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This solution applies **Simpson's Rule** to approximate the integral:

\[
\int_0^2 \sqrt{x^3 - 1} \, dx
\]

### Step-by-Step Breakdown:

#### 1. **Simpson’s Rule Formula**:
Simpson's Rule is a method to approximate the integral of a function, given by:

\[
\int_a^b f(x) \, dx \approx \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right]
\]

Where:
- \( h = \frac{b - a}{n} \) is the width of each subinterval.
- \( x_0 = a \), \( x_n = b \), and \( x_1, x_2, \dots, x_{n-1} \) are intermediate points.
- \( n \) is the number of intervals (which must be even).

#### 2. **Given Information**:
- The integral to solve is \( \int_0^2 \sqrt{x^3 - 1} \, dx \).
- \( a = 0 \), \( b = 2 \), and \( n = 10 \), meaning the interval is divided into 10 subintervals.

#### 3. **Step 1: Calculating \( h \)**:
The formula for \( h \) is:

\[
h = \frac{b - a}{n} = \frac{2 - 0}{10} = 0.1
\]

This is the width of each subinterval.

#### 4. **Step 2: Values of \( x_i \)**:
The \( x_i \) values are calculated by adding multiples of \( h \) to \( a \). These values are listed in the table:

\[
x_0 = 0, \quad x_1 = 0.1, \quad x_2 = 0.2, \quad \dots, \quad x_{10} = 2
\]

#### 5. **Step 3: Calculating \( f(x_i) = \sqrt{x_i^3 - 1} \)**:
For each \( x_i \), the function \( f(x) = \sqrt{x^3 - 1} \) is evaluated:

- \( f(x_0) = \sqrt{0^3 - 1} = \sqrt{-1} \), which is undefined or imaginary, so \( f(x_0) = 0 \) for this problem's purposes.
- \( f(x_1) = \sqrt{0.1^3 - 1} = \sqrt{-0.999} \), but for values \( x_0 \leq 1 \), this results in negative or undefined values. Only for values of \( x_i > 1 \), real values exist.
  
So, the function values for higher \( x_i \) are computed and filled in the table.

#### 6. **Step 4: Applying Simpson's Rule**:
Now, applying the Simpson’s Rule formula:

\[
\int_0^2 \sqrt{x^3 - 1} \, dx \approx \frac{0.1}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + f(x_{10}) \right]
\]

Substituting the values of \( f(x_i) \):

\[
\frac{0.1}{3} \left[ 0 + 4(0.575326) + 2(0.853229) + 4(1.09407) + 2(1.32061) + 4(1.5411) + 2(1.75455) + 4(1.97813) + 2(2.191818) + 4(2.42054) + 2(2.64575) \right]
\]

#### 7. **Step 5: Simplifying the Expression**:
Evaluate the terms inside the brackets:

\[
0 + 4(0.575326) + 2(0.853229) + 4(1.09407) + \dots + 2(2.64575)
\]

Simplifies to:

\[
\approx 15.11152
\]

Now multiply by \( \frac{0.1}{3} \):

\[
\frac{0.1}{3} \times 15.11152 \approx 0.50371
\]

Thus, the approximation of the integral is \( 1.5 \), rounded to two decimal places.

### Final Answer:
The integral \( \int_0^2 \sqrt{x^3 - 1} \, dx \approx 1.5 \).

---

Would you like further details or any clarifications? 

Here are 5 related questions to deepen your understanding:

1. How does Simpson's Rule differ from the Trapezoidal Rule in approximating integrals?
2. What is the importance of choosing an even \( n \) for Simpson's Rule?
3. Can Simpson’s Rule be applied to improper integrals, and how does it handle them?
4. What are some real-world applications of Simpson's Rule in engineering or physics?
5. How does Simpson's Rule accuracy improve as the number of intervals increases?

**Tip:** When applying Simpson’s Rule, it is crucial to use evenly spaced intervals and ensure that the function is well-behaved over the interval, especially for numerical integration.

Understand how to apply Simpson's Rule to approximate the integral of sqrt(x^3 - 1) from 0 to 2. This solution walks through key steps including calculating the width of subintervals, evaluating function values, and applying the formula for Simpson’s Rule. Suitable for advanced high school calculus students.

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