设f(x)在(一∞,+∞)内连续,以T为周期,则 (1)∫aa+Tf(x)dx=∫0Tf(x)dx(a为任意实数); (2)∫0xf(t)dt以T为周期←→∫0Tf(x)dx=0; (3)∫f(x)dx(即f(x)的全体原函数)周期为T←→∫0Tf(x)d

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问题 设f(x)在(一∞,+∞)内连续,以T为周期,则
(1)∫aa+Tf(x)dx=∫0Tf(x)dx(a为任意实数);
(2)∫0xf(t)dt以T为周期←→∫0Tf(x)dx=0;
(3)∫f(x)dx(即f(x)的全体原函数)周期为T←→∫0Tf(x)dx=0.

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答案(1)[*]∫aa+Tf(x)dx=f(a+T)一f(a)=0 →∫aa+Tf(x)dx=∫aa+Tf(x)dx|a=0=∫aa+Tf(x)dx. (2)∫axf(t)dt以T为周期←→∫0x+Tf(t)dt一∫0xf(t)dt=∫xx+Tf(t)dt[*]∫0Tf(t)dt=0. (3)只需注意∫f(x)dx=∫0xf(t)dt+C,∫0xf(t)dt是f(x)的一个原函数.

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