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,
问题描述：
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,
答
Let x = a + (b-a) y, where 0(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
答
99
答
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