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!
how calculate integral $\int \dot{x} \; dx$
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1@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 $t$. – 2012-02-01
2 Answers
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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0@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. – 2012-02-02
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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1Can someone down vote this answer, it is wrong – 2012-02-01