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.