A) \[\int_{0}^{a}{f(x)dx}\]
B) \[\int_{0}^{a}{g(x)dx}\]
C) \[\int_{0}^{a}{[g(x)-f(x)]dx}\]
D) \[\int_{0}^{a}{[g(x)+f(x)]dx}\]
Correct Answer: A
Solution :
Given that \[f(x)=f(a-x)\] ?(i) and \[g(x)+g(a-x)=2\] ?(ii) Now, let \[I=\int_{0}^{a}{f(x)g(x)dx}\] \[=\int_{0}^{a}{f(a-x)g(a-x)dx}\] \[\Rightarrow \] \[I=\int_{0}^{a}{f(x)[2-g(x)]dx}\] [using (i) and (ii)] \[=\int_{0}^{a}{2f(x)dx-\int_{0}^{a}{f(x)g(x)dx}}\] \[\Rightarrow \] \[I=\int_{0}^{a}{2f(x)dx-I}\] \[\Rightarrow \] \[2I=\int_{0}^{a}{2f(x)dx}\] \[\Rightarrow \] \[I=\int_{0}^{a}{f(x)dx}\] Note:\[\int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx.}}\]
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