Question

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

\[\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]

 

Answer 1

\[Let\ I = \int_0^{2a} f\left( x \right) d x\]
\[\text{By Additive property}\]
\[I = \int_0^a f\left( x \right) d x + \int_a^{2a} f\left( x \right) d x\]
\[\text{Consider the integral} \int_a^{2a} f\left( x \right) d x\]
\[Let \ x = 2a - t, \text{then }dx = - dt\]
\[When\ x = a . t = a, x = 2a, t = 0\]
\[Hence\, \int_a^{2a} f\left( x \right) d x = - \int_a^0 f\left( 2a - t \right) d t\]
\[ = \int_0^a f\left( 2a - t \right) d t = \int_0^a f\left( 2a - x \right) d x \]
Therefore
\[I = \int_0^a f\left( x \right) d x + \int_0^a f\left( 2a - x \right) d x \]
\[ = \int_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]
\[\text{Hence, proved} .\]