To evaluate the integral $\int_{1}^{5} \frac{8}{w^2} \, dw$ using the Fundamental Theorem of Calculus, follow these steps:

1. Rewrite the integrand:

   Rewrite $\frac{8}{w^2}$ as $8w^{-2}$.

   $\int_{1}^{5} 8w^{-2} \, dw$

2. Find the antiderivative:

   The antiderivative of $8w^{-2}$ is found by using the power rule for integration. The power rule states that $\int w^n \, dw = \frac{w^{n+1}}{n+1} + C$ for $n \neq -1$.

   Here, $n = -2$:

   $\int 8w^{-2} \, dw = 8 \int w^{-2} \, dw = 8 \left( \frac{w^{-1}}{-1} \right) + C = -8w^{-1} + C$

   So, the antiderivative of $8w^{-2}$ is $-\frac{8}{w} + C$.

3. Apply the Fundamental Theorem of Calculus:

   According to the Fundamental Theorem of Calculus, if $F(w)$ is an antiderivative of $f(w)$, then: $\int_{a}^{b} f(w) \, dw = F(b) - F(a)$

   Here, $f(w) = 8w^{-2}$, so $F(w) = -\frac{8}{w}$.

   We need to evaluate $F$ at the upper and lower limits of the integral: $F(5) = -\frac{8}{5}$ and $F(1) = -8$

4. Compute the definite integral: $\int_{1}^{5} \frac{8}{w^2} \, dw = F(5) - F(1)$ $= \left( -\frac{8}{5} \right) - (-8)$ $= -\frac{8}{5} + 8$ $= -\frac{8}{5} + \frac{40}{5}$ $= \frac{40 - 8}{5}$ $= \frac{32}{5}$

Thus, the value of the integral is: $\int_{1}^{5} \frac{8}{w^2} \, dw = \frac{32}{5}$

Would you like further details or have any questions about this solution?

Here are some related questions you might find interesting:

1. How do you integrate functions with negative exponents?
2. What are the steps to apply the Fundamental Theorem of Calculus?
3. How do you convert a fraction with a polynomial denominator to a negative exponent form?
4. Why is it important to rewrite integrands in a simpler form before integrating?
5. What is the difference between definite and indefinite integrals?
6. How do you handle integrals with rational functions?
7. What are some common mistakes when integrating functions with exponents?
8. How do you interpret the result of a definite integral in terms of area?

Tip: When working with integrals, always simplify the integrand as much as possible before finding the antiderivative. This can make the process much easier and help avoid mistakes.