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I know $\int 2x \,dx = x^2 + C$ (by the power rule) but why does the following proof not give the same answer?

\begin{align*} \int 2x \,dx &= \int \underbrace{(2 + 2 + 2 + \dots + 2)}_{x \text{ times}} \, dx \\ &= \underbrace{\int{2} \, dx + \int{2} \, dx + \dots \ + \int{2}_ \, dx}_{x \text{ times}}\\ &= 2x + 2x + \dots + 2x + C \\ &= 2x \times x + C \\ &= 2x^2 + C \end{align*}

(And I have the same question for this false proof that $\int{2^x} \, dx = 2^{x}x+ C$)

\begin{align*} \int{2^x} \,dx &= \int \underbrace{(2 \cdot 2 \cdot 2 \cdot \dots \cdot 2)}_{x \text{ times}} \cdot 1 \, dx \\ &= 2 \cdot \int \underbrace{(2 \cdot 2 \cdot 2 \cdot \dots \cdot 2)}_{(x-1) \text{ times}} \cdot 1 \, dx && (\text{Constant Multipule Rule})\\ &= 2^2 \cdot \int \underbrace{(2 \cdot 2 \cdot 2 \cdot \dots \cdot 2)}_{(x-2) \text{ times}} \cdot 1 \, dx && (\text{Constant Multipule Rule})\\ &= 2^x \cdot \int{1} \, dx \\ &= 2^{x}x+ C \\ \end{align*}

I suspect that it has something to do with not being able to:

  1. Change integral of sums to sums of integrals for an arbitrary $x$, and

  2. Remove a constant out of an integral if there are variable numbers of those constants.

But I'm not sure why these do not hold. If this is the reason, is there a theorem stating it?

Thanks in advance!

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    $\begingroup$ $x$ is a variable. You can't take it out of the integral. If you do want to convert an integral into a sum: this can be dome, but you have to be more careful. $\endgroup$
    – M. Wind
    Jun 29, 2022 at 4:31
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    $\begingroup$ Similar: math.stackexchange.com/q/164444 and math.stackexchange.com/q/1096 $\endgroup$
    – Gary
    Jun 29, 2022 at 4:32
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    $\begingroup$ @sloth FYI, using an Approach0 search, I found the AoPS thread Spot the error in calculus which deals with basically the same question as the $2$nd link of math.stackexchange.com/q/164444/602049 in Gary's comment above. $\endgroup$ Jun 29, 2022 at 4:39
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    $\begingroup$ Something more general...from algebra. I do not think there is a definition of adding something $s$ times, where $s$ is not a naturall number. $\endgroup$
    – dmtri
    Jun 29, 2022 at 4:45
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    $\begingroup$ Reviewers: this Question does not duplicate any of the above 3 linked suggestions because, cast as an integration question, it requires an answer that is not merely incidentally different from what the other 3 questions require. I know because I tried answering those instead and just leaving a pointer here, but wasn't able to shoehorn it. $\endgroup$
    – ryang
    Jun 29, 2022 at 5:35

3 Answers 3

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The first integral is wrong because $x\in\mathbb{R}$ need not be a natural number. So the multiplication cannot be split as you did. Similar reasoning applies to the second case, that is to say, since $x$ need not be a natural number, the interpretation for $2^{x}$ is wrong.

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The above comments and answer point out that you can't take a variable out of an integral, and that $x$ is not a natural number, but in fact, in the given examples, the error could be more fundamental.

Let $f:\mathbb Z \to \mathbb R$ and $g:\mathbb Z \to \mathbb R$ such that $f(x)=2x$ and $g(x)=2^x.$ Giving benefit of the doubt that the author indeed meant to work in $\mathbb Z,$ that is, that $$\int 2x \,\mathrm dx=\int f,\\ \int 2^x \,\mathrm dx=\int g,$$ then both integrals immediately equal $0,$ since the domain of integration in each case is a set of isolated points.

If the intention, however, was for interval integration domains, then it is of course invalid to assert that $2x=\underbrace{(2 + 2 + 2 + \dots + 2)}_{x \text{ times}}.$

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An integral might run from $x=0$ to $x=X$.

In $\int2xdx$, the function is increasing as you go along. So the first $2$ is there all the way as x goes from 1 to X, but the second $2$ doesn't start until x is 2, and the final $2$ doesn't appear until x has already reached X.
In $\int2dx+\int2dx+...$, all the $2$s are there all the way as x goes from 0 to X.

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