Integral of directional derivative.

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I have just learned about a proof for why the gradient "shows the direction of steepest ascent" involving:

$\partial_{v_0} f(x_0) = (f \circ \gamma)^{\prime}(t_0) = \langle\nabla f(x_0), v_0\rangle$

with: $\gamma (t_0) = x_o, \gamma^{\prime}(t_0)=v_0$

which I now understand. A little later in the script, it is presented that:

$f(x_1) - f(x_0) = \int^b_a \langle \nabla f (\gamma(t)), \gamma^{\prime}(t)\rangle dt$.

for every $C^1$-curve $\gamma:[a,b] \rightarrow U \subset \mathbb{R}^n$ with $\gamma(a) = x_0, \gamma(b) = x_1$, which is without proof or definition. Does anybody have an intuition or explanation as to why this holds true or maybe just a link to where I can read more?

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By the chain rule, the derivative of $t \mapsto f(\gamma(t))$ is $$f^\prime(\gamma(t)) \circ \gamma^\prime(t)=\langle \nabla f (\gamma(t)), \gamma^{\prime}(t)\rangle$$

Which leads to the required result applying the Funtamental theorem of calculus.