \cdot \infty = \infty$

Thus, the integral:

$\int_0^{\infty} x \cdot 8 \, dx$

diverges to infinity.

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Would you like further clarification on improper integrals or convergence/divergence concepts?

Here are some related questions to explore:

1. What are the conditions for an improper integral to converge?
2. How does the comparison test help in determining the convergence of an integral?
3. Can you explain the concept of infinite limits of integration in more detail?
4. How can we handle integrals with singularities (i.e., undefined points within the limits)?
5. What are some common techniques for evaluating integrals with infinity as a limit?

Tip: Always check for convergence before attempting to evaluate an integral with infinite limits or undefined points.