Find the order for the following quadrature formulae:
1) $\displaystyle \int_0^1 f(t)dt=\frac{1}{6}f(0)+\frac{2}{3}f(\frac{1}{2})+\frac{1}{6}f(1) $
2) $\displaystyle \int_0^1 f(t)dt=\frac{1}{8}f(0)+\frac{3}{8}f(\frac{1}{3})+\frac{3}{8}f(\frac{2}{3})+\frac{1}{8}f(1) $
My attempt does not seem quite right because when I extend $f(t)$ to its Taylor extension, I am not sure if the point I chose to be evaluated at , that is $t=0$, is right or wrong. I would appreciate any help or hints with that.
Thank you.
The easy answer is that both rules reproduce $$\int_0^11\,dx=\frac16+\frac23+\frac16=\frac18+\frac38+\frac38+\frac18=1$$ $$\int_0^1x\,dx=\frac16(0)+\frac23\left(\frac12\right)+\frac16(1)=\frac18(0)+\frac38\left(\frac13\right)+\frac38\left(\frac23\right)+\frac18(1)=\frac12$$ $$\int_0^1x^2\,dx=\frac16(0)^2+\frac23\left(\frac12\right)^2+\frac16(1)^2=\frac18(0)^2+\frac38\left(\frac13\right)^2+\frac38\left(\frac23\right)^2+\frac18(1)^2=\frac13$$ $$\int_0^1x^3\,dx=\frac16(0)^3+\frac23\left(\frac12\right)^3+\frac16(1)^3=\frac18(0)^3+\frac38\left(\frac13\right)^3+\frac38\left(\frac23\right)^3+\frac18(1)^3=\frac14$$ But they both get $4^{\text{th}}$ order polynomials wrong: $$\int_0^1\left(x^4-2x^3+\frac54x^2-\frac14x\right)dx=-\frac1{120}\ne\frac16(0)+\frac23\left(0\right)+\frac16(0)=0$$ and $$\int_0^1\left(x^4-2x^3+\frac{11}9x^2-\frac29x\right)dx=-\frac1{270}\ne\frac18(0)+\frac38\left(0\right)+\frac38\left(0\right)+\frac18(0)=0$$ So they are of order $3$.
The hard part is to prove the remainder formula for the second rule, Simpson's $3/8$ Rule. If we assume $f(x)$ has a continuous $4^{\text{th}}$ derivative and $F(x)$ is any function such that $F^{\prime}(x)=f(x)$ then $$\begin{align}F(1)-F(0)&=\int_0^1f(x)dx=-\left.(1-x)f(x)\right|_0^1+\int_0^1(1-x)f^{\prime}(x)dx\\ &=f(0)-\left.\frac12(1-x)^2f^{\prime}(x)\right|_0^1+\frac12\int_0^1(1-x)^2f^{\prime\prime}(x)dx\\ &=f(0)+\frac12f^{\prime}(0)+\frac16f^{\prime\prime}(0)+\frac1{24}f^{\prime\prime\prime}(0)+\frac1{24}\int_0^1(1-x)^4f^{(4)}(x)dx\end{align}$$ After a few more integrations by parts. Also $$\begin{align}f\left(\frac13\right)-f(0)&=\int_0^{1/3}f^{\prime}(x)dx=-\left.\left(\frac13-x\right)f^{\prime}(x)\right|_0^{1/3}+\int_0^{1/3}\left(\frac13-x\right)f^{\prime}(x)dx\\ &=\frac13f^{\prime}(0)+\frac12\left(\frac13\right)^2f^{\prime\prime}(0)+\frac16\left(\frac13\right)^3f^{\prime\prime\prime}(0)+\frac16\int_0^{1/3}\left(\frac13-x\right)^3f^{(4)}(x)dx\end{align}$$ $$\begin{align}f\left(\frac23\right)-f(0)&=\int_0^{2/3}f^{\prime}(x)dx=-\left.\left(\frac23-x\right)f^{\prime}(x)\right|_0^{2/3}+\int_0^{2/3}\left(\frac23-x\right)f^{\prime}(x)dx\\ &=\frac23f^{\prime}(0)+\frac12\left(\frac23\right)^2f^{\prime\prime}(0)+\frac16\left(\frac23\right)^3f^{\prime\prime\prime}(0)+\frac16\int_0^{2/3}\left(\frac23-x\right)^3f^{(4)}(x)dx\end{align}$$ $$f(1)-f(0)=f^{\prime}(0)+\frac12f^{\prime\prime}(0)+\frac16f^{\prime\prime\prime}(0)+\frac16\int_0^1(1-x)^3f^{(4)}(x)dx$$ When we add everything up we get $$\begin{align}&F(1)-F(0)-\frac18f(0)-\frac38f\left(\frac13\right)-\frac38f\left(\frac23\right)-\frac18f(1)\\ &\quad\quad=\left(1-\frac18-\frac38-\frac38-\frac18\right)f(0)+\left(\frac12-\frac18-\frac14-\frac18\right)f^{\prime}(0)\\ &\quad\quad\quad+\left(\frac16-\frac1{48}-\frac1{12}-\frac1{16}\right)f^{\prime\prime}(0)+\left(\frac1{24}-\frac1{432}-\frac1{54}-\frac1{48}\right)f^{\prime\prime\prime}(0)\\ &\quad\quad\quad+\int_0^1K(x)f^{(4)}(x)dx\end{align}$$ Since all those fractions add up to zero, the remainder is given by that integral with the Peano kernel. With a little algebra for each subinterval we get $$K(x)=\begin{cases}\frac1{48}x^3(2x-1)&0\le x\le\frac13\\ \frac1{432}(3x^2-3x+1)(6x^2-6x+1)&\frac13\le x\le\frac23\\ \frac1{48}(1-x)^3(1-2x)&\frac13\le x\le1\end{cases}$$ A quick check shows that $K(x)\le0$ for $0\le x\le1$ so we can say that $$\max_{x\in(0,1)}(f^{(4)}(x))\int_0^1K(x)dx\le\int_0^1K(x)f^{(4)}(x)dx\le\min_{x\in(0,1)}(f^{(4)}(x))\int_0^1K(x)dx$$ The intermediate value theorem assures us that there is some $0<\xi<1$ such that $$\int_0^1K(x)f^{(4)}(x)dx=f^{(4)}(\xi)\int_0^1K(x)dx$$ Now, $$\int_0^{1/3}K(x)dx=\int_0^{1/3}\frac1{48}x^3(2x-1)dx=-\frac7{233280}$$ $$\int_{1/3}^{2/3}K(x)dx=\int_{1/3}^{2/3}\frac1{48}\left(2x^4-4x^3+3x^2-x+\frac19\right)dx=-\frac{11}{116640}$$ $$\int_{2/3}^1K(x)dx=\int_{2/3}^1\frac1{48}(1-x)^3(1-2x)dx=-\frac7{233280}$$ So $$\int_0^1K(x)dx=-\frac1{6480}$$ And we have $$\int_0^1f(x)dx-\frac18f(0)-\frac38f\left(\frac13\right)-\frac38f\left(\frac23\right)-\frac18f(1)=-\frac1{6840}f^{(4)}(\xi)$$ For some $\xi\in(0,1)$. This checks with our previous calculation in that $-\frac1{120}=-\frac1{6840}\cdot4!$