Exrtensão: Regra de Simpson's

Simpson's Rule is a method of approximating integrals in which one approximates a curve by a parabola.

Suppose that we integrate a function $f$ over an interval $[a,b]$.

We would like to find the parabola that passes through the points: $$(a,f(a))$$ $$(b,f(b))$$ $$(\frac{a+b}{2},f(\frac{a+b}{2}))$$ Translate this function so that the points are: $$(-k,f(-k))$$ $$(0,f(0))$$ $$(k,f(k))$$ In this case, $k=\frac{b-a}{2}$.

Suppose that our parabolic approximation is $p(x)=Ax^2+Bx+C$. Calculating the area under the curve $p(x)$, $$\int_{-k}^{k}p(x)dx=\int_{-k}^{k}(Ax^2+Bx+C)dx$$ $$\int_{-k}^{k}p(x)dx=2\int_{0}^{k}(Ax^2+C)dx$$ $$\int_{-k}^{k}p(x)dx=\frac{k}{3}(2Ak^2+6C)$$ Now notice that $f(k)=Ak^2+Bk+C$, $f(0)=C$, and $f(-k)=Ak^2-Bk+C$.

Thus, we show the following: $$f(-k)+4f(0)+f(k)=2Ak^2+6C$$ $$\int_{-k}^{k}p(x)dx=\frac{k}{3}[f(-k)+4f(0)+f(k)]$$ Translating this function back to the interval $[a,b]$, $$\int_{a}^{b}p(x)dx=\frac{b-a}{6}[f(a)+4f(\frac{a+b}{2})+f(b)]$$ If one divides the interval $[a,b]$ into $2n$ parts, one may take $n$ parabolas: the $k$th parabola passes through the $x$-values $x_{2k-2}$, $x_{2k-1}$, and $x_{2k}$ for $k=1,2,...,n$.

If we call this integral approximation $S_{2n}$, one obtains: $$S_{2n}=\frac{b-a}{6n}(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_{2n-2}+4y_{2n-1}+y_{2n})$$ One may also find this through the weighted average of the trapezoidal and midpoint approximations: $$S_{2n}=\frac{1}{3}T_n+\frac{2}{3}M_n$$ The error bound in the Simpson's approximation is: $$E_S\le\frac{K(b-a)^5}{180n^4}$$ where $|f^{(4)}(x)|\le K$ for $a\le x\le b$.