A) \[\frac{b-a}{2}\]
B) \[\frac{a-b}{2}\]
C) \[\frac{a}{2}\]
D) \[\frac{b}{2}\]
Correct Answer: A
Solution :
Key Idea: \[\int_{a}^{b}{f(x)}dx=\int_{a}^{b}{f(a+b-x)dx}\] Let \[I=\int_{a}^{b}{\frac{f(x)}{f(x)+f(a+b-x)}dx}\] ... (i) \[\Rightarrow \] \[I=\int_{a}^{b}{\frac{f(a+b-x)}{f(x)+f(a+b-x)}dx}\] ? (ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{a}^{b}{\frac{f(x)+f(a+b-x)}{f(x)+f(a+b-x)}dx}\] \[=\int_{a}^{b}{1dx=[x]_{a}^{b}}\] \[\Rightarrow \] \[I=\frac{b-a}{2}\]
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