We can't find integrals of some fumulas like $sinx^2$, but we can get a approximate value using methods like the simpson method.
Since only one order-two equation exits which passes 3 points, we divide the equation into even parts and approximate the integral value as if it is a order-two equation.
In Thomas's Calculus, I learned that the error of the Simpson equation is $\frac{(M(b-a)^5)}{(180n^4)}$. M being the upper bound of the 4th derivative of the given function.
However if the function is order-three equation the M is always 0 since the 4th derivative is always 0. This means that the error using the simpson method is 0. However, it cant be 0 since we estimated the function as if it is order two.
Any explanation about this?
Simpson's formula is $exact$ for cubics.
Let $p(t)$ be a polynomial of degree $\leq 3.$ For fixed $M$ and for $x\geq 0$ let $I(x)=\int_{M-x}^{M+x}p(t)dt$ and let $S(x)=\frac {2x}{6}(p(M-x)+4p(M)+p(M+x)).$ Let $E(x)=I(x)-S(x).$ Then $E(x)$ is a polynomial of degree $\leq 4.$ We can show that $E(x)=0$ for all $x$ by showing that $E(0)=E'(0)=E''(0)=E'''(0)=E''''(0)=0.$
It is immediate from the definition of $E(x)$ that $E(0)=0.$
We have $E'(x)=$ $$(p(M+x)+p(M-x))-\frac {1}{3}(p(M-x)+4p(M)+p(M+x))-\frac {x}{3}(p'(M+x)-p'(M-x))=$$ $$=\frac {2}{3}(p(M+x)-2p(M)+p(M-x))-\frac {x}{3}(p'(M+x)-p'(M-x)).$$ Putting $x=0$ into this gives $E'(0)=0.$
We have $E''(x)=$ $$\frac {2}{3}(p'(M+x)-p'(M-x)-\frac {1}{3}(p'(M+x)-p'(M-x))-\frac {x}{3}(p''(M+x)+p''(M-x))=$$ $$=\frac {1}{3}(p'(M+x)-p'(M-x))-\frac {x}{3}(p''(M+x)+p''(M-x)).$$ Putting $x=0$ into this gives $E''(0)=0.$
We have $E'''(x)=$ $\frac {1}{3}(p''(M+x)+p''(M-x))-\frac {1}{3}(p''(M+x)+p''(M-x))-\frac {x}{3}(p'''(M+x)-p'''(M-x))$ $$=\frac {x}{3}(p'''(M+x)-p'''(M-x)).$$ But $p$ is a polynomial of degree $\leq 3$ so $p'''$ is a constant, so $p'''(M+x)-p'''(M-x)=0$ for all $x.$ So $E'''(x)=0$ for all $x. $ Of course this implies $E'''(0)=E''''(0)=0.$
Remark. There is higher-degree approximation using only $3$ points.. For $a<b$ let $m=(b+a)/2$ and $d=(b-a)/2.$ Let $J=\frac {(b-a)}{18}(5f(m-d\sqrt {3/5}\;)+8f(m)+5f(m+d\sqrt {3/5}\;)).$ If $f$ is a polynomial of degree $\leq 5$ then $J=\int_a^bf(t)dt.$