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) \[\frac{a+b}{2}\int_{a}^{b}{f(a+b+x)\,dx}\]
Correct Answer: B
Solution :
\[\int_{a}^{b}{x\,f(x)dx=}\int_{a}^{b}{(a+b-x)\,f(a+b-x)\,dx}\] Let \[l=\int_{a}^{b}{x\,f(x)\,dx}\] ... (i) \[\Rightarrow \] \[l=\int_{a}^{b}{(a+b-x)\,f(a+b-x)dx}\] \[\Rightarrow \] \[l=\int_{a}^{b}{(a+b-x)f(x)dx}\] \[[\because f(a+b-x)=f(x)\], given] \[\Rightarrow \] \[l=(a+b)\int_{a}^{b}{f(x)dx-\int_{a}^{b}{x\,f(x)dx}}\] \[\Rightarrow \] \[l=(a+b)\,\int_{a}^{b}{f(x)dx-l}\] [from Eq. (i)] \[\Rightarrow \] \[2\,l=(a+b)\int_{a}^{b}{f(x)dx}\] \[\Rightarrow \] \[l=\left( \frac{a+b}{2} \right)\,\int_{a}^{b}{f(x)\,dx}\]
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