A) \[3{{R}_{1}}=2{{R}_{2}}\]
B) \[2{{R}_{1}}=3{{R}_{2}}\]
C) \[{{R}_{1}}={{R}_{2}}\]
D) \[{{R}_{1}}=2{{R}_{2}}\]
Correct Answer: D
Solution :
We have \[{{\operatorname{R}}_{2}}=\int\limits_{-2}^{3}{x\,f\left( x \right)dx}=\int\limits_{a}^{b}{\left( 1-x \right)f\left( 1-x \right)dx}\]\[\left[ using\int\limits_{a}^{b}{f\left( x \right)dx=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}} \right]\]\[\Rightarrow {{\operatorname{R}}_{2}}=\int\limits_{-2}^{3}{\left( 1-x \right)f\left( x \right)dx}\] \[\left( \because f\left( x \right)=f\left( 1-x \right)\operatorname{on}\left[ -2,3 \right] \right)\]\[\therefore {{R}_{2}}+{{R}_{2}}=\int\limits_{-2}^{3}{x\,f\left( x \right)dx+\int\limits_{-2}^{3}{\left( 1-x \right)f\left( x \right)dx}}\] \[=\int\limits_{-2}^{3}{f\left( x \right)dx={{R}_{1}}}\]\[=2{{R}_{2}}={{R}_{1}}\]You need to login to perform this action.
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