If $$f\left( {2a - x} \right) = f\left( x \right)$$ and $$\int_0^a {f\left( x \right)dx = \lambda } $$ then $$\int_0^{2a} {f\left( x \right)dx} $$ is :
A.
$$2\lambda $$
B.
$$\lambda $$
C.
0
D.
none of these
Answer :
$$2\lambda $$
Solution :
$$\eqalign{
& \int_0^{2a} {f\left( x \right)dx} = \int_0^a {f\left( x \right)dx} + \int_a^{2a} {f\left( x \right)dx} \cr
& {\text{Putting }}x = 2a - z{\text{ in the second integrand,}} \cr
& \int_0^{2a} {f\left( x \right)dx} = \int_0^a {f\left( x \right)dx} + \int_a^0 {f\left( {2a - z} \right)\left( { - dz} \right)} \cr
& = \int_0^a {f\left( x \right)dx} + \int_0^a {f\left( {2a - z} \right)dz} \cr
& = \int_0^a {f\left( x \right)dx} + \int_0^a {f\left( z \right)dz} \cr
& = \lambda + \lambda \cr
& = 2\lambda \cr} $$
Releted MCQ Question on Calculus >> Application of Integration
Releted Question 1
The area bounded by the curves $$y = f\left( x \right),$$ the $$x$$-axis and the ordinates $$x = 1$$ and $$x = b$$ is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$ Then $$f\left( x \right)$$ is-