Let f : R to R be a function such that f(2 - | vedclass

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(C) Given $f(2 - x) = f(2 + x)$,the function is symmetric about $x = 2$. Given $f(4 - x) = f(4 + x)$,the function is symmetric about $x = 4$. Replacin

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$100$

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(C) Given $f(2 - x) = f(2 + x)$,the function is symmetric about $x = 2$. Given $f(4 - x) = f(4 + x)$,the function is symmetric about $x = 4$. Replacing $x$ with $x - 2$ in the second equation: $f(4 - (x - 2)) = f(4 + (x - 2)) \Rightarrow f(6 - x) = f(2 + x)$. Since $f(2 - x) = f(2 + x)$,we have $f(6 - x) = f(2 - x)$. Let $t = 2 - x$,then $f(4 + t) = f(t)$,which shows that $f(x)$ is a periodic function with period $T = 4$. We are given $\int_{0}^{2} f(x) dx = 5$. Since $f(2 - x) = f(2 + x)$,$\int_{0}^{4} f(x) dx = \int_{0}^{2} f(x) dx + \int_{2}^{4} f(x) dx = 2 \int_{0}^{2} f(x) dx = 2 	imes 5 = 10$. Now,$\int_{10}^{50} f(x) dx = \frac{50 - 10}{4} \int_{0}^{4} f(x) dx = 10 	imes 10 = 100$.

Let $f : R \to R$ be a function such that $f(2 - x) = f(2 + x)$ and $f(4 - x) = f(4 + x)$,for all $x \in R$. If $\int_{0}^{2} f(x) dx = 5$,then the value of $\int_{10}^{50} f(x) dx$ is:

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