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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.

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  • $\begingroup$ On a related note, simplify $\frac d {dx}(\frac 1 2 v^2)$ $\endgroup$
    – PM 2Ring
    Commented Feb 3, 2019 at 9:02
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    $\begingroup$ $(Vdv) $ is not equal to $(x+c)$ $\endgroup$
    – SmarthBansal
    Commented Feb 3, 2019 at 12:01

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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

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

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  • $\begingroup$ I see, just to clarify: should we never shift variables when integrating? In your example it seems algebraically possible $\endgroup$
    – Frinko
    Commented Feb 3, 2019 at 9:35
  • $\begingroup$ 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 $\endgroup$
    – B. Brekke
    Commented Feb 3, 2019 at 9:42

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