If we know that $\int_{a}^b t(x)=h \sum_{k=1}^2 dk * t(a+kh)+O(h^m)$ where $h=\frac{b-a}{3}$, how do we find the coefficient d1, d2 and m in the equation?
Answer says that d1=3/2, d2=3/2, m=3
I tried to substitute x=a+y in the integral, and then expand it in h on the left hand side, expand on the right hand side, then equate them. However got stuck...
$\int_{a}^b t(x) dx=\int_{0}^{b-a} t(a+y) dy$
$=\int_{0}^{b-a} t(a+h)+t'(a+h)*(y-h)+(1/2)*t''(a+h)*(y-h)^2+O((y-h)^3) dy$
$=...$
Could anyone help and be a bit specific please? Thanks in advance,
Set $c_1=a+h=\frac{2a+b}3$ and $c_2=a+2h=\frac{a+2b}3$. Using $t(x)=x-c_1$ one gets $$ \int_a^bt(x)dx=\frac{b^2-a^2}2-c_1(b-a)=\frac{(b-a)^2}6 $$ and $$ h(d_1t(c_1)+d_2f(c_2))=hd_2(c_2-c_1)=\frac{b-a}3·d_2\frac{b-a}3=d_2\frac{(b-a)^2}9 $$ giving rise to $d_2=\frac32$. The same goes for $d_1$.
Finally explore $t(x)=(x-c_1)(x-c_2)$.