Question
If f(x) is a continuous function defind on [-a, a], then prove that:
$\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\big\{\text{f(x)}+\text{f}(-\text{x})\big\}\text{dx}$ 

Answer

Let $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}$
By Additive property
$\text{I}=\int\limits^0_{-\text{a}}\text{f(x)}\text{dx}+\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
Let $\text{x}=-\text{t},$ then $\text{dx}=-\text{dt}$
When $\text{x}=-\text{a},\text{ t}=\text{a},\text{ x}=0,\text{ t}=0$
Hence, $\int\limits^0_{-\text{a}}\text{f(x)}\text{dx}=-\int\limits^0_{\text{a}}\text{f}(-\text{t})\text{dt}$
$=\int\limits_0^{\text{a}}\text{f}(-\text{t})\text{dt}=\int\limits_0^{\text{a}}\text{f}(-\text{x})\text{dx}$ (Changing the varible)
Therefore,
$\text{I}=\int\limits_0^{\text{a}}\text{f}(-\text{x})\text{dx}+\int\limits_0^{\text{a}}\text{f}(\text{x})\text{dx}$
$=\int\limits^{\text{a}}_0\big\{\text{f(x)}+\text{f}(-\text{x})\big\}\text{dx}$
Hence, proved.

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