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The given integral is:

\[
\int_{1}^{3} 2t^{-3} \, dt
\]

To solve this, follow these steps:

### Step 1: Apply the power rule for integration.
The general power rule for integration is:
\[
\int t^n \, dt = \frac{t^{n+1}}{n+1} + C \quad \text{(for } n \neq -1\text{)}
\]
Here, we have \( t^{-3} \), so:

\[
\int t^{-3} \, dt = \frac{t^{-3+1}}{-3+1} = \frac{t^{-2}}{-2}
\]

Multiplying by 2 (since the integrand is \( 2t^{-3} \)):

\[
\int 2t^{-3} \, dt = 2 \times \frac{t^{-2}}{-2} = -t^{-2}
\]

### Step 2: Evaluate the definite integral from 1 to 3.
Now, we need to compute:

\[
\left[-t^{-2}\right]_{1}^{3} = -\left(\frac{1}{3^2}\right) + \left(\frac{1}{1^2}\right)
\]

This becomes:

\[
-\left(\frac{1}{9}\right) + 1 = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}
\]

### Final Answer:
The value of the integral is:

\[
\frac{8}{9}
\]

Let me know if you need more details or clarification.

Here are some related questions you may find useful:
1. How does the power rule for integration work in general?
2. What is the difference between definite and indefinite integrals?
3. How can you apply integration to physical problems, such as computing area or work?
4. What happens if the integrand has a power of -1, like \( t^{-1} \)?
5. How can we solve integrals involving exponential functions?

**Tip:** Always simplify the integrand, if possible, before integrating to make calculations easier!

Learn how to solve the definite integral ∫[1 to 3] 2t^-3 dt step by step using the power rule for integration. This problem covers key concepts from integral calculus, including definite integrals and the fundamental theorem of calculus. Suitable for university-level students.

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