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 $\int_0^2 \sqrt{x^3 - 1} \, dx \approx 1.5$.

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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.