Question 705056

Function $f(x) = \frac{2x}{1+x^2}$, $h=0.5$. Grid points $x_0=0, x_1=0.5, x_2=1$. $y_0 = f(0)=0, y_1 = f(0.5) = \frac{1}{1.25} = 0.8$, $y_2 = f(1)=1$. Simpson's rule $I \approx \frac{h}{3}(y_0 + 4y_1 + y_2) = \frac{0.5}{3}(0 + 4(0.8) + 1) = \frac{0.5}{3}(4.2) = 0.7$. Exact integral $I = [\ln(1+x^2)]_0^1 = \ln(2) \approx 0.6931$. Difference $= 0.7 - 0.6931 = 0.0069 \approx 0.007$.