5
$\begingroup$

first, I must apologize for somewhat misleading a title.

To save both your and my time, I will go straight to the point.

By definition, an indefinite integral, or a primitive, or an antiderivative of a (some condition) function $f(x)$ is any $F(x)$ such that $F'(x)=f(x)$. All well and good.

Because any other primitive can be written as $F(x)+C$ for some constant $C$ (and this requires a proof), if we were to denote by $\int f(x)dx$ an antiderivative of $f(x)$, then \begin{equation} \int f(x)dx = F(x)+C. \end{equation}

Fine. But here is the part that every textbook seems to have no problem with, but bugs me greatly: Often they say that integrate both sides of the following equation: \begin{equation} f(x)=g(x),\end{equation}

to obtain \begin{equation} \int f(x)dx = \int g(x)dx. \end{equation}

This looks like an ABSOLUTE nonsense to be for the following reason: IF both sides of the previous equation are TRULLY equal, then surely

\begin{equation} \int f(x)dx - \int g(x)dx =0. \end{equation}

But

\begin{equation} \int f(x)dx - \int g(x)dx =\int (f(x)-g(x))dx = \int 0dx, \end{equation}

which then equals $C$, any constant. Surely this is not necessarily 0!

So in short, this is my question: IS IT, STRICTLY SPEAKING, LEGAL, TO TAKE THE ANTIDERIVATIVE OF BOTH SIDES OF AN EQUATION?

$\endgroup$
  • 2
    $\begingroup$ You can either redefine $=$ to mean "equal up to an arbitrary constant" or define $\int f(x)dx$ as a class of functions, then $\int f(x)dx = \int g(x)dx$ says that two classes are equivalent. In either of those cases equality holds. If you define $\int f(x)dx$ to be $F(x)+C$ for some arbitrary $F(x)$ and $C$, then no, $\int f(x)dx$ does not equal $\int g(x)dx$ (though I would argue that definitions which depend on arbitrary choices are not good definitions). Thus the exact definitions of each symbol is important here. $\endgroup$ – user137731 Oct 10 '15 at 19:29
0
$\begingroup$

The problem here might be the notation $\int f(x)\,dx$. Does it mean one primitive? All primitives? Something else?

If we for a while agree that $F$ is a primitive of $f$ on an interval $I$ and $G$ is a primitive of $g$ on the same interval $I$, and it holds that $f(x)=g(x)$ for all $x\in I$, then we can be sure that $F(x)=G(x)+C$ for all $x\in I$, where $C$ is some constant.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Whatever those calc textbooks claim to be... Say, in many cases, like wiki, they say: any such function whose derivative equals the the function of interest. This is why I find it very annoying, for by this definition, indefinite integral found today may as well differ from those I found yesterday, day after yesterday, a year ago, etc. $\endgroup$ – user134070 Oct 10 '15 at 19:28
  • $\begingroup$ I agree that the notation is a bit irritating. Or rather that it might mean different things for different people. But do you really feel this is a big problem when working with primitives? $\endgroup$ – mickep Oct 10 '15 at 19:33
  • $\begingroup$ No, not really, but some "elementary proofs" such as that of integration by parts that make some sly use of this fact (that you can take the antiderivative of both sides and still end up with an equality) do. $\endgroup$ – user134070 Oct 10 '15 at 19:39
  • $\begingroup$ I would not be angry if the down voter explained him/herself. $\endgroup$ – mickep Oct 10 '15 at 19:44
0
$\begingroup$

Paul's Online Math Notes has an excellent section on the Constant of Integration. It shows how two apparently different results of integrating the same equation differ only by the Constant. Since we often write the Constant simply as $C$, it's easy to miss the meaning of this term.

While this discussion doesn't deal precisely with the question at hand, it is germane in as much as it deals with nomenclature of apparently different functions while integrating, and the resulting Constant.

Read the whole discussion here.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't see how does this address the original question. $\endgroup$ – dbanet Oct 10 '15 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.