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This image provides a solution for evaluating the definite integral of \(\int_1^2 \sqrt{x^3 - 1} \, dx\) using the **Rectangular (Midpoint) Method** with \(n = 10\) subintervals. Here’s the step-by-step explanation:

### Step 1: Set up the Problem
The integral is:
\[
\int_1^2 \sqrt{x^3 - 1} \, dx
\]
Given that \(n = 10\), we will use the midpoint method to approximate this value.

### Step 2: Calculate the Width of Each Subinterval (h)
The width \(h\) of each subinterval is calculated by:
\[
h = \frac{b - a}{n} = \frac{2 - 1}{10} = 0.1
\]
So each subinterval has a width of \(0.1\).

### Step 3: Identify the Midpoints
Next, the midpoints of the subintervals are determined. These are the \(x\)-values where the height of each rectangle is evaluated. The subintervals are:
- \([1, 1.1], [1.1, 1.2], \dots, [1.9, 2]\)

The midpoints \(x_i\) for each subinterval are listed as:
\[
1.05, 1.15, 1.25, 1.35, 1.45, 1.55, 1.65, 1.75, 1.85, 1.95
\]

### Step 4: Evaluate the Function at Each Midpoint
For each midpoint \(x_i\), the function \(f(x) = \sqrt{x^3 - 1}\) is evaluated:
- \(f(1.05) = \sqrt{(1.05)^3 - 1} = 0.39702\)
- \(f(1.15) = \sqrt{(1.15)^3 - 1} = 0.72117\)
- \(f(1.25) = 0.976281\)
- \(f(1.35) = 1.20846\)
- Continue calculating for all midpoints...

### Step 5: Apply the Midpoint Rule Formula
The midpoint rule formula for approximating the integral is:
\[
\int_a^b f(x) \, dx \approx h \left(f(x_1) + f(x_2) + \dots + f(x_{10})\right)
\]
Substitute the values:
\[
\int_1^2 \sqrt{x^3 - 1} \, dx \approx 0.1 \times (0.39702 + 0.72117 + 0.976281 + 1.20846 + 1.4313 + 1.65042 + 1.86872 + 2.08791 + 2.30903 + 2.5376)
\]

### Step 6: Final Approximation
Summing the function values gives approximately 15.1836. Thus, the integral is approximately:
\[
\int_1^2 \sqrt{x^3 - 1} \, dx \approx 0.1 \times 15.1836 = 1.51836 \approx 1.52
\]

The final approximation for the integral using the midpoint method is **1.52**.

---

Would you like more details or have any questions?

### Related Questions:
1. What is the Midpoint Rule and how does it compare to other methods like Trapezoidal Rule?
2. How do we know the number of subintervals \(n\) to use for an accurate approximation?
3. How would the result change if we used \(n = 20\) instead of \(n = 10\)?
4. Can the Midpoint Rule be applied to other types of integrals (e.g., trigonometric or exponential)?
5. What are the advantages and disadvantages of using numerical methods like this?

### Tip:
The accuracy of numerical methods often improves with a higher number of subintervals, but computation becomes more intensive.

Explain the solution to the integral \(\int_1^2 \sqrt{x^3 - 1} \, dx\) using the Midpoint Method with 10 subintervals.

Step-by-step solution for approximating the integral \(\int_1^2 \sqrt{x^3 - 1} \, dx\) using the Midpoint Method with 10 subintervals. Includes key formulas for the Midpoint Rule and detailed calculations for each subinterval, resulting in an approximate value of 1.52.

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