# 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

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• 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

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