If for all real triplets (a, b, c), ƒ(x) = a + bx + cx²; then $\int\limits_0^1 {f(x)dx}$ is equal to :

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