Let me begin with an example. If we were to integrate the indefinite integral of $(2x)^2$ with respect to $x$ with u-substitution, we would first say that $u=2x$ and therefore $du=2dx$. In order to substitute $du$ in for $dx$, we have to multiply the inside of the integral by $2$ and the outside by $0.5$. Then we have the indefinite integral of $u^2 du$. We say this equals $(u^3)/6 + C$ which gives us a final answer of $(4x^3)/3 + C$.
I'm able to do u-substitution, but I am confused as to one aspect of it. In the above example the only way we can substitute in $du$ is if the $dx$ and $(2x)^2$ in the original integral were being multiplied. This is what confuses me: why are $f(x)$ and $dx$ being multiplied together in the indefinite integral.
Now, I understand why in a definite integral $dx$ and $f(x)$ are multiplied. I believe it is because the definition of the definite integral from $a$ to $b$ is the limit as "$n$" goes to infinity of the sum of "$n$" rectangles under the curve, each with area $f(x)$ times $dx$. Notice in the definition $f(x)$ and $dx$ are multiplied so in a definite integral $f(x)$ and $dx$ should also be multiplied.
However, why in the indefinite integral are $f(x)$ and $dx$ multiplied?
Note: I had origianlly asked this question on "Questions about U-Substitution and Integration" but I felt my original question there was not accurately portraying what I was asking.
Thank you for helping.