$\begin{aligned} & \text { Let } I=\int_0^1 f(k-1+x) d x \\ & \Rightarrow \quad I=\int_{k-1}^k f(t) d t, \text { where } t=k-1+x \Rightarrow \quad I=\int_{k-1}^k f(x) d x \\ & \therefore \sum_{k=1}^n \int_{k-1}^k f(x) d x=\int_0^1 f(x) d x+\int_1^2 f(x) d x+\ldots .+\int_{n-1}^n f(x) d x \\ & =\int_0^n f(x) d x\end{aligned}$