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If $f(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are continuous functions satisfying $f(x)=f(a-x)$ and $g(x)+g(a-x)=2$, then what is
$\int_{0}^{a} f(x) g(x) d x$ equal to?
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$\int_{0}^{a} f(x) g(x) d x$ equal to?
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Verified Answer
The correct answer is:
$\int_{0}^{a} f(x) d x$
$\begin{aligned} I &=\int_{0}^{a} f(x) \cdot g(x) d x \\ & I=\int_{0}^{a} f(a-x) \cdot g(a-x) d x \\ & I=\int_{0}^{a} f(x) \cdot[2-g(x)] d x \end{aligned}$
$\therefore f(a-x)=f(x)$ and $g(a-x)=2 g(x)$
$\mathrm{I}-\int_{0}^{\mathrm{a}} 2 . \mathrm{f}(\mathrm{x}) \cdot \mathrm{d} \mathrm{x}-\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{s}) \cdot \mathrm{d} \mathrm{x}$
$\mathrm{I}=2 \cdot \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x})-\mathrm{I}$
$\Rightarrow \not z^{\prime} I=\not^{\prime} \int_{0}^{a} f(x) d x$
$\therefore I=\int_{0}^{a} f(x) d x$
$\therefore f(a-x)=f(x)$ and $g(a-x)=2 g(x)$
$\mathrm{I}-\int_{0}^{\mathrm{a}} 2 . \mathrm{f}(\mathrm{x}) \cdot \mathrm{d} \mathrm{x}-\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{s}) \cdot \mathrm{d} \mathrm{x}$
$\mathrm{I}=2 \cdot \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x})-\mathrm{I}$
$\Rightarrow \not z^{\prime} I=\not^{\prime} \int_{0}^{a} f(x) d x$
$\therefore I=\int_{0}^{a} f(x) d x$
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