Question

If `f` is an integrable function such that f(2a − x) = f(x), then prove that

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

 

Answer 1

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

Hence Proved.