- #36
pwsnafu
Science Advisor
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Tim77 said:In this case using the wrong procedure leads to the correct answer but you technically can't pull the function out of the integral.
This is false. You can pull constant functions out.
Let ##r \in \mathbb{R}## then define ##f_r (x) = r## and so
##\int f_r(x) \, dx = \int f_r(x) \cdot f_1(x) \, dx = \int r \cdot f_1(x) \, dx = r \int f_1(x) \, dx = f_r(x) \int f_1(x) \, dx##
QED.
You can't do it with non-constant functions, but this thread is not about non-constants and everything you have written about non-constant functions is irrelevant to this thread.
The form f(x)dx is required for integration to have meaning you can't separate it even if f(x)= a and especially if f(x)=0
This is only true for differential forms. For other areas of analysis the dx is nothing more than notation. And even in differential forms ##\int \omega## is a common notation.
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