# What is wrong with my proof that $\int 2x dx= 2x^2$ by writing $2x=\underbrace{2+2+\cdots+2}_{x\;\text{times}}$?

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?

• $x$ 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. Jun 29, 2022 at 4:31
• – Gary
Jun 29, 2022 at 4:32
• @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. Jun 29, 2022 at 4:39
• 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. Jun 29, 2022 at 4:45
• 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. Jun 29, 2022 at 5:35

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.
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 $$2x=\underbrace{(2 + 2 + 2 + \dots + 2)}_{x \text{ times}}.$$
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.