Given, $f(x)=a+b x+c x^{2}$
$\therefore \int_{0}^{1} f(x) d x=\int_{0}^{1}\left(a+b x+c x^{2}\right) d x$
$=\left[a x+\frac{b x^{2}}{2}+\frac{c x^{3}}{3}\right]_{0}^{1}$
$=a+\frac{b}{2}+\frac{c}{3}$
Here, $f(0)=a, f\left(\frac{1}{2}\right)=a+\frac{b}{2}+\frac{c}{4}$
and $f(1)=a+b+c$
Now, $\frac{f(0)+4 f\left(\frac{1}{2}\right)+f(1)}{6}$
$=\frac{a+4\left(a+\frac{b}{2}+\frac{c}{4}\right)+a+b+c}{6}$
$=\frac{a+4\left(\frac{4 a+2 b+c}{4}\right)+a+b+c}{6}$
$=\frac{a+4 a+2 b+c+a+b+c}{6}=\frac{6 a+3 b+2 c}{6}$
$=a+\frac{b}{2}+\frac{c}{3}$
$\therefore$ From Eqs. (i) and (ii), we get
$\int_{0}^{1} f(x) d x=\frac{f(0)+4 f\left(\frac{1}{2}\right)+f(1)}{6}$