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If I have a function $f(t)$, and an indefinite integral of this function, $g(x) = \int f(t)\, dt$, is there any way I can express $g(x)$ as a definite integral whose bounds depend on $x$?

I thought I saw somewhere that I could express it as $$g(x) = \int_a^x f(t)\, dt$$ but I don't know what $a$ is. Is this representation correct? Or is there a better way to express $g$?

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  • $\begingroup$ The second representation is correct. The first one doesn't represent $g$ as a function of $x$. $\endgroup$ – user23793 Dec 30 '18 at 14:50
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That fixed but arbitrary lower limit is essentially the "arbitrary constant" you use when finding antiderivatives, because $$ g(x) = \int_a^x f(t)\, dt = \int_b^x f(t)\, dt + \int_a^b f(t)\, dt . $$ The second term in that sum is just a number.

The fundamental theorem of calculus says that both integrals with upper limit $x$ have derivative $f$ with respect to $x$.

The "indefinite integral" with no limits specified refers to the whole set of antiderivatives, not any particular one of them.

Edit in response to comment.

A loose common way to describe the indefinite integral is to say $$ \int f(t)\, dt = F(x) + c $$ for a function $F$ whose derivative is $f$ and any real number $c$. That's really a set of functions, one for each value of $c$.

In my answer you can take $$ F(x) = \int_b^x f(t)\, dt \quad \text{ and } \quad c = \int_a^b f(t)\, dt $$ to get the $g(x)$ in the question.

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  • $\begingroup$ I'm almost following what you're saying, but not quite. I understand that both integrals with upper limit $x$ have derivative $f$ with respect to $x$, but how does that make the constant integral the arbitrary constant? $\endgroup$ – clabe45 Dec 30 '18 at 16:52
  • $\begingroup$ @clabe45 See my edit. $\endgroup$ – Ethan Bolker Dec 30 '18 at 19:02
  • $\begingroup$ Ahh that makes more sense now. $\endgroup$ – clabe45 Dec 31 '18 at 2:17
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No, not always.

In fact, the definite integral with variable bounds can be and is strictly more restrictive than the indefinite integral, in the most general case where both can be taken as sensibly existing to their fullest extent that we can say one defines the same kind of function as the other.

While it is true that if we have two functions $F_1$ and $F_2$ defined by definite integrals of the same function with different bounds, i.e.

$$F_1(x) := \int_{a_1}^{x} f(t)\ dt\ \mathrm{and}\ F_2(x) := \int_{a_2}^{x} f(t)\ dt$$

with $a_1 \ne a_2$, then we will have that $F_1 - F_2$ is a constant function, the range of possible such constants may be limited. A simple counterexample is to take $f(x) := \sin(x)$. Now the integral from any lower bound $a$ to $x$ is just $\int_a^x \sin(t)\ dt = \cos(a) - \cos(x)$, but here $\cos(a)$ - the "constant" of our integration - can only range in $[-1, 1]$ (and the difference of any two such "constants" only within $[-2, 2]$), but the constant in $\int \sin(x)\ dx = -\cos(x) + C$ can be any real number whatsoever, even one far outside this range, say assign $C := \mbox{Graham's number}$. Thus the integrals with variable bounds are not exhaustive of the full family of antiderivatives, which is what is represented by the indefinite integral, and if we are really pedantic should be truly written as the set it is: in "reality", not in the informalities of school calculus,

$$\int f(x)\ dx = \left\{ F(x) + C : C \in \mathbb{R} \right\}$$

where $F'(x) = f(x)$. Actually, if we want to be really correct - and in fact this is important but only in cases where the definite integral technically doesn't make sense as evaluable between arbitrary pairs of points - we should take

$$\int f(x)\ dx = \left\{ F(x) \in \mathbb{R}^\mathbb{R} : \mbox{$F'$ exists and $F'(x) = f(x)$}\ \right\}$$.

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  • $\begingroup$ +1 I thought about going to this level of sophistication in my much shorter answer but decided the OP needed just the central points I made. Now they can see both views. $\endgroup$ – Ethan Bolker Dec 30 '18 at 15:12
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    $\begingroup$ Very pedantic answer that is 100% valid and makes a distinction I had never thought about. +1 from me too. Heck, I'd give it a +5 if it was possible/allowed. $\endgroup$ – JonathanZ Dec 30 '18 at 15:14
  • $\begingroup$ I'm having a hard time understanding this, as I'm new to Calculus, but it's ok. $\endgroup$ – clabe45 Dec 30 '18 at 18:16

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