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The integral presented is:

\[
\int_1^2 \frac{3}{t^4} \, dt
\]

### Solution:

1. Rewrite the integrand \(\frac{3}{t^4}\) as \(3t^{-4}\) for easier integration:
   \[
   \int_1^2 3t^{-4} \, dt
   \]

2. Apply the power rule of integration: 
   \[
   \int t^n \, dt = \frac{t^{n+1}}{n+1}, \quad \text{where } n \neq -1
   \]

   For \(n = -4\), the integral becomes:
   \[
   3 \cdot \frac{t^{-3}}{-3} = -t^{-3}
   \]

3. Now, evaluate the definite integral:
   \[
   -t^{-3} \Big|_1^2
   \]

4. Substituting the limits:
   \[
   -\left( \frac{1}{2^3} \right) + \left( \frac{1}{1^3} \right)
   = -\left( \frac{1}{8} \right) + 1
   = 1 - \frac{1}{8}
   = \frac{8}{8} - \frac{1}{8}
   = \frac{7}{8}
   \]

Thus, the value of the integral is \(\frac{7}{8}\).

Would you like further details or explanations?

Here are 5 related questions to expand on this topic:
1. How does the power rule for integration differ when dealing with positive vs. negative exponents?
2. How would the result change if the limits of integration were \([1, 3]\) instead of \([1, 2]\)?
3. How can substitution be used in cases where the exponent is more complex, such as \(t^{-4} + t^2\)?
4. What are the implications of indefinite integrals versus definite integrals in solving such problems?
5. How would you approach a problem where the integrand involves both \(t^{-4}\) and a logarithmic function?

**Tip**: Always rewrite terms in simpler forms (e.g., changing fractions into powers) before attempting integration. This simplifies the application of rules like the power rule.

Evaluate the integral: ∫₁² (3 / t⁴) dt.

This problem covers the evaluation of the definite integral ∫₁² (3 / t⁴) dt using the power rule for integration. It provides a step-by-step approach to finding the solution, including rewriting the integrand and applying limits. This type of problem is suitable for undergraduate calculus students or advanced high school learners.

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