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    (Original post by cloisters)

    I hope you understand my confusion and any help would be appreciated.
    Not entirely sure what's confusing you, sorry!

    But: A is defined as \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 x_n are N equidistant points spaced throughout [a,b] and \delta x = x_n - x_{n-1}.

    If your confusion is the indefinite integral, then it is just \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) from a general starting point to a given point x.

    (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 \int_a^b y dx is the area under the curve from a to b: what is \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 \int ydx=\inf (infinitely large) for say y=e^{-x}?

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

    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 \int y dx will never be infinitely large.

    (Original post by cloisters)
    Sorry, just being honest here but I still don't get it: wouldn't \int_a^x f(x) dx=f(a,x)? I do sort of understand what you mean by \int f(x) dx being the area under the curve from an arbitrary point to x but I'm completely failing to connect this to the fact that \int f(x)dx is a one-variable function.
    Yes, indeed! Which is precisely why when you integrate a function say f(x) = x you don't just get something involving x, you also get an arbitrary constant - which appears precisely because of that arbitrary lower bound.

    So, by the definition: \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 \frac{x^2}{2} + c.

    Although it's a bit weird writing f(a,x) since a is a constant. Kind of like referring to x^2 + 1 = f(x,1).

    (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!
    You're welcome!
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