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

Answer

$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{f}\big(\text{x}^2\big)\text{dx}$
Here, $\text{g(x)}=\text{f}\big(\text{x}^2\big)$
$\Rightarrow\text{g}(-\text{x})=\text{f}(-\text{x})^2=\text{f}(\text{x})^2=\text{g(x)}\text{ i.e., g(x) is even}$
Therefore,
$\text{I}=2\int\limits^{\text{a}}_{0}\text{f}(\text{x}^2)\text{dx}$ $\bigg[\text{Using}\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{g}(\text{x})\text{dx}=\int\limits^{\text{a}}_{0}\text{f}(\text{x})\text{dx}\text{ when g(x) is even}\bigg]$

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