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I want to find the coefficients $A_0,A_1,C_0,C_1$ so that this integration formula integrates exactly polynomials of degree $3$ or less. Prove that this formula is not exact for polynomaials of degree $4$. $$\int_a^bf(x)=A_0f(a)+A_1f(b)+C_0f''(a)+C_1f''(b)$$ I know that if this integration formula has to integrate exactly polynomials of degree 3 or less, it must fulfill the system $$\begin{cases}b-a=A_0+A_1\\ b^2-a^2=2(aA_0+bA_1)\\ b^3-a^3=3(a^2A_0+b^2A_1+2C_0+2C_1)\\ b^4-a^4=4(a^3A_0+b^3A_0+48aC_0+48bC_1)\end{cases}$$

However that is not enough, since this system has $4$ equations and $6$ variables. I would add more equations and impose exactitude for $x^4$ and $x^5$ but I can't do that since this integration formula can't be exact for polynomials of degree greater than $3$. How do I follow?

mathie12
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    You only have $4$ variables for each fixed interval $[a,b]$. I presume no rule can integrate exactly on every interval. – qualcuno Jun 16 '18 at 19:33
  • $B_0=A_1{{{}}}$? – Angina Seng Jun 16 '18 at 19:34
  • @Guido A Then I should just pick an interval, for example $[0,1]$ and work there? – mathie12 Jun 16 '18 at 19:36
  • I think that'd make the linear system easier to handle. Iirc, you can find a rule for integration in $[0,1]$ and it will induce one on $[a,b]$ with the same exactness. – qualcuno Jun 16 '18 at 19:39
  • Don't consider $a$ and $b$ unknown. Then you can solve the system of equations and get $C_0=C_1=\frac{1}{24} (a-b)^3$ and $A_0=A_1=\frac{b-a}{2}$. – Oppenede Jun 16 '18 at 19:44

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