Given is
$$ I(f) = w_0 f(-1) +w_1 f(0)+w_2 f(1) + w_3 f(2) $$ I need to calculate the $w_i$ so its exact value of $\int _0^1 f dx$ for polynoms with grade 3.
Now i want to show that $|\int _0^1 f dx-I(f)| < \frac{11}{720} c$
How do i show that estimation and how does c look. What has to be given about f so this works?
We can get the weights by integration of a judicious selection of cubics: $$\int_0^1x(x-1)(x-2)dx=\left.\left[\frac14x^4-x^3+x^2\right]\right|_0^1=\frac14=-6w_0$$ $$\int_0^1(x+1)(x-1)(x-2)dx=\left.\left[\frac14x^4-\frac23x^3-\frac12x^2+2x\right]\right|_0^1=\frac{13}{12}=2w_1$$ $$\int_0^1(x+1)x(x-2)dx=\left.\left[\frac14x^4-\frac13x^3-x^2\right]\right|_0^1=-\frac{13}{12}=-2w_2$$ $$\int_0^1(x+1)x(x-1)dx=\left.\left[\frac14x^4-\frac12x^2\right]\right|_0^1=-\frac14=6w_3$$ Thus our formula is $$I(f)=-\frac1{24}f(-1)+\frac{13}{24}f(0)+\frac{13}{24}f(1)-\frac1{24}f(2)$$ Since it's exact for all cubics, it's also exact for the polynomial $p_3(x)$ which interpolates $f(x)$ at $x=-1$, $x=0$, $x=1$, and $x=2$. Choose any $y\in(0,1)$ and let $$e(x)=f(x)-p_3(x)-\frac{\pi(x)}{\pi(y)}\left(f(y)-p_3(y)\right)$$ Where $\pi(x)=(x+1)x(x-1)(x-2)$. Then by construction, $e(-1)=e(0)=e(y)=e(1)=e(2)=0$, so by $4$ applications of Rolle's theorem, for any $y\in(0,1)$ there is some $u\in(-1,2)$ such that $$e^{(4)}(u)=f^{(4)}(u)-\frac{24}{\pi(y)}\left(f(y)-p_3(y)\right)=0$$ Solving for $p_3(y)$, $$p_3(y)=f(y)-\frac1{24}\pi(y)f^{(4)}(u(y))$$ Thus $$I(f)=\int_0^1p_3(x)dx=\int_0^1f(x)dx-\frac1{24}\int_0^1\pi(x)f^{(4)}(u(x))dx$$ Since $\pi(x)>0$ for $x\in(0,1)$, $$\begin{align}\frac1{24}\inf_{y\in(-1,2)}f^{(4)}(y)\int_0^1\pi(x)dx&=\frac1{24}\inf_{y\in(-1,2)}f^{(4)}(y)\left.\left[\frac15x^5-\frac12x^4-\frac13x^3+x^2\right]\right|_0^1\\ &=\frac{11}{720}\inf_{x\in(-1,2)}f^{(4)}(x)\\ &\le\frac1{24}\int_0^1\pi(x)f^{(4)}(u(x))dx\\ &=\int_0^1f(x)dx-I(f)\\ &\le\frac{11}{720}\sup_{x\in(-1,2)}f^{(4)}(x)\end{align}$$ Thus, assuming continuity of $f^{(4)}(x)$ in $(-1,2)$, by the intermediate value theorem there must be some $\xi\in(-1,2)$ such that $$\int_0^1f(x)dx=I(f)+\frac{11}{720}f^{(4)}(\xi)$$ so $$\left|\int_0^1f(x)dx-I(f)\right|\le\frac{11}{720}\sup_{x\in(-1,2)}\left|f^{(4)}(x)\right|$$