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I'm currently studying Calculus from Stewart's book, and for the The Fundamental Theorem of Calculus Pt. 1, he defined a function $g(x) = \int_0^x f(t) dt$ which represented the area under $f(x)$ from $0$ up to $x$ and proved that $g(x)$ is the antiderivative of $f(x)$ and, in this case, if I plugged in an $x$ for $g(x)$, it would give me the area under the curve from $0$ til that $x$ since $g(x) = \int_0^x f(t) dt$

However, for an arbitrary function $f(x)$, if I found the antiderivative and plugged in an $x$, it would give me the area under the curve of $f(x)$ from which point upto $x$?

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It depends on your choice of antiderivative.

In what you write, $g(x)$ is an antiderivative of $f(x)$. Indeed, any function of the form $g(x)+c$, where $c$ is a constant would still be an antiderivative for $f$.

Your $g(x)$ is that one antiderivative that satisfies the relation $g(0)=0$. There are infinitely many choices you can make, but once you require an antiderivative $h(x)$ to satisfy $h(x)=c'$, where $c'$ is a constant, you get a unique function.

Also, from this you know that $h(x)=g(x)+c'$. So, a priori, the antiderivative evaluated at a point $x_0$ does not have a natural interpretation as an area.

I think I am having troubles explaining myself on this, so if you any question that could help me clarify this to you, please ask.

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  • $\begingroup$ Your answer would be much more helpful to OP if you could add a diagram to illustrate your point. $\endgroup$ – User2956 Nov 15 '15 at 3:40
  • $\begingroup$ @Pulzz I don't know what you mean by "diagram" in this context. $\endgroup$ – Silvia Ghinassi Nov 15 '15 at 3:48
  • $\begingroup$ I was thinking maybe a diagram(s) showing different anti-derivatives of a function that differ by a constant and the area under the f(x) curve that each of them represent. $\endgroup$ – User2956 Nov 15 '15 at 4:38
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    $\begingroup$ But that would be misleading; as I said one should not think of antiderivatives as areas. $\endgroup$ – Silvia Ghinassi Nov 15 '15 at 16:10
  • $\begingroup$ Yes, but an antiderivative + C gives a unique function that represents the area. And showing antiderivatives with different constants and the area they represent might help $\endgroup$ – User2956 Nov 16 '15 at 17:47

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