设f(x)在(-∞,+∞)内连续,以T为周期,试证明: (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T∫0Tf(x)dx

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问题 设f(x)在(-∞,+∞)内连续,以T为周期,试证明:
(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T<=>∫0Tf(x)dx=0.

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答案(1) ∫aa+Tf(x)d=∫a0f(x)dx+∫0Tf(x)dx+∫TT+af(x)dx, 其中∫TT+af(x)dx=∫TT+af(x-T)dx[*]∫0af(s)ds=∫0af(x)dx. 代入上式得∫aa+Tf(x)=∫a0f(x)dx+∫0Tf(x)dx+∫0af(x)dx=∫0Tf(x)dx. (2)∫0xf(t)dt以T为周期<=>∫0x+Tf(t)dt-∫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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