If for all real triplets

$\int_{0}^{1}\left(a+b x+c x^{2}\right) d x=a x+\frac{b x^{2}}{2}+\left.\frac{c x^{3}}{3}\right|_{0} ^{1}=a+\frac{b}{2}+\frac{c}{3}$

Now, $f(1)=a+b+c, f(0)=a$ and $f\left(\frac{1}{2}\right)=a+\frac{b}{2}+\frac{c}{4}$

Now, $\frac{1}{6}\left(f(1)+f(0)+4 f\left(\frac{1}{2}\right)\right)$

$=\frac{1}{6}\left(a+b+c+a+4\left(a+\frac{b}{2}+\frac{c}{4}\right)\right)$

$=\frac{1}{6}(6 a+3 b+2 c)=a+\frac{b}{2}+\frac{c}{3}$

Hence, $\int_{0}^{1} f(x)=\frac{1}{6}\left\{f(0)+f(1)+4 f\left(\frac{1}{2}\right)\right\}$