JEE Main 20209 Jan, Morning shiftMathematicsMediumIntegral Calculus369× since 2002If for all real triplets (a, b, c), ƒ(x) = a + bx + cx²; then ∫01f(x)dx\int\limits_0^1 {f(x)dx}0∫1f(x)dx is equal to :AOption A. 16{f(0)+f(1)+4f(12)}{1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}61{f(0)+f(1)+4f(21)}BOption B. 2{3f(1)+2f(12)}2\left\{ 3{f(1) + 2f\left( {{1 \over 2}} \right)} \right\}2{3f(1)+2f(21)}COption C. 13{f(0)+f(12)}{1 \over 3}\left\{ {f(0) + f\left( {{1 \over 2}} \right)} \right\}31{f(0)+f(21)}DOption D. 12{f(1)+3f(12)}{1 \over 2}\left\{ {f(1) + 3f\left( {{1 \over 2}} \right)} \right\}21{f(1)+3f(21)}CheckSkip · show solution
