Question

$\int (x-1^1{C}_1{x}^2{+}^{11}{C}_2{x}^3{-}^{11}{C}_3{x}^4\dots \dots {-}^{11}{C}_{11}{x}^{12})dx$ equals

• A. $\frac{(1-x{)}^{12}}{12}-\frac{(1-x{)}^{11}}{11}+c$
• B. $\frac{(1-x{)}^{13}}{13}-\frac{(1-x{)}^{12}}{12}+c$
• C. $\frac{(1-x{)}^{11}}{11}-\frac{(1-x{)}^{12}}{12}+c$
• D. $\frac{(1-x{)}^{12}}{12}-\frac{(1-x{)}^{12}}{12}-\frac{(1-x{)}^{13}}{13}+c$

Answer 1

The correct option is A $\frac{(1-x{)}^{12}}{12}-\frac{(1-x{)}^{11}}{11}+c$

$\int x[{}^{11}{C}_0{-}^{11}{C}_{1}x{+}^{11}{C}_2{x}^2+.........{-}^{11}{C}_0{x}^{11}]dx=\int x(1-x{)}^{11}dx=x(\frac{(1-x{)}^{12}}{-12})-\int (\frac{(1-x{)}^{12}}{-12})dx=(\frac{-x}{12}\left)\right(1-x{)}^{12}+\left(\frac{1}{12}\right)\left(\frac{(1-x{)}^{13}}{-13}\right)=\left(\frac{(1-x{)}^{12}}{12}\right)[-x-(\frac{1-x}{13}\left)\right]=\left(\frac{(1-x{)}^{12}}{12}\right)(-12x-1)+C$