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If $\int_{-a}^a f(x) d x=\int_0^a f(x) d x+\int_0^a g(x) d x$ then $g(x)=$
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Verified Answer
The correct answer is:
$f(-x)$
Since, $\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x$
$$
\begin{aligned}
& =\int_0^a f(x) d x=f \int_0^a f(x) d x \\
& =\int_0^a f(x) d x=f \int_0^a f(-x) d x
\end{aligned}
$$
Here $f(-x)=f(x)$ is an even function
$$
\Rightarrow \mathrm{g}(\mathrm{x})=\mathrm{f}(-\mathrm{x})
$$
$$
\begin{aligned}
& =\int_0^a f(x) d x=f \int_0^a f(x) d x \\
& =\int_0^a f(x) d x=f \int_0^a f(-x) d x
\end{aligned}
$$
Here $f(-x)=f(x)$ is an even function
$$
\Rightarrow \mathrm{g}(\mathrm{x})=\mathrm{f}(-\mathrm{x})
$$
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