Taking the Lagrange interpolation formular for a second order polynomial $$f_2(x)=L_0(x)f(x_0)+L_1(x)f(x_1)+L_2(x)f(x_2)$$ Where $$\begin{align}L_0(x)&=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\\L_1(x)&=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\\L_2(x)&=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\end{align}$$ Therefore, $$f_2(x)=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}f(x_0)+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}f(x_1)+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}f(x_2)$$ The area of $f(x)$ under the interval $[x_0,x_2]$ is given by $$\begin{align}\int_{x_0}^{x_2}f(x)dx&\approx\int_{x_0}^{x_2}\left[L_0(x)f(x_0)+L_1(x)f(x_1)+L_2(x)f(x_2)\right]dx\\\int_{x_0}^{x_2}f(x)dx&\approx\left[\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}f(x_0)+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}f(x_1)+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}f(x_2)\right]dx\end{align}$$
After integrating I reached this$$\int_{x_0}^{x_2}f(x)dx\approx\left[\frac{\frac{x^3}{3}-\frac{x^2(x_2+x_1)}{2}+(x_1x_2)x}{(x_0-x_1)(x_0-x_2)}f(x_0)+\frac{\frac{x^3}{3}-\frac{x^2(x_2+x_0)}{2}+(x_0x_2)x}{(x_1-x_0)(x_1-x_2)}f(x_1)+\frac{\frac{x^3}{3}-\frac{x^2(x_0+x_1)}{2}+(x_1x_0)x}{(x_2-x_1)(x_2-x_0)}f(x_2)\right]$$ but after this I am unable to get the answer. Can anyone help me, please?
The Wikipedia article on Simpson's rule also gives your deviation, but suppresses the calculations that you could not master. Maybe you forgot to input that $x_0=a$, $x_1={1\over2}(a+b)$, $x_2=b$.
At any rate I would suggest to use another approach. We may assume $x_0=-1$, $x_1=0$, $x_2=1$ and verify easily that Simpson's rule gives the correct integral values for the three parabolas $y=ax^2+bx+c$ having $y=1$ at one of the three points and $y=0$ at the other two points. This and the "linearity of everything" proves your claim.