Question about integration shouldn't be too difficult

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The discussion focuses on performing integration with respect to a variable that is dependent on time, specifically transforming the integral of a function f(φ(t)) into a new form g(φ(t)) (dφ/dt) dt. The participants confirm that it is valid to cancel dt and integrate with respect to φ, provided the limits of integration are adjusted accordingly. An example is provided, illustrating the calculation of mass M for a rod of length L with uniform mass density D, demonstrating the application of integration by substitution.

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  • Knowledge of functions and their dependencies, particularly in relation to time.
  • Basic concepts of mass density and its mathematical representation.
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Really, I should know the answer to this, but...

Suppose I'm trying to perform an integration with respect to [itex]t[/itex]:

[tex] \int_0^T f(\phi(t)) dt[/tex]

So my function [itex]f[/itex] is explicitly a function of [itex]\phi[/itex], and [itex]\phi[/itex] depends on time [itex]t[/itex]. But then suppose I end up being able to write the integral as

[tex] \int_0^T g(\phi(t)) \frac{d \phi}{dt} dt.<br /> [/itex]<br /> <br /> Can I just cancel the [itex]dt[/itex] and perform an integral with respect to [itex]\phi[/itex]? If so, I need to change the limits of integration, right?[/tex]
 
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Yes, this is my understanding.

Example: Find the mass of a rod of length L and uniform mass density D.

M = integral of dm

D = dm/dx

so D dx/dm = 1 and

M = integral of (1 dm) = integral of (D dx/dm dm) = integral of(D dx)

Since you're now working in distance-space, you just switch the limit to the distance-space limit, namely, 0 -> L.
 
Also known as http://en.wikipedia.org/wiki/Integration_by_substitution" .
 
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