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To give some context: This integral occurs in the solution of the homogenous $1+1$ dimensional wave equation with certain boundary conditions. $x$ represents the space coordinate, $t$ the time coordinate and $c>0$. I have the following seemingly trivial question. I'm a bit unsure about the following integral

$$F(x,t) = \int_{x-ct}^{x+ct} f(y) dy$$

Is it that simple?

$$\int_{x-ct}^{x+ct} f(y) dy = F(x+ct) - F(x-ct)$$

where $F'(\xi) = f(\xi)$. I'm unsure, since the boundaries are functions $g_1(x,t)$ and $g_2(x,t)$. Is the following application of the fundamental theorem of diff. and int. true?

$$\int_{g_1(x,t)}^{g_2(x,t)} f(y) dy \overset{\hat{}}{=} \int_{a}^{b} f(y) dy = F(b) - F(a) = F(g_2(x,t)) - F(g_1(x,t))$$

My argument would be that with respect to the integration variable $y$ the boundaries are constants, so $y \in [x-ct,x+ct] \overset{\hat{}}{=} [a,b]$. What unsettles me a bit is, that we get a function depending on both $x$ and $t$ out of this integral. I mean, from a physical point of view, $x\pm ct$ is "just" a space coordinate, i.e. has the dimension of space, so $F$ actually depends only on the space coordinate of that form, i.e. $F(x,t) \equiv F(x(t))$, right? Still not entirely sure about this.

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I think it is a mistake to write the statement of the Fundamental Theorem in the usual way, using the symbol $F$. My concern is that $F$ already has a (possibly different) meaning defined in the problem statement. If two functions are given the same name, it is all too easy to mistakenly try to equate them, which is an error I think you fell into.

Since the letters we use to label the functions in the Fundamental Theorem are arbitrary, let's just choose different names, as follows:

$$ \int_{a}^{b} g(y) dy = G(b) - G(a), $$

where $G'(y) = g(y)$.

Now it is clear enough that $f(y)$ in the formula

$$F(x,t) = \int_{x-ct}^{x+ct} f(y) dy$$

corresponds to $g(y)$ in our (relabeled) version of the FTC. So we need to correspond the $G$ in our FTC to some function such that $G'(y) = f(y)$ in our particular problem. That is,

$$ G(y) = \int f(y) dy, $$

setting $G(y)$ equal to some antiderivative of $f$. (The constant of integration is not required here, because any single antiderivative of $f$ is sufficient for our needs; we do not need the ability to produce every possible antiderivative, as we would for the general solution of an indefinite integral.)

The FTC then tells us that

$$\int_{x-ct}^{x+ct} f(y) dy = G(x+ct) - G(x-ct).$$

Plugging this back into the definition of $F(x,t)$, we see that

$$F(x,t) = G(x+ct) - G(x-ct).$$

That's it: the definition of the desired function as a function of two variables, $x$ and $t$. Unless there is some additional constraint not mentioned in the question, You can hold $x$ constant and vary $t$, and this may cause the value of $F(x,t)$ to vary, so we cannot say just yet that $F(x,t)$ can be expressed as a function of $x$ alone.

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  • $\begingroup$ Thank you, sir! Yes, sometimes one can lose track. :-) $\endgroup$ – root Jan 9 '16 at 19:55

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