A) \[\frac{a+b}{2}\int_{a}^{b}{f(b-x)\,dx}\]
B) \[\frac{a+b}{2}\int_{a}^{b}{f(x)\,dx}\]
C) \[\frac{b-a}{2}\int_{a}^{b}{f(x)\,dx}\]
D) None of these
Correct Answer: B
Solution :
Since \[I=\int_{a}^{b}{xf(x)dx=\int_{a}^{b}{(a+b-x)f(a+b-x)dx}}\] Þ \[I=\int_{a}^{b}{(a+b)}f(x)dx-\int_{a}^{b}{xf(x)dx}\] \[\left\{ \because f(a+b-x)=f(x)\text{given} \right\}\] Þ \[2I=(a+b)\int_{a}^{b}{f(x)dx}\]Þ \[I=\int_{a}^{b}{x\,f(x)dx=\frac{a+b}{2}\int_{a}^{b}{f(x)dx}}\].
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