Why is there a "dx" in indefinite integrals?

[vaclav_gregor]

Hello. I don't really understand why there is a dx in indefinite integrals. If it was just a matter of notation, I wouldn't care, but since we can substitute different things instead of dx for example when doing u-substitution, I wanted to know what does it mean. Why are we allowed to do all these things with the dx? In definite integrals, it makes sence, because they are like a sum of infinitely small rectangles with height f(x) and width dx, so I guess we can substitute different things instead of dx, because the rectangles are going to be the same. But with indefinite integrals, we are not calculating area. I hope you understand my question, my english is not very good. If it's not clear, please tell me and I will try to explain it better. Thanks for your answers

 

[tkhunny]

If the context is clear, you don't need one. It's not actually a thing. It might be function of sorts.

Try this.

\(\displaystyle \int\;2x\; dx = \int\;d\left(x^{2}\right) = x^2 + C\)

Try that again?

\(\displaystyle \int 2x^{3}\;dx = \int x^{2}\;d\left(x^{2}\right) = \dfrac{\left(x^{2}\right)^{2}}{2} + C = \dfrac{x^{4}}{2} + C\)

You may recognize that as an alternate notation for a u-substitution.

Or, more standardly, when 'x' is the only variable we're talking about:

\(\displaystyle \int 2x = x^{2}+C\)

A little more unusual

\(\displaystyle \int \; dx \; 2x = x^{2} + C\)

And, this collection wouldn't be nearly as fun without:

\(\displaystyle \int \; 2x\;dy = 2xy + C(x)\)

Don't get too hung up on it. Let the notation help you. If you find it confusing you, perhaps you're taking it too seriously.

How about Integration by Parts?

\(\displaystyle \int log(x)\;dx = x\cdot \;log(x) - \int x\; d\left(log(x)\right) = x\cdot log(x) - \int x\cdot \dfrac{dx}{x} = x\cdot log(x) - \int \;dx = x\cdot log(x) - x + C\)

It may not need to be nearly as rigid as you are trying to define it.

 

[vaclav_gregor]

If the context is clear, you don't need one. It's not actually a thing. It might be function of sorts.

Try this.

\(\displaystyle \int\;2x\; dx = \int\;d\left(x^{2}\right) = x^2 + C\)

I don't understand how you got this: \(\displaystyle \int\;d\left(x^{2}\right) = x^2 + C\). If we take the derivative of \(\displaystyle x^2 + C \), we should get the function that is in the integral expression, but if we take the derivative of \(\displaystyle x^2 \), we get \(\displaystyle dx^2/dx = 2x \) and neither of those is equal to the expression \(\displaystyle d(x^2) \) - which is in the integral expression. I know this is probably really obvious to you, but I'm bit slower, so if you could explain it bit more it would help me a lot. Anyone thank you for your original answer.

 

[tkhunny]

Typically, \(\displaystyle d\left(x^{2}\right)\) is taken to mean \(\displaystyle 2x\;dx\)

If you were to define, \(\displaystyle u = x^{2}\), what would you produce for \(\displaystyle du\)?

Some things can be just for notational convenience. Personally, I always hated drawing a separate tableau for u-substitution, but especially for Integration By Parts. This avoids the separate demonstration.

Keep in mind that there may be other definitions of integration, besides little quadrilaterals.