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Let $x$ depend on $t$. $\dot{x}$ is derivative $x$ over $t$. I want to calculate the integral $\int \dot{x} \; dx$. I asked similar question about differentiation here. Any thoughts and ideas are appreciated. Thank you!

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  • $\begingroup$ The case $\int \dot x dt$ is easier. $\endgroup$ – Jon Feb 1 '12 at 16:55
  • $\begingroup$ well, $\int \dot{x} dx = \int \dot{x}^2 dt$ $\endgroup$ – ashim Feb 1 '12 at 17:06
  • $\begingroup$ According to W.A. this integral has no solution in terms of elementary functions... $\endgroup$ – Peđa Terzić Feb 1 '12 at 17:09
  • $\begingroup$ @pedja: That was I meant. $\endgroup$ – Jon Feb 1 '12 at 17:14
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    $\begingroup$ @MichaelHardy capuloca's comment is saying that a sum of infinitesimal ratios $\frac{dx}{dt}\cdot dx$ is the same as a sum of infinitesimal ratios $\frac{dx}{dt}\frac{dx}{dt}\cdot dt$. And it is helpful for calculation if you have $x$ explictly in terms of $t$. $\endgroup$ – alex.jordan Feb 1 '12 at 22:16
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The integral $\int \dot x dx$ cannot be evaluated explicitely unless the form of the function $x(t)$ is also given. This can be easily understood in the following way

$$\int \dot x dx=\int (\dot x)^2 dt$$

that cannot be furtherly explicited. This kind of computations generally come out from studies on mechanics with dissipative systems. If you have a differential equation like

$$\ddot x=-\dot x+F(x)$$

you can multiply both sides by $\dot x$ and integrating obtain

$$\int dt \frac{d}{dt}\left(\frac{{\dot x}^2}{2}\right)-\int dx F(x)=-\int \dot x dx$$

that cannot be reduced anymore even if lhs can be expressed through an energy integral.

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  • $\begingroup$ can you explain more about the last equation, plz. Why from that you conclude that there is no solution? Thanks $\endgroup$ – ashim Feb 2 '12 at 3:18
  • $\begingroup$ @capoluca: I am not implying that there is no solution but that, in order to evaluate your integral, you will need to know $x(t)$ explicitly, solving the original ode. In this way you are able to evaluate the integral on the rhs that is your problem. $\endgroup$ – Jon Feb 2 '12 at 7:35
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You can write it as $\int \frac{dx}{dt}dx$ which, assuming appropriate smoothness conditions on $\dot{x}$ is the same as $\frac{d}{dt}\int x dx = \frac{d}{dt} (\frac{x^2}{2} + C) = x\dot{x}$

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  • $\begingroup$ could explain details plz, why can we put derivation over t outside of the integral? and by the way you don't have constant in the answer, however, you should have one. $\endgroup$ – ashim Feb 1 '12 at 16:57
  • $\begingroup$ Moreover, if you take derivative of your answer you would not get $\dot{x}$ $\endgroup$ – ashim Feb 1 '12 at 17:01
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    $\begingroup$ Can someone down vote this answer, it is wrong $\endgroup$ – ashim Feb 1 '12 at 17:19

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