Show that if $f: D\to \mathbb R^m$ is $\mathcal C^1$, then for $x,y \in D$ we have that $f(x)-f(y)=\int_0^1 df(tx+(1-t)y)\cdot (x-y)\ dt.$

Question

Let $D\subset \mathbb R^n$ be a bounded convex set. Show that if $f: D\to \mathbb R^m$ is $\mathcal C^1$, then for $x,y \in D$ we have that $$f(x)-f(y)=\int_0^1 df(tx+(1-t)y)\cdot (x-y)\ dt.$$

Any hints on to get started on?

Answer 1

Hint:$$f(x) - f(y) = f(tx + (1-t)y)\bigg|_{t = 0}^1 = \int_0^1 \frac{d}{dt}[f(tx +(1-t)y)]\, dt = \cdots$$

Answer 2

$H:[0,1]\rightarrow \mathbb{R}^n$ by $H(t)=f(tx+(1-t)y)$, $H$ is differentiable and $H'(t)=df.(tx+(1-t)y).(x-y)$, $H(1)-H(0)=\int_0^1H'(t)dt=f(x)-f(y)=\int_0^1df.(tx+(1-t)y).(x-y)$.