I'm struggling to understand the geometrical/physical meaning behind why $\int_{v_0}^{v_f} dv=v$, especially how this relates to the area under the curve.
I was recently looking at the following simple derivation of kinematics equations and got stuck on step 3: \begin{equation} \frac{dv_x}{dt}=a_x \tag{1} \end{equation} \begin{equation} \int_{v_0}^{v_1}dv_x=\int_{t_0}^{t_1} a_x \, dt \tag{2} \end{equation} \begin{equation} v_x(t_1)-v_x(t_0)=\int_{t_0}^{t_f} a_x \, dt \tag{3} \end{equation}
I understand the mathematical reason (it is equivalent to $\int 1 \,dv$ and consequently, the antiderivative is $v$) and a reason having to do with adding up an infinite number of small velocity differences ($dv$) that add to the total velocity, but am struggling to make the connection between the area under the function $f(v)=1$ and velocity itself?
Pictures of area under the graphs (representing the two integrals)
In other words, I get how the area under the curve of the $a_x(t)$ graph represents velocity (you are multiplying $a_x$ by a small difference in time, which will yield velocity if you add it up), but how do we use this same logic to reason why the area under $f(y)=1$ (what does this function even represent?) results in our velocity?
Hopefully this wasn't too confusing and I am more than willing to provide extra clarification. Thank you so much for your help!