设fx在[a,a+1)连续,则lim(x->a+)x^2/(x-a)定积分(a到x)f(t)dt=

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设fx在[a,a+1)连续,则lim(x->a+)x^2/(x-a)定积分(a到x)f(t)dt=设fx在[a,a+1)连续,则lim(x->a+)x^2/(x-a)定积分(a到x)f(t)dt=设f

设fx在[a,a+1)连续,则lim(x->a+)x^2/(x-a)定积分(a到x)f(t)dt=
设fx在[a,a+1)连续,则lim(x->a+)x^2/(x-a)定积分(a到x)f(t)dt=