Question
If f(2a - x) = -f(x), prove that $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=0$

Answer

Let $\text{I}=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=0$
Using 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 $\text{x}=\text{a},\text{t}=\text{a}$ and $\text{x}=2\text{a},\text{t}=0$
Threrfore,
$=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=-\int\limits^0_\text{a}\text{f}(2\text{a}-\text{t})\text{dt}$
$=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{t})\text{dt}$
$=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$ (Chang ing the variable)
We have
$\text{f}(2\text{a}-\text{x})=-\text{f(x)}$
Therefore,
$\text{I}=\int\limits^\text{a}_0\text{f(x)}\text{dx}-\int\limits^\text{a}_0\text{f(x)}\text{dx}=0$ 

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