A) \[2\lambda -\mu \]
B) \[\lambda +\mu \]
C) \[\mu -\lambda \]
D) \[\lambda -2\mu \]
E) None of the above
Correct Answer: B
Solution :
Given, \[\int_{0}^{a}{f(2a-x)}dx=\mu \] and \[\int_{0}^{a}{f(x)}dx=\lambda \] Now, using properties of definite integral \[\int_{0}^{2a}{f(x)}dx=\int_{0}^{a}{f(x)}dx+\int_{0}^{a}{f(2a-x)dx}\] \[\Rightarrow \]\[\int_{0}^{2a}{f(x)}dx=\lambda +\mu \]
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