设f(x)在[0,1]上连续,且∫01f(x)dx=A,求∫01[∫11f(t)dt+(1一x)f(x)]dx.

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问题 设f(x)在[0,1]上连续,且∫01f(x)dx=A,求∫01[∫11f(t)dt+(1一x)f(x)]dx.

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答案令φ(x)=∫x1f(t)dt,则φ’(x)=—f(x),φ(0)=∫01f(t)dt=A.于是, ∫01[∫x1f(t)dt+(1—x)f(x)]dx=∫01φ(x)+(x—1)φ’(x)]dx =∫01[[(x一1)φ(x)]’dx=(x—1)φ(x)|01 =φ(0)=A.

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