To solve the definite integral

$\int_{1}^{3} |5x - 4| \, dx$

we need to handle the absolute value carefully by determining where the expression inside the absolute value, $5x - 4$, changes sign.

1. Find where $5x - 4 = 0$: $5x - 4 = 0 \Rightarrow x = \frac{4}{5} = 0.8$

Since $0.8$ is outside the interval $[1, 3]$, we can observe that $5x - 4$ is positive within this interval (since $5x - 4 > 0$ for $x > 0.8$). Therefore, we can drop the absolute value for this interval, and the integral becomes:

$\int_{1}^{3} (5x - 4) \, dx$

2. Integrate $5x - 4$:

   The antiderivative of $5x - 4$ is: $\int (5x - 4) \, dx = \frac{5x^2}{2} - 4x$

3. Evaluate the definite integral: Substitute $x = 3$ and $x = 1$: [ \left[ \frac{5 \cdot 3^2}{2} - 4 \cdot 3 \right] - \left[ \frac{5 \cdot 1^2}{2} - 4 \cdot 1 \right] ]

   Simplify each part: $= \left( \frac{5 \cdot 9}{2} - 12 \right) - \left( \frac{5 \cdot 1}{2} - 4 \right)$ $= \left( \frac{45}{2} - 12 \right) - \left( \frac{5}{2} - 4 \right)$ $= \left( 22.5 - 12 \right) - \left( 2.5 - 4 \right)$ $= 10.5 + 1.5 = 12$

So, the answer to the integral is:

$\boxed{12}$

If you have any questions or would like further details, let me know!

Additional Related Questions:

1. How would the solution change if the limits of integration were from $0$ to $3$?
2. What is the effect of absolute values on definite integrals in general?
3. Can you explain how to integrate functions with absolute values over intervals with sign changes?
4. How would the answer differ if the function inside the absolute value were negative over part of the interval?
5. Could this problem be solved by breaking it into piecewise integrals?

Tip: When dealing with absolute values in integrals, check where the expression inside the absolute value changes sign within the integration limits. This can simplify your work.