Question
Let $$\int_0^a {f\left( x \right)dx} = \lambda $$ and $$\int_0^a {f\left( {2a - x} \right)dx} = \mu .$$ Then $$\int_0^{2a} {f\left( x \right)dx} $$ is equal to :
A.
$$\lambda + \mu $$
B.
$$\lambda - \mu $$
C.
$$2\lambda - \mu $$
D.
$$\lambda - 2\mu $$
Answer :
$$\lambda - \mu $$
Solution :
$$\eqalign{
& \int_0^a {f\left( x \right)} dx = \lambda \cr
& f\left( a \right) - f\left( 0 \right) = \lambda \to 1 \cr
& \int_0^a {f\left( {2a - x} \right)dx} = \mu \cr
& f\left( a \right) - f\left( {2a} \right) = \mu \to 2 \cr
& {\text{from 1 and 2}} \cr
& \,\,\,\,f\left( a \right) - f\left( 0 \right) = \lambda \cr
& \,\,\,\,f\left( a \right) - f\left( {2a} \right) = \mu \cr
& + \,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \cr
& \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \cr
& f\left( {2a} \right) - f\left( 0 \right) = \lambda - \mu \cr
& \int_0^{2a} {f\left( x \right)} dx = \lambda - \mu \cr} $$