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(edited 7 years ago)
Reply 1
Avatar for Zacken
Zacken
22
Original post by cloisters


I hope you understand my confusion and any help would be appreciated. :smile:


Not entirely sure what's confusing you, sorry!

But: AA is defined as ∫abf(x) dx=lim⁡δx→0∑n=0N−1f(xn)δx\displaystyle \int_a^b f(x) \, \mathrm{d}x = \lim_{\delta x \to 0} \sum_{n=0}^{N-1} f(x_n) \delta x (i.e: summing up small rectangles underneath the curve and taking the width tending to 0). So xnx_n are NN equidistant points spaced throughout [a,b][a,b] and δx=xn−xn−1\delta x = x_n - x_{n-1}.


If your confusion is the indefinite integral, then it is just ∫f(x) dx=∫xf(t) dt\int f(x) \, \mathrm{d}x = \int^x f(t) \, \mathrm{d}t where the bottom limit is arbitrary, but it still comes down to area underneath a curve f(t)f(t) from a general starting point to a given point xx.
Reply 2
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Zacken
22
Original post by cloisters
Hi, sorry I should have made myself clearer: the latter is what is confusing me, in other words what does an indefinite integral represent? I can see that ∫abydx\int_a^b y dx is the area under the curve from aa to bb: what is ∫ydx\int y dx? I have tried to understand your post but I have some questions: if it is the lower bound is arbitrary for an indefinite integral then wouldn't ∫ydx=inf⁡\int ydx=\inf (infinitely large) for say y=e−xy=e^{-x}?

Thanks


Arbitrary just means that it doesn't matter. So if you want define ∫y dx=∫axf(t) dt\int y \, \mathrm{d}x = \int_a^x f(t) \, \mathrm{d}t, i.e: the integral from any point you want aa to a general point xx.

Not sure what you mean by the infinitely large bit, we're not quite working with improper integrals just yet, we're trying to wrap our heads around normal integrals, so the lower arbitrary bound has to be a real number (so not infinity) which means that ∫ydx\int y dx will never be infinitely large.
Reply 3
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Zacken
22
Original post by cloisters
Sorry, just being honest here but I still don't get it: wouldn't ∫axf(x)dx=f(a,x)\int_a^x f(x) dx=f(a,x)? I do sort of understand what you mean by ∫f(x)dx\int f(x) dx being the area under the curve from an arbitrary point to xx but I'm completely failing to connect this to the fact that ∫f(x)dx\int f(x)dx is a one-variable function.


Yes, indeed! Which is precisely why when you integrate a function say f(x)=xf(x) = x you don't just get something involving xx, you also get an arbitrary constant - which appears precisely because of that arbitrary lower bound. :smile:

So, by the definition: ∫x dx=∫axt dt=t22]ax=x22−a22\int x \, \mathrm{d}x = \int_a^x t \, \mathrm{d}t = \frac{t^2}{2} \bigg]_a^x = \frac{x^2}{2} - \frac{a^2}{2}, but the latter is just an arbitrary constant, so the same thing as x22+c\frac{x^2}{2} + c.

Although it's a bit weird writing f(a,x)f(a,x) since aa is a constant. Kind of like referring to x2+1=f(x,1)x^2 + 1 = f(x,1).
Reply 4
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Zacken
22
Original post by cloisters
Ah completely forgot about the constant - should've followed cgp's advice and got a +C stamp lol. But very interesting, I never thought of +C as being related to the indefinite integral like that although it makes complete sense in hindsight. Thank you very much! :smile:


You're welcome!

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