A) \[\frac{{{\left( 2011 \right)}^{2}}}{2}\]
B) \[\frac{\left( 2011 \right)\left( 2012 \right)\left( 4023 \right)}{2}\]
C) 0
D) 2010
Correct Answer: C
Solution :
Here \[I=\int\limits_{1}^{2011}{(x-1)(x-2)(x-3)....(x-2011)dx}\] \[\therefore \]\[I=\int\limits_{1}^{2011}{\begin{align} & (2012-x-1)(2012-x-2)(2012-x-3) \\ & ......(2012-x-2011)dx \\ \end{align}}\] \[\left[ \because \,\,\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx}} \right]\] \[\therefore \] \[I=\int\limits_{1}^{2011}{(2011-x)(2010-x)(2009-x)\cdot \cdot \cdot (1-x)dx}\]\[\therefore \] \[I=\int\limits_{1}^{2011}{{{(-1)}^{2011}}(x-2011)(x-2010)\cdot \cdot \cdot (x-1)dx}\] \[=-\int\limits_{1}^{2011}{(x-1)(x-2)\cdot \cdot \cdot (x-2011)dx}\] \[=-I\] \[\Rightarrow \,2I=0\] \[\Rightarrow \,I=0\].
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