We are tasked with evaluating the following integral:

$\int_1^9 (x^2 - 1) \, dx$

Step 1: Separate the terms

We can break the integral into two parts:

$\int_1^9 x^2 \, dx - \int_1^9 1 \, dx$

Step 2: Integrate each term

1. For $\int x^2 \, dx$, the antiderivative of $x^2$ is $\frac{x^3}{3}$.
2. For $\int 1 \, dx$, the antiderivative of $1$ is $x$.

Thus, the integral becomes:

[ \left[ \frac{x^3}{3} \right]_1^9 - \left[ x \right]_1^9 ]

Step 3: Evaluate the limits

Now, substitute the upper and lower limits into the antiderivatives:

1. For $\frac{x^3}{3}$: $\left(\frac{9^3}{3}\right) - \left(\frac{1^3}{3}\right) = \frac{729}{3} - \frac{1}{3} = \frac{728}{3}$

2. For $x$: $9 - 1 = 8$

Step 4: Combine the results

Now, subtract the results of the two integrals:

$\frac{728}{3} - 8 = \frac{728}{3} - \frac{24}{3} = \frac{704}{3}$

Final Answer:

$\int_1^9 (x^2 - 1) \, dx = \frac{704}{3}$

Would you like more details on any part of the solution or any related questions?

Here are 5 related questions you might find useful:

1. How do you apply the power rule to integrate polynomials like $x^2$?
2. What is the difference between definite and indefinite integrals?
3. How can you verify the result of a definite integral using numerical methods or a graphing calculator?
4. What happens if the limits of integration are switched (e.g., from 9 to 1)?
5. Can the same method be used for more complex polynomials or rational functions?

Tip: When evaluating definite integrals, always remember to find the antiderivatives first, then apply the limits.