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Chapter 7: Problem 12

Find the indicated Midpoint Rule approximations to the following integrals. \(\int_{1}^{9} x^{3} d x\) using \(n=1,2,\) and 4 subintervals

Short Answer

Expert verified
Answer: For 1 subinterval (n=1), the approximation M_1 is 1000. For 2 subintervals (n=2), the approximation M_2 is 1480. For 4 subintervals (n=4), the approximation M_4 is 1600.

Step by step solution

01

Midpoint Rule Formula

The midpoint rule is a numerical method for approximating definite integrals. The formula for the midpoint rule is given by: $$\int_{a}^{b} f(x) d x \approx M_n = \sum_{i=1}^{n}\Delta x \cdot f\left(\frac{x_{i} + x_{i-1}}{2}\right)$$ Where \(M_n\) is the approximation with \(n\) subintervals, \(\Delta x = \frac{b-a}{n}\), \(x_i = a + i \cdot \Delta x\), and \(f(x)\) is the function to be integrated.
02

Midpoint Rule with 1 Subinterval (n=1)

For the given function \(f(x) = x^3\), and the given interval [1,9], let's apply the midpoint rule with 1 subinterval: a = 1 b = 9 n = 1 So, we need to calculate 螖x as follows: $$\Delta x = \frac{b-a}{n} = \frac{9-1}{1} = 8$$ Now, let's find the function value at the midpoint: $$f\left(\frac{x_1 + x_0}{2}\right) = f\left(\frac{1 + 9}{2}\right) = f(5) = 5^3 = 125$$ Now, let's calculate the approximation using the midpoint rule: $$M_1 = \Delta x \cdot f\left(\frac{x_1 + x_0}{2}\right) = (8)(125) = 1000$$
03

Midpoint Rule with 2 Subintervals (n=2)

Now, let's apply the midpoint rule with 2 subintervals: a = 1 b = 9 n = 2 We need to calculate 螖x as follows: $$\Delta x = \frac{b-a}{n} = \frac{9-1}{2} = 4$$ We need to find the function value at the midpoints for the subintervals [1,5] and [5,9]: $$f\left(\frac{x_1 + x_0}{2}\right) = f\left(\frac{1+5}{2}\right) = f(3) = 3^3 = 27$$ $$f\left(\frac{x_2 + x_1}{2}\right) = f\left(\frac{5+9}{2}\right) = f(7) = 7^3 = 343$$ Now, we calculate the approximation using the midpoint rule: $$M_2 = \Delta x \cdot \left(f\left(\frac{x_1 + x_0}{2}\right) + f\left(\frac{x_2 + x_1}{2}\right)\right) = (4)(27+343) = 1480$$
04

Midpoint Rule with 4 Subintervals (n=4)

Lastly, let's apply the midpoint rule with 4 subintervals: a = 1 b = 9 n = 4 We need to calculate 螖x as follows: $$\Delta x = \frac{b-a}{n} = \frac{9-1}{4} = 2$$ We need to find the function value at the midpoints for all subintervals: f(2) = 2^3 = 8 f(4) = 4^3 = 64 f(6) = 6^3 = 216 f(8) = 8^3 = 512 Now, we calculate the approximation using the midpoint rule: $$M_4 = \Delta x \cdot (f(2)+f(4)+f(6)+f(8)) = (2)(8+64+216+512) = 1600$$ We got the following approximations using the midpoint rule: 1 subinterval (n=1): M_1 = 1000 2 subintervals (n=2): M_2 = 1480 4 subintervals (n=4): M_4 = 1600

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