Question

1.设f(x)在区间[0,1]上连续,证明：∫ 0到派 xf(sinx)dx=派∫0到派/2 f（sinx)dx？

Answer 1

证明：
令t=π-x,则x∈[0,π]时,t∈[π,0] dx=-dt
则I=∫(0→π) xf(sinx) dx
=-∫(π→0) (π-t)f(sin(π-t)) dt
=-∫(π→0) (π-t)f(sint) dt
=∫(0→π)(π-t)f(sint) dt
=∫(0→π)πf(sint) dt-∫(0→π)tf(sint)dt
=∫(0→π)πf(sinx) dx-I
∴I=(1/2)∫(0→π)πf(sinx) dx=π∫(0→π/2)f(sinx) dx
证毕,10,