Wednesday, 29 October 2014

calculus - $dx$ being a desginator (with respect to $x$) or being a term?

I am confused as to what $dx$ truly is. I am doing some u-substitution problems and this is what I came across:



$$\int 2x(x-1)^{1/2}\,dx$$




$u=x-1$ and therefore $du=1$



when we substitute we get:
$$2 \int (u^{3/2}+u^{1/2}) \, du$$
(here the du simply replaces the dx because our variable changed)



In another example:
$$\int 4x^5(x^2+1)^{1/3} \, dx$$
$$u=x^3+1$$

$$du=3x^2$$
therefore it becomes:
$$\int 4(u-1)(du/3)(u)^{1/3}\,$$
-here my teacher didn't put $d$x at the end, she just left it off
So my question is this: why is it that sometimes $dx$ and $du$ are treated as values that can be multiplied to other terms in the integrand and sometimes they are simply treated as a command (do "blank" with respect to $x$, or $u$ or whatever is used)?

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