In the 10th grade, meaning few months back, I've studied the potential and Kinetic energy (I was ignorant about the importance of calculus), but When I learned calculus, and the Constant of integration which is the initial conditions in physics. I remembered that also in the expression of potential energy we have : $$E_p=mgz+c$$ Where $c=0$ at $z_0$, this is the antiderivative of $mg$, obviously $$\int mg dz=mgz+C$$ and that's the expression of the energy also in the Kinetic one : $$E_{K}=\frac{1}2 mv^2$$
And when we differentiate it we get : $$\frac{dE_K}{dv} =mv=p$$Where $p$ is the momentum.

My question is: How this operation 'Integration' can produce the 'Energy' expression?

  • 2
    $\begingroup$ Does this answer your question? Why is the potential energy equal to the negative integral of a force? $\endgroup$
    – user258881
    Jul 23, 2020 at 10:04
  • $\begingroup$ The question, which is linked as "containing the answer", focusses on the sign, while this question focusses on the integration. Reading the linked answers I don't feel that the OPs questions has been answered. $\endgroup$
    – Semoi
    Jul 23, 2020 at 17:01

5 Answers 5


Integration and differentiation are always joint in physical laws. So, we can

  • either say that energy (work) is the integration of the force along the taken path, $E = \int F \cdot dx$ -- which tells us that the energy is given by the area under "the force curve",
  • or we state that the force is the derivative of work with respect to position, $F = \frac{dE}{dx}$ -- which tells us that the tangent to "the energy curve" yields the instantaneous force.

In physics we have such definitions all over the place -- e.g. check the
relationships between position, velocity, and acceleration, \begin{align} a &= \frac{dv}{dt} \hspace{1em} \Leftrightarrow \hspace{1em} v = \int a dt \\ v &= \frac{ds}{dt} \hspace{1em} \Leftrightarrow \hspace{1em} s = \int v dt \end{align} Personally, I prefer the differential form, because to me it is more natural to ask "how does one quantity change if I change another quantity a little bit?". However, in the end these are just definitions used to describe nature.


The reason that bodies feel a force in a particular direction is because the potential energy of the system is reduced when the bodies move in that direction.

Example, a ball will roll down a hill because the lowest potential energy state for the ball is at the bottom of the hill ($U=mgh$ is minimized by reducing $h$).

In a one dimensional problem, this can be stated mathematically as $F = -\frac{d}{dx} U(x)$ or by integrating both sides

$$ \int \limits_{x_0}^{x_f} F dx = -\left[U(x_f) - U(x_0)\right] = -\Delta U. $$

Again using the example of a body near the surface of the earth, feeling a roughly constant force $mg$.

$$\Delta U = -\int \limits_{h_0}^{h_f} mg ~dh = -mg\Delta h.$$

In higher dimensions, the same concept holds, but now $\vec{F} = -\nabla U(\vec{x})$, and we must do a integral over some path in order to recover the change in $U$.

So the reason that "integrals of force 'give energy' " is precisely because forces arise from local potential energy differences (at least classical conservative forces, the type we typically study in physics). In fact, you can think of forces as another view into the same information as the potential energy function.


Here's an attempt to explain this :

Like all the energies, the potential energy is defined with its variation with respect to $d\vec{z}$ thus the following expression : $$dU_0=-\vec{P}.d\vec{z}$$ Check here for a full explanation.

For the constant, as you said it depends on the initial conditions. In physics 'the integration constant' is frequently depends on it, here's an example : $$\boldsymbol{v}=\int \boldsymbol{a}\ dt=\boldsymbol{a}t+c$$ And $c$ can be determined, putting $t=0$ we get instantly $c=\boldsymbol{v}_0$

For the kinetic energy, I highly recommend you to research the 'vis viva' introduced by Leibniz, it's kind of a historatical research, but I gave you the Wikipedia link because it involves an overview about it.

Good luck


I'm not exactly sure what you are asking but I'll take a crack at it:

Work, W, done by a force, F, moving an object is given by:

$$ W = \int Fdx$$

Using $F = ma = m\frac{dv}{dt}$ we have

$$ W = \int m\frac{dv}{dt}dx $$

As $v = \frac{dx}{dt}$,we have

$$ W = \int m v dv = \frac12 mv^2$$


when u take the Integral it is with respect to position, when you to the derivative it was in respect to velocity ,, dz vs dv this is what differentiates the 2 different forms of energy one with the respect to an outside force, and the internally contained energy contained by the object , with respect to its "perceived" velocity