Official solution
f(x) = a + bx + cx2
\[\int\limits_0^1 {f\left( x \right)dx} \] = \[\left[ {ax + {{b{x^2}} \over 2} + {{c{x^3}} \over 3}} \right]_0^1\]
= \[{a + {b \over 2} + {c \over 3}}\]
= \[{1 \over 6}\left[ {6a + 3b + c} \right]\]
f(1) = a + b + c
f(0) = a
\[f\left( {{1 \over 2}} \right) = a + {b \over 2} + {c \over 4}\]
By checking each option you have to find the solution.
\[{1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}\]
= \[{1 \over 6}\left[ {a + a + b + c + 4\left( {a + {b \over 2} + {c \over 4}} \right)} \right]\]
= \[{1 \over 6}\left[ {6a + 3b + c} \right]\]
\[ \therefore \] Option (A) is correct option.