What does the area under $p=mv$ mean? The kinetic energy equation
$E_k=\frac{1}{2}mv^2$
seems to fit as an integral of the momentum equation
$p=mv$
But i can't figure out how to relate the meaning of the area under the momentum equation line to the meaning of energy.
 A: That's not actually the correct definition of kinetic energy. You're describing kinetic energy as
$$ K = \int p \operatorname{d}v.$$
The actual definition is
$$ K = \int v \operatorname{d}p.$$
Since they both yield $K = \frac{1}{2}mv^2$, it may seem like a distinction without a difference. When you get into relativity, though, the second one gives you the correct relativistic kinetic energy
$$K = \sqrt{(mc^2)^2 + (pc)^2} - mc^2 = mc^2 \left[\frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} - 1\right].$$
The definition of kinetic energy comes from the work-energy theorem. Start from Newton's second law
$$\vec{F} = \frac{\operatorname{d}\vec{p}}{\operatorname{d}t}. $$
Dot both sides with $\vec{v}$ and integrate $\operatorname{d}t$  to get
$$\int \vec{F}\cdot \operatorname{d}\vec{x} = \int \vec{v}\cdot \frac{\operatorname{d}\vec{p}}{\operatorname{d} t} \operatorname{d}t.$$
On the left hand side I used $\vec{v} = \frac{\operatorname{d}\vec{x}}{\operatorname{d} t}$ to change it into a path integral. A similar trick works on the right hand side to get $K = \int \vec{v}\cdot \operatorname{d}\vec{p}$.
