A) Independent of \[f\]
B) Independent of \[g\]
C) Independent of both f and g
D) None of these
Correct Answer: B
Solution :
| \[\int\limits_{-a}^{a}{xf(x)\,\,\,dx=\int\limits_{0}^{a}{f(x)\,dx+\int\limits_{0}^{a}{f(-x)dx}}}\] |
| \[\therefore \int\limits_{-a}^{a}{\frac{f(x)}{{{b}^{g(x)}}+1}dx=\int\limits_{0}^{a}{\frac{f(x)}{{{b}^{g(x)}}+1}}}dx+\int\limits_{0}^{a}{\frac{f(-x)}{{{b}^{g(-x)}}+1}dx}\] |
| \[=\int\limits_{0}^{a}{\frac{f(x)}{{{b}^{g(x)\,}}+1}dx=\int\limits_{0}^{a}{\frac{f(x)}{{{b}^{-g(x)}}+1}dx}}\] |
| (\[\because f\]is a even and g is odd) |
| \[=\int\limits_{0}^{a}{f(x)dx}\]Which is independent of g. |
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