If $f(x)$ is a polynomial in $x$ of third degree and: $$u_{-1}=\int_{-3}^{-1}f(x)dx\ ;\ u_{0}=\int_{-1}^{1}f(x)dx\ ; u_{1}=\int_{1}^{3}f(x)dx$$
then show that $$f(0) = \frac{1}{2}\Bigg(u_0-\frac{\Delta^2u_{-1}}{12}\Bigg)$$
I attempted this question by assuming that the function $f(x)$ is of the following form: $$f(x) = a + bx + cx^3$$
In this case I obtained the following values and was able to arrive at the correct solution: $$ \begin{align} u_{-1}&=\int_{-3}^{-1}f(x)dx\\&=\int_{-3}^{-1}(a+bx+cx^3)dx\\&=\Bigg[ax+b\frac{x^2}{2} + c\frac{x^4}{4}\Bigg]_{-3}^{-1}=2a - 4b - 20c \end{align} $$ $$ \begin{align} u_0 = \int_{-1}^{1}f(x)dx = \int_{-1}^{1}(a+bx+cx^3)dx=2a \end{align} $$ $$ \begin{align} u_{-1}&=\int_{1}^{3}f(x)dx\\&=\int_{1}^{3}(a+bx+cx^2)dx\\&=\Bigg[ax+b\frac{x^2}{2} + c\frac{x^4}{4}\Bigg]_{1}^{3}=2a + 4b + 20c \end{align} $$ Now, $$ \begin{align} RHS\ =\ \frac{1}{2}\Bigg[u_0-\frac{\Delta^2u_{-1}}{12}\Bigg]&=\frac{1}{2}\Bigg[u_0-\frac{1}{24}(u_{-1}-2u_{0}+u_{1})\Bigg]\\&=\frac{1}{2}\big[2a-\frac{1}{24}(4a-4a)\big] \\&= a = f(0) = LHS \end{align} $$ However, I was unable to prove in case I assumed that $f(x)$ was of the form: $$ f(x) = a + bx + cx^2 + dx^3 $$ Will my proof be sufficient for the given question or is it necessary to also show for the latter form?
This is a straightforward power-rule calculation: $$ u_{-1} = \int _{-3}^{1} f(x)\,dx = 2 a-4 b+\frac{26 c}{3}-20 d; $$ $$ u_0 = \int _{-1}^1 f(x)\,dx = 2 a+\frac{2 c}{3}; $$ $$ u_1 = \int _1^3 f(x)\,dx = 2 a+4 b+\frac{26 c}{3}+20 d $$With your notation, $$ \Delta^2 u_{-1} = \frac{1}{2}\left(u_{-1}-2u_0+u_1\right) = 8c; $$so we have $$ \frac{1}{2}\left(2a+\frac{2}{3}c - \frac{1}{12}\cdot 8c\right) = a = f(0), $$as was to be shown.