91Ó°ÊÓ

Consider the integral \ \ \ \[ \int_{0}^{1} \frac{1}{1+x^{2}} dx \] \ \ Our function is \ \ \( f(x) = \frac{1}{1+x^{2}} \), \ \ and the interval of integration is from \( a = 0 \) to \( b = 1 \). We use \( n = 6 \) subintervals.

The width of each subinterval (\( \Delta x \)) can be calculated as: \ \ \ \[ \Delta x = \frac{b-a}{n} = \frac{1-0}{6} = \frac{1}{6} \]

The general formula for the Trapezoid Rule is: \ \ \ \[ T = \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n) \right] \] \ \ With \( \Delta x = \frac{1}{6} \), and \( x_i = a + i \Delta x \): \ \ \ \( x_0 = 0, x_1 = \frac{1}{6}, x_2 = \frac{2}{6}, x_3 = \frac{3}{6}, x_4 = \frac{4}{6}, x_5 = \frac{5}{6}, x_6 = 1 \) \ \ \( T = \frac{1}{12} \left[ f(0) + 2f(\frac{1}{6}) + 2f(\frac{2}{6}) + 2f(\frac{3}{6}) + 2f(\frac{4}{6}) + 2f(\frac{5}{6}) + f(1) \right] \) \ \ Compute each term: \ \ \( f(0) = 1, f(\frac{1}{6}) = 0.9866, f(\frac{2}{6}) = 0.9428, f(\frac{3}{6}) = 0.8944, f(\frac{4}{6}) = 0.8497, f(\frac{5}{6}) = 0.8101, f(1) = 0.5 \) \ \ \( T = \frac{1}{12} \left[ 1 + 2(0.9866) + 2(0.9428) + 2(0.8944) + 2(0.8497) + 2(0.8101) + 0.5 \right] = \frac{1}{12} (1 + 1.9732 + 1.8856 + 1.7888 + 1.6994 + 1.6202 + 0.5) = \frac{1}{12} (10.4672) \) = 0.8723

The general formula for Simpson's Rule is: \ \ \ \[ S = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + f(x_6) \right] \] \ \ \( S = \frac{1}{18} \left[ f(0) + 4f(\frac{1}{6}) + 2f(\frac{2}{6}) + 4f(\frac{3}{6}) + 2f(\frac{4}{6}) + 4f(\frac{5}{6}) + f(1) \right] \) \ \ Substitute the values: \ \ \( S = \frac{1}{18} \left[ 1 + 4(0.9866) + 2(0.9428) + 4(0.8944) + 2(0.8497) + 4(0.8101) + 0.5 \right] = \frac{1}{18} (1 + 3.9464 + 1.8856 + 3.5776 + 1.6994 + 3.2404 + 0.5) = \frac{1}{18} (16.8494) \) = 0.9361