Question
If f is an integrable function such that f(2a - x) = f(x), then prove that:
$\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=2\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$ 

Answer

Let $\text{I}=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
By Additive property
$\text{I}=\int\limits^{\text{a}}_0\text{f(x)}\text{dx}+\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Consider the integral $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Let $\text{x}=2\text{a}-\text{t},$ then $\text{dx}=-\text{dt}$
When $\text{x}=\text{a},\text{ t}=\text{a},\text{ x}=2\text{x},\text{t}=0$
Hence $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=-\int\limits^{0}_\text{a}\text{f(2a}-\text{t})\text{dt}$
$=\int\limits_{0}^\text{a}\text{f(2a}-\text{t})\text{dt}$
$=\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}$ (Changeing the varible)
Therefore,
$\text{I}=\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}$
$=\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(x})\text{dx}$ $\Bigg[\text{Given}\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}\Bigg]$
$=2\int\limits_{0}^\text{a}\text{f(x})\text{dx}$

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