设f(x) 在[a,b] 上连续,证明∫(下限为a,上限为b)f(x)=(b-a)∫(下限为0,上限为1)f[a+(b-a)x]dx,

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设f(x)在[a,b]上连续,证明∫(下限为a,上限为b)f(x)=(b-a)∫(下限为0,上限为1)f[a+(b-a)x]dx,设f(x)在[a,b]上连续,证明∫(下限为a,上限为b)f(x)=(

设f(x) 在[a,b] 上连续,证明∫(下限为a,上限为b)f(x)=(b-a)∫(下限为0,上限为1)f[a+(b-a)x]dx,
设f(x) 在[a,b] 上连续,证明∫(下限为a,上限为b)f(x)=(b-a)∫(下限为0,上限为1)f[a+(b-a)x]dx,

设f(x) 在[a,b] 上连续,证明∫(下限为a,上限为b)f(x)=(b-a)∫(下限为0,上限为1)f[a+(b-a)x]dx,
难道不是直接一个变量代换就搞定了么?
Let x = a + (b-a) y, where 0

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Let x = a + (b-a) y, where 0<=y<=1, so that a<=x<=b.
(b-a)∫(下限为0,上限为1)f[a+(b-a)x]dx
= (b-a)∫(下限为0,上限为1)f[a+(b-a)y]dy
= ∫(下限为0,上限为1)f[a+(b-a)y]d[a+(b-a)y]
= ∫(下限为a,上限为b)f(x)dx --- plug in x = a + (b-a) y