B Justification for cancelling dx in an integral

[SamRoss]

TL;DR Summary

Confusion with cancelling two dx "terms" as if they were simple fractions.

In Paul Nahin's book Inside Interesting Integrals, on pg. 113, he writes the following line (actually he wrote a more complicated function inside the integral where I have simply written f(x))...

$\int_0^\phi \frac {d} {dx} f(x) dx = \int_0^\phi d{f(x)} = f(x) |_0^\phi$

It looks like the dx in the derivative symbol is cancelling with the dx from the integral, leaving only a $d$. I have always been somewhat confused when derivatives are treated as fractions. How is what is being done here justified? Also, what is the meaning of $df(x)$? Finally, how does one get from the middle expression to the final expression? It looks like an example of integration and differentiation being inverses of each other but I'm not sure.

 

[fresh_42]

In Paul Nahin's book Inside Interesting Integrals, on pg. 113, he writes the following line (actually he wrote a more complicated function inside the integral where I have simply written f(x))...

$\int_0^\phi \frac {d} {dx} f(x) dx = \int_0^\phi d{f(x)} = f(x) |_0^\phi$

It looks like the dx in the derivative symbol is cancelling with the dx from the integral, leaving only a $d$. I have always been somewhat confused when derivatives are treated as fractions. How is what is being done here justified? Also, what is the meaning of $df(x)$? Finally, how does one get from the middle expression to the final expression? It looks like an example of integration and differentiation being inverses of each other but I'm not sure.

A formal proof why this solppiness can be done is a bit of work to do as it involves two limits (Riemann integration and differential quotient).

The question about $df(x)$ is easier: Just set $y=f(x)$ then $\int df(x) =\int dy = \int 1\cdot dy = y +C= f(x)+C$.

 

[SamRoss]

In Paul Nahin's book Inside Interesting Integrals, on pg. 113, he writes the following line (actually he wrote a more complicated function inside the integral where I have simply written f(x))...

$\int_0^\phi \frac {d} {dx} f(x) dx = \int_0^\phi d{f(x)} = f(x) |_0^\phi$

It looks like the dx in the derivative symbol is cancelling with the dx from the integral, leaving only a $d$. I have always been somewhat confused when derivatives are treated as fractions. How is what is being done here justified? Also, what is the meaning of $df(x)$? Finally, how does one get from the middle expression to the final expression? It looks like an example of integration and differentiation being inverses of each other but I'm not sure.

Now that I think about it, does the second expression follow from the first simply by the definition of a differential, which I think is $df = f'(x)dx$ ? In that case, can the second expression be skipped altogether so we can just go from the first to the last by the first fundamental theorem of calculus?

 

[fresh_42]

Now that I think about it, does the second expression follow from the first simply by the definition of a differential, which I think is $df = f'(x)dx$ ? In that case, can the second expression be skipped altogether so we can just go from the first to the last by the first fundamental theorem of calculus?

Yes, that's an idea, although it would only be a rewording.

 

[WWGD]

If you want to read a bit more, lookup differential forms, closed and exact forms. d is the differential operator; ( somewhat confusing), the differential of a differentiable function is a differential form.