How to show $f(x) = f(y) + \int\limits_0^1 \nabla f(y + t(x-y))^T(x-y)dt$?

Question

I am familiar with the following result.

Suppose $f$ is continuously differentiable, then there exists $t \in (0,1)$ such that $$f(x) = f(y) + \nabla f(y + t(x-y))^T(x-y)$$

However, I recently read an article (Page 267 of Text, Page 40 of PDF) whereby the author seemingly used the following result,

$$f(x) = f(y) + \int\limits_0^1 \nabla f(y + t(x-y))^T(x-y)dt$$

It is not apparently how $$\int\limits_0^1 \nabla f(y + t(x-y))^T(x-y)dt = \nabla f(y + t(x-y))^T(x-y)$$

Does anyone know how to show $f(x) = f(y) + \int\limits_0^1 \nabla f(y + t(x-y))^T(x-y)dt$?

Answer 1

Let $g(t) = f(y + t(x-y))$. Then by the chain rule, $g'(t) = \nabla f(y + t(x-y))^\top (x-y)$. The statement is simply the fundamental theorem of calculus: $$g(1) - g(0) = \int_0^1 g'(t) \, dt.$$

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The other result is the mean value theorem: there exists $t \in (0,1)$ such that $$\frac{g(1) - g(0)}{1-0} = g'(t).$$