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I am reading notes for my Calculus 2 class and am confused why these two functions are not equivalent " $x$ is not a dummy variable, for example, $$\int 2xdx = x^2+C$$ and $$\int 2tdt = t^2 + C$$ are functions of different variables, so they are not equal. " Is it because the arbitrary C's can be different?

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  • $\begingroup$ What do you mean by "equivalent"? $\endgroup$ – Future Jan 11 '16 at 2:26
  • $\begingroup$ In the book it just says "they are not equal". Is equivalent the wrong word to associate with this? $\endgroup$ – Biomed Boy Jan 11 '16 at 2:30
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    $\begingroup$ They can be equal, assuming that $x=t$ and, for come arbitrary constant $c$ within the domain of both the functions share the same value. What this means is that if the first function is $f(x)$ and the second is $g(t)$, then $f(c)=g(c)$. This second condition is often known as the "initial condition". In general the two expression are not equal though $\endgroup$ – Brevan Ellefsen Jan 11 '16 at 2:59
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    $\begingroup$ @RyanWisth With Calc 2 as context I would say that the author is more likely referencing the second condition, that the initial conditions are not the same. The two are note necessarily equal though... for example, $x$ may be a vector function of $s$ and $r$ and $t$ may be a vector function of $p$ and $q$. This is more Calc 3 though, and I think for your purposes this is irrelevant. What context did you find this in? (Note that an example where these could be considered the same is where the first is the function $f(x)$ and the second is simply a substitution with $x=t$ for all $x$ and $t$. ) $\endgroup$ – Brevan Ellefsen Jan 11 '16 at 3:20
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    $\begingroup$ @RyanWisth That sounds like the most likely explanation for that context! $\endgroup$ – Brevan Ellefsen Jan 11 '16 at 3:30
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You're right on both counts; the arbitrary $C$'s can be different, and $x$ and $t$ can be different. The author is referring only to $x$ and $t$, from what I can tell. I would say that the way the book puts this is a bit confusing. (At least, it confused me). Consider

$$ \begin{aligned} f(x) &= x^2 + 1\\ g(t) &= t^2 + 1\\ \end{aligned} $$

These two functions are equal, because the $x$'s and $t$'s are what your book would call "dummy variables" -- variables that exist nowhere except inside the function.

When you use definite integration, the variable you're integrating over goes away, and you get a number. When you use indefinite integration, the variable doesn't go away. So the author says "they are not equal" because their values still depend on $x$ and $t$, which might not be equal. But they are equal (or "equivalent") as functions.

For example, $f$ and $g$ are equal as functions, but their values are different if into $f$ I insert $x = 1$ and into $g$ I insert $t = 2$.

A separate issue, which you are right to bring up, is that the $C$ in the first equation is not the same $C$ in the second.

Here's how I think about indefinite vs. definite integration. When viewed as functions in their own right, machines with inputs and outputs, they differ totally. Indefinite integration takes in a function and spits out a lot of functions (all of its antiderivatives, differing from each other by choice of constant $C$):

$$\int:: \textbf{a function} \rightarrow \textbf{all of its antiderivatives}$$

while definite integration takes in a function, a start point, and an end point, and gives you a number:

$$\int_a^b:: (\textbf{a function}, a, b) \rightarrow \textbf{a number}$$

What the author is trying to express is that indefinite integration gives you back a function, while definite integration gives you back a number.

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  • $\begingroup$ a beautiful wording of what I was trying to express to the OP. +1 $\endgroup$ – Brevan Ellefsen Jan 11 '16 at 4:13
  • $\begingroup$ @BrevanEllefsen: Thank you! If only these things didn't have almost the same symbol. $\endgroup$ – Eli Rose Jan 11 '16 at 9:39
  • $\begingroup$ Thank you, this answer cleared everything up! $\endgroup$ – Biomed Boy Jan 19 '16 at 18:34
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As discussed in the comments, the two expressions are likely not stated to be equal because the initial conditions are not know. For example, take the trig identity $\tan^2 x + 1 = \sec^2 x$. If we were to take the derivative of these functions we would get the same answer; however, once we take the antiderivative again we could get that $\tan^2 x + C_1 = \sec^2 x + C_2$. Information concerning constants is lost during integration.

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Either $x$ or $t$ is just a token representing numbers, so they are equivalent.

Let's take the expression $x+1$ (or $t+1$) for instance: Letting $x=1.5$ is equivalent to letting $t=1.5$, because both of them make the expression evaluate to $1.5+1=2.5$.

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What we see here is an abuse of notation ingrained in our way of doing calculus since centuries. If ${\rm sq}:t\mapsto t^2$ is the squaring function on ${\mathbb R}$ nobody would denote the set of all functions differing from ${\rm sq}$ by an additive constant by $\{{\rm sq}+c\>|\>c\in{\mathbb R}\}$. Instead we all write $t^2+C$, or $x^2+C$, for this set of functions.

Similarly, given any function $x\mapsto f(x)$ $(x\in I)$, e.g. $f(x):=2x$ $(x\in{\mathbb R})$, the typographical picture $\int f(x)\>dx$ by definition denotes the set of all functions $$t\mapsto F(t)\quad(t\in I)\tag{1}$$ satisfying $F'(x)=f(x)$ for all $x\in I$. We might as well have written $u\mapsto F(u)$ $(u\in I)$ in $(1)$. Indeed it is customary in integral tables to denote the independent variable on the right hand side again by $x$.

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