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## Homework Statement

Calculate ##\lim_{x\to 1} \frac{1}{x-1} \int_{1}^{f(x)} sin(\pi t^2) dt##. f is differentiable in the neighbourhood of point ##x=1## and ##f(1)=1##.

## Homework Equations

If ##f## is continuous on a closed interval ##[a,b]##, then there exists ##ξ∈]a,b[## such that

##\int_{a}^{b} f(x)dx=f(ξ)(b-a)##.

( First Mean Value Theorem for Integrals)

## The Attempt at a Solution

I think the trick is to use the First Mean Value Theorem for Integrals. Obviously ##\sin(\pi t^2)## is continuous everywhere and so for a fixed ##x≠1##; ##\frac{1}{x-1} \int_{1}^{x} sin(\pi t^2) dt=\frac{1}{x-1}(x-1)\sin(\pi ξ^2)=\sin(\pi ξ^2)##. Therefore ##\lim_{x\to 1} \frac{1}{x-1} \int_{1}^{x} sin(\pi t^2) dt=\lim_{x\to 1} sin(\pi ξ^2)=0 ##

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