Midpoint Rule Approximation for Integral ∫(4 to 12) x^2 dx with n = 4

Math Problem Statement

Use the Midpoint Rule with n = 4 to approximate the integral: ∫(4 to 12) x^2 dx.

To approximate the given integral $\int_{4}^{12} x^2 \, dx$ using the Midpoint Rule with $n = 4$ , we follow these steps:

The interval is $[4, 12]$ . The width of each subinterval ( $\Delta x$ ) is: $\Delta x = \frac{b - a}{n} = \frac{12 - 4}{4} = 2.$

The subintervals are: [ [4, 6], [6, 8], [8, 10], [10, 12]. ]

The midpoints are: $x_1 = 5, \quad x_2 = 7, \quad x_3 = 9, \quad x_4 = 11.$

The function is $f(x) = x^2$ . Evaluating at the midpoints:

f(7) = 7^2 = 49, \quad f(9) = 9^2 = 81, \quad f(11) = 11^2 = 121.$$ ### Step 4: Apply the Midpoint Rule formula The Midpoint Rule formula is: $$M_4 = \Delta x \cdot \sum_{i=1}^{4} f(x_i).$$ Substituting the values: $$M_4 = 2 \cdot \left[f(5) + f(7) + f(9) + f(11)\right]$$ $$M_4 = 2 \cdot \left[25 + 49 + 81 + 121\right]$$ $$M_4 = 2 \cdot 276 = 552.$$ ### Final Answer: $$M_4 = 552$$ Let me know if you want further clarification or a detailed breakdown of any step! ### Follow-up Questions: 1. What is the exact value of the integral for comparison? 2. How does the error of the Midpoint Rule depend on the number of subintervals $$n$$? 3. Can we use other numerical methods, like the Trapezoidal Rule, for this problem? How would it compare? 4. What happens to the approximation if $$n$$ increases to a larger value? 5. How does the Midpoint Rule handle functions with rapid changes or discontinuities? ### Tip: When using numerical integration methods, increasing $$n$$ (the number of subintervals) generally reduces the error and improves the accuracy of the approximation!

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Numerical Integration
Midpoint Rule
Polynomials

Midpoint Rule Formula: M_n = Δx * Σf(x_i)
Subinterval width: Δx = (b - a)/n

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