设f(x)是以T为周期的连续函数,即f(x+T)=f(x),

问题描述:

设f(x)是以T为周期的连续函数,即f(x+T)=f(x),
则,对于任意a,有∫(a,a+T)f(x)d(x)=∫(T,0)f(x)d(x),如何证明啊,

∫(a,a+T)f(x)d(x)=∫(a,0)f(x)d(x)+∫(0,T)f(x)d(x)+∫(T,a+T)f(x)d(x)上式右边最后一个积分中,令x=T+t,有∫(T,a+T)f(x)d(x)=∫(0,a)f(T+t)d(t)=∫(0,a)f(t)d(t)=-∫(a,0)f(x)d(x)代入得证...