Question about integration shouldn't be too difficult

  • #1

Main Question or Discussion Point

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.
[/itex]

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?
 

Answers and Replies

  • #2
287
0
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.
 
  • #3
316
0
Also known as http://en.wikipedia.org/wiki/Integration_by_substitution" [Broken].
 
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