# If kinetic energy is mass times the integral of velocity, isn't it just a product of mass times distance? [closed]

I'm still learning Calculus at the moment and I'm currently on integration. The moment I realized the "$$1/2$$" and square value in $$v^2$$ are just products of integration, can't one just use integrated $$v$$, assume $$m$$ is a constant, and hence say $$KE$$ is really just mass multiplied by its position? e.g. $$KE = m * (x + C)$$?

I know something's not right, after all $$KE$$ is the energy of a moving mass, but I'd like to know of other reasons why this won't work too.

## closed as unclear what you're asking by user191954, niels nielsen, ZeroTheHero, Buzz, Kyle KanosFeb 4 at 10:59

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• On a related note, simplify $\frac d {dx}(\frac 1 2 v^2)$ – PM 2Ring Feb 3 at 9:02
• $(Vdv)$ is not equal to $(x+c)$ – SmarthBansal Feb 3 at 12:01

## 1 Answer

Watch out for which variable you are integrating in!

$$W=\int \vec{F}\cdot d\vec{x}$$

$$W=m\int\vec{a}\cdot d\vec{x}$$

$$W=m\int \frac{d\vec{v}}{dt}\cdot d\vec{x}$$

$$W=m\int \frac{d\vec{v}}{dt}\cdot \frac{d\vec{x}}{dt} dt$$

$$W=m\int \vec{v}\cdot d\vec{v}$$

This is where the kinetic energy is just the integral of the velocity. Note that the integration is in the variable $$v$$. I believe the wrong result comes from doing the integration

$$W=m\int \frac{d\vec{x}}{dt} dt$$

but this is wrong, we should not integrate in the variable $$t$$.

Choosing the correct integral we obtain as expected

$$W=\frac{1}{2}mv^2$$

• I see, just to clarify: should we never shift variables when integrating? In your example it seems algebraically possible – Frinko Feb 3 at 9:35
• I'm not sure I understand what you mean. You can alway change the integration variable. dx=du*dx/du, but you have to remember the factor dx/du for the integrals to be the same – B. Brekke Feb 3 at 9:42