Error analysis in numerical integration 
Error analysis for basic Simpson's rule: $$\int_a^{a+2h}f(x)\:dx\approx \frac{h}{3}[f(a)+4f(a+h)+f(a+2h)]$$
  The error term in Simpson's rule can be established by using the Taylor series such that 
  $$f(a+h)=f+hf'+\frac{1}{2!}h^2f''+\frac{1}{3!}h^3f'''+\cdots$$
$$f(a+2h)=f+2hf'+2h^2f''+\frac{4}{3}h^3f'''+\cdots$$
$$\frac h3[f(a)+4f(a+h)+f(a+2h)]=2hf+2h^2f'+\frac 43h^3f''+\frac 23h^4f'''+\cdots\quad (1)$$
  Let $$F(x)=\int_a^xf(t)\:dt$$ The taylor series for $F(a+2h)$ is 
  $$F(a+2h)=F(a)+2hF'(a)+2h^2F''(a)+\frac{4}{3}h^3F'''(a)+\cdots$$
  By the FTC, $F'=f,F''=f',$ and so on. Observe that $F(a)=0$,
  $$\int_a^{a+2h}f(x)\:dx=0+2hf+2h^2f'+\frac{4}{3}h^3f''+\cdots\qquad(2)$$ Subtracting $(1)$ from $(2)$,
   $$\int_a^{a+2h}f(x)\:dx-\frac h3[f(a)+4f(a+h)+f(a+2h)]=-\frac{h^5}{90}f^{(4)}+\cdots$$
Error analysis for composite Simpson's rule:
  Suppose that the interval $[a,b]$ is subdivided into an even number of subintervals, say n,each of width $h=\frac{b-a}{n}$. Then the partition points are $x_i=a+ih,0\leq i\leq n$. Now we have,
  $$\int_a^bf(x)\:dx=\sum_{i=1}^{\frac n2}\int_{a+2(i-1)h}^{a+2ih}f(x)\:dx$$
  Using basic Simpson's Rule,
  $$\approx \sum_{i=1}^{\frac n2}\frac h3[f(a+2(i-1)h)+4f(a+(2i-1)h)+f(a+2ih)]-\frac{1}{90}h^5\sum_{i=1}^{\frac n2}f_{a\leq\delta\leq b}^{(4)}(\delta)$$

But how to get the error term like mentioned in book $-\frac{b-a}{180}h^4f^{(4)}(\delta)?$
I feel there must be another way to derive those formula by easily as I found great response from my previous Post.
I heartily thank if anyone explain any easy approach in details.

The kernel for the error of the Simpson rule in one segment $[0,2]$ is
$\newcommand{\one}{{1\!\!1}}$
$$
k_0(s)=\one_{[0,2]}(x)-\frac13(δ_0(x)+4δ_1(x)+δ_2(x))
$$
This can be integrated using the zero boundary condition from the left, $k_{i+1}(x)=\int_0^xk_i(s)ds$.
\begin{align}
k_1(x)&=\one_{[0,2]}(x)\cdot x -\frac13(u(x)+4u(x-1)+u(x-2))
\\
&=\one_{[0,2]}(x)⋅(x-1)+\tfrac23\one_{[0,1]}(x)-\tfrac23\one_{[1,2]}(x)
\\
k_2(x)&=\tfrac12\one_{[0,2]}(x)⋅\left[(x-1)^2+\tfrac13\right]+\tfrac23\one_{[0,1]}(x)⋅(x-1)-\tfrac23\one_{[1,2]}(x)⋅(x-1)
\\[.5em]
k_3(x)&=\tfrac16\one_{[0,2]}(x)⋅\left[(x-1)^3+(x-1)\right]+\tfrac13\one_{[0,1]}(x)⋅(x-1)^2-\tfrac13\one_{[1,2]}(x)⋅(x-1)^2
\\[.5em]
k_4(x)&=\tfrac1{24}\one_{[0,2]}(x)⋅\left[(x-1)^4+2(x-1)^2-\tfrac13\right]+\tfrac19\one_{[0,1]}(x)⋅(x-1)^3-\tfrac19\one_{[1,2]}(x)⋅(x-1)^3
\\[.5em]
k_5(x)&=\tfrac1{120}\one_{[0,2]}(x)⋅\left[(x-1)^5+\tfrac{10}3(x-1)^3-\tfrac53(x-1)\right]+\tfrac1{36}\one_{[0,1]}(x)⋅(x-1)^4-\tfrac1{36}\one_{[1,2]}(x)⋅(x-1)^4
\\[.5em]
\end{align}
While in $k_3$ the values at the boundary are zero directly as a result of the symmetric construction of the integral function, the same symmetric construction leads to values of $\pm\frac1{180}$ of opposing signs in $k_5$, so that this can not be repaired via an integration constant. 
In consequence one gets 
\begin{align}
\int_{a}^{a+2h}f(x)\,dx-\frac{h}3(f(a)+4f(a+h)+f(a+2h))
&=h\int_0^2 f(a+sh)k_0(s)\,ds
\\
&=h^5\int_0^2f^{(4)}(a+sh)k_4(s)\,ds
\\
&=h^4\int_a^{a+2h}f^{(4)}(x)k_4\left(\frac{x-a}h\right)\,dx
\end{align}
$k_4$ is non-negative over the interval $[0,2]$. Now think of $k_4$ as periodically continued, then the error of the composite method is, applying the weighted mean value theorem,
\begin{align}
h^4\int_{x_0}^{x_{2n}}f^{(4)}(x)k_4\left(\frac{x-x_0}h\right)\,dx
&=
h^4f^{(4)}(ξ)\int_{x_0}^{x_{2n}}k_4\left(\frac{x-x_0}h\right)\,dx
\\&=
h^4f^{(4)}(ξ)\,(nh\,[k_5(2)-k_5(0)])
=-\frac{h^4(b-a)}{180}f^{(4)}(ξ)
\end{align}
