Question 574959
Simpson's rule states that


*[tex \LARGE \int_{a}^{b} f(x) dx \approx \frac{b-a}{6}[f(a) + 4f(\frac{a+b}{2}) + f(b)]], in this case,


*[tex \LARGE \int_{0}^{\frac{\pi}{2}} \cos x dx \approx \frac{\frac{\pi}{2} - 0}{6}[\cos 0 + 4\cos(\frac{\pi}{4}) + \cos(\frac{\pi}{2})]]


*[tex \LARGE = \frac{\pi}{12}(1 + 2\sqrt{2}) \approx 1.0023]


If you were to actually integrate, you would get


*[tex \LARGE \int_{0}^{\frac{\pi}{2}} \cos x dx = \sin x] evaluated at x = pi/2, subtract the value obtained at x = 0. The end result is sin (pi/2) - sin 0, or 1.