\int_{1}^{5} {-\frac{2}{x^3}} \; \textrm{d}x \,.

To solve the integral 

\[
\int_{1}^{5} {-\frac{2}{x^3}} \; \textrm{d}x \,,
\]

we proceed as follows:

### Step 1: Simplify the integrand
The integrand is \(-\frac{2}{x^3}\), which can be rewritten as \(-2x^{-3}\).

### Step 2: Integrate using the power rule
The power rule for integration states that:

\[
\int x^n \, \textrm{d}x = \frac{x^{n+1}}{n+1} + C \quad \text{for} \, n \neq -1.
\]

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

\[
\int x^{-3} \, \textrm{d}x = \frac{x^{-3+1}}{-3+1} = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}.
\]

Now, apply this to the integral:

\[
\int -2x^{-3} \, \textrm{d}x = -2 \cdot \left(-\frac{1}{2x^2}\right) = \frac{1}{x^2}.
\]

### Step 3: Evaluate the definite integral
Now, we evaluate this from \(x = 1\) to \(x = 5\):

\[
\int_{1}^{5} {-\frac{2}{x^3}} \; \textrm{d}x = \left[\frac{1}{x^2}\right]_{1}^{5}.
\]

Substitute the limits:

\[
\left[\frac{1}{x^2}\right]_{1}^{5} = \frac{1}{5^2} - \frac{1}{1^2} = \frac{1}{25} - 1 = \frac{1}{25} - \frac{25}{25} = -\frac{24}{25}.
\]

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

\[
-\frac{24}{25}.
\]

Would you like more details on this process or have any other questions?

Here are some related questions you might find interesting:
1. What is the power rule for integration and how does it apply to different exponents?
2. How do you integrate functions with negative exponents?
3. What are some methods for evaluating definite integrals?
4. How does the evaluation of definite integrals differ from indefinite integrals?
5. What are some real-world applications of integrals involving functions with negative powers?

**Tip:** When integrating functions with negative exponents, always check to ensure that the power rule is correctly applied, especially when dealing with definite integrals.

Learn how to evaluate the definite integral of -2/x^3 from 1 to 5 using the power rule for integration. This problem explores the concept of definite integrals and their application, suitable for advanced high school students and calculus learners.

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