设f(x)在(一∞,+∞)内连续,以T为周期,证明: (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

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问题 设f(x)在(一∞,+∞)内连续,以T为周期,证明:
    (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T0Tf(x)dx=0.

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答案(1)[*]=f(a+T)一f(a)=0, 故 ∫aa+Tf(x)dx=∫aa+Tf(x)dx|a=0=∫0Tf(x)dx. (2)∫0xf(t)dt以T为周期[*]∫0x+Tf(t)dt—∫0xf(t)dt=∫0x+Tf(t)dt[*]∫0Tf(t)dt=0. (3)由∫f(x)dx=∫0xf(t)dt+C,易知此命题成立.

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