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Question Number 130448 by physicstutes last updated on 25/Jan/21

Given that f (x) = f (π - x), prove that \int_{0}^{π} x f (x) d x = \frac{π}{2} \int_{0}^{π} f (x) d x .

please what are different methods to approach this question ?

Answered by TheSupreme last updated on 25/Jan/21

\int_{0}^{π} x f (x) d x

u = π - x

d u = - d x

\int_{π}^{0} (π - u) f (π - u) (- d u) = \int_{0}^{π} x f (x)

\int_{0}^{π} π f (π - u) d u - \int_{0}^{π} u f (π - u) d u = \int_{0}^{π} x f (x) d x

t h a n, f (u) = f (π - u); s e t I = \int_{0}^{π} x f (x) d x

\int_{0}^{π} π f (u) d u = 2 \int_{0}^{π} u f (u) d u

\int_{0}^{π} \frac{π}{2} f (u) d u = \int_{0}^{π} u f (u) d u

Commented by physicstutes last updated on 25/Jan/21

thanks sir