Let f(x) be a function satisfying f(x) + f(pi | vedclass

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(D) Let $I = \int_{0}^{\pi} f(x) \sin x \, dx$  $(1)$Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$,we get:$I = \int_{0}^{

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$\pi^2$

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(D) Let $I = \int_{0}^{\pi} f(x) \sin x \, dx$  $(1)$Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$,we get:$I = \int_{0}^{\pi} f(\pi - x) \sin(\pi - x) \, dx$Since $\sin(\pi - x) = \sin x$,this becomes:$I = \int_{0}^{\pi} f(\pi - x) \sin x \, dx$  $(2)$Adding equations $(1)$ and $(2)$:$2I = \int_{0}^{\pi} [f(x) + f(\pi - x)] \sin x \, dx$Given that $f(x) + f(\pi - x) = \pi^2$,we substitute this into the integral:$2I = \int_{0}^{\pi} \pi^2 \sin x \, dx$$2I = \pi^2 \int_{0}^{\pi} \sin x \, dx$$2I = \pi^2 [-\cos x]_{0}^{\pi}$$2I = \pi^2 [-(\cos \pi - \cos 0)]$$2I = \pi^2 [-(-1 - 1)]$$2I = \pi^2 [2]$$2I = 2\pi^2$$I = \pi^2$

Let $f(x)$ be a function satisfying $f(x) + f(\pi - x) = \pi^2, \forall x \in R$. Then $\int_{0}^{\pi} f(x) \sin x \, dx$ is equal to $...........$.

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