Let f ( x ) be a function satisfying f ( x ) + f ( - x ) = 2 , x . Then 0 f ( x ) sin x ? d x is equal to [2023]

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Solution

Q.
$Let f (x) be a function satisfying f (x) + f (π - x) = π^{2}, \forall x \in ℝ .$ $Then \int_{0}^{π} f (x) \sin x d x is equal to$ [2023]

1 $π^{2}$

2 $2 π^{2}$

3 $\frac{π^{2}}{2}$

4 $\frac{π^{2}}{4}$

Ans.
(1)

$Let I = \int_{0}^{π} f (x) \sin x d x ...(i)$

$\Rightarrow I = \int_{0}^{π} f (π - x) \sin (π - x) d x$

$\Rightarrow I = \int_{0}^{π} f (π - x) \sin x d x ...(ii)$

$Adding (i) and (ii), we get$

$2 I = \int_{0}^{π} [f (x) + f (π - x)] \sin x d x$

$\Rightarrow I = \frac{π^{2}}{2} \int_{0}^{π} \sin x d x$ $(∵ f (x) + f (π - x) = π^{2})$

$= \frac{π^{2}}{2} {[- \cos x]}_{0}^{π}$ $= \frac{π^{2}}{2} [2] = π^{2}$

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