Question

If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

Answer 1

\[Let\ I = \int_0^{2a} f\left( x \right) d x\]
Using additive property
\[I = \int_0^a f\left( x \right) d x + \int_a^{2a} f\left( x \right) d x\]
\[\text{Consider the integral} \int_a^{2a} f\left( x \right) d x\]
\[\text{Let }x = 2a - t, \text{Then }dx = - dt\]
\[\text{When }x = a, t = a\text{ and }x = 2a, t = 0\]
Therefore,
\[ \int_a^{2a} f\left( x \right) d x = - \int_a^0 f\left( 2a - t \right) d t\]
\[ = \int_0^a f\left( 2a - t \right) d t\]
\[ = \int_0^a f\left( 2a - x \right) dx ................\left( \text{changing the variable} \right)\]
\[\text{We have }f\left( 2a - x \right) = - f\left( x \right)\]
Therefore,
\[I = \int_0^a f\left( x \right) d x - \int_0^a f\left( x \right) d x = 0\]