Q. If $\int_{0}^{a}{f(2a - x)dx = \mu\ \ and\ \int_{0}^{a}{f(x)\ dx = \lambda,}}$ then $\int_{0}^{2a}{f(x)dx}$ equals
A) $2\lambda - \mu$
B) $\lambda + \mu$
C) $\mu - \lambda$
D) $\lambda - 2\mu$
Explanation
(b)
We have, $\int_{0}^{a}{f\ (2a - x)dx = \mu}$
and $\int_{0}^{a}{f(x)dx = \lambda}$
Now, using properties of definite integral
$\int_{0}^{2a}{f(x)dx = \int_{0}^{a}{f(x)dx + \int_{0}^{a}{f(2a - x)dx}}}$
$\Rightarrow \ \ \int_{0}^{2a}{f(x)dx = \lambda + \mu}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.