设f(x)连续,则{∫0xsin[∫0tf(u)du]dt}=___________.

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问题 设f(x)连续,则{∫0xsin[∫0tf(u)du]dt}=___________.

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答案sin2[∫0xf(u)du]

解析0xsin2[∫0xf(u)du]dt是形如∫0xφ(t)dt形式的变上限积分,由
=φ(x)=sin2[∫0xf(u)du].
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