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The given problem asks to determine whether the integral 

\[
\int_1^\infty 2x^{-3} \, dx
\]

is convergent or divergent, and to evaluate it if it converges.

### Step 1: Check for Convergence or Divergence
This is an improper integral because it has an infinite upper limit. To determine if it converges, we need to evaluate the integral.

### Step 2: Set up the integral for evaluation
First, express the integral:

\[
I = \int_1^\infty 2x^{-3} \, dx
\]

The integral can be rewritten as:

\[
I = 2 \int_1^\infty x^{-3} \, dx
\]

### Step 3: Evaluate the integral
To integrate \( x^{-3} \), recall that the integral of \( x^n \) is:

\[
\int x^n \, dx = \frac{x^{n+1}}{n+1} \quad \text{for} \quad n \neq -1
\]

Here, \( n = -3 \), so we have:

\[
\int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}
\]

Now, apply the limits of integration from \( 1 \) to \( \infty \):

\[
I = 2 \left[ -\frac{1}{2x^2} \right]_1^\infty
\]

### Step 4: Apply the limits
First, evaluate the expression at \( \infty \) and \( 1 \):

- As \( x \to \infty \), \( \frac{1}{x^2} \to 0 \), so:

\[
-\frac{1}{2x^2} \Big|_{x=\infty} = 0
\]

- At \( x = 1 \):

\[
-\frac{1}{2x^2} \Big|_{x=1} = -\frac{1}{2}
\]

Thus, the integral becomes:

\[
I = 2 \left( 0 - \left(-\frac{1}{2}\right) \right) = 2 \times \frac{1}{2} = 1
\]

### Step 5: Conclusion
Since the integral evaluates to a finite number, the integral converges, and its value is:

\[
\boxed{1}
\]

Let me know if you'd like more details or have any questions!

Here are 5 related questions for further practice:
1. What is the convergence condition for the integral \( \int_1^\infty x^{-p} \, dx \)?
2. How do you test for convergence of improper integrals with infinite limits?
3. What happens if the exponent in the integrand were \( -2 \) instead of \( -3 \)?
4. How do you compute improper integrals with a singularity at a finite point?
5. How does substitution affect the evaluation of improper integrals?

**Tip:** When dealing with improper integrals, always check for both infinite limits and discontinuities to decide if the integral converges.

Determine whether the integral \( \int_1^\infty 2x^{-3} dx \) is convergent or divergent. Evaluate if the integral is convergent.

Learn how to determine whether the improper integral \( \int_1^\infty 2x^{-3} dx \) is convergent. This problem involves evaluating an improper integral with infinite limits using basic integration techniques. Suitable for college-level calculus students.

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