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I recently saw someone post an equation like this:

$$\int_a^b {dx \over f(x)}$$

Does this just mean $\int_a^b {1 \over f(x)} dx$? As far as I know, the $dx$ isn't an actual term and just notational convenience to declare the argument of the function that's integrated so we don't have to define it seperately.

Relatedly, assuming I have a function $f$, can I represent the antiderivative as $\int f$?

schuelermine
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  • For the first question, yes, both mean the same thing, just a notational choice. – quasi May 21 '20 at 14:20
  • So long as you don't do anything obviously wrong if $dx$ were treated as any other type of variable, no, they mean the same thing. It might be seen as improper by some, but not inherently wrong per se. – JMoravitz May 21 '20 at 14:22
  • For the second question, the presence of $dx$ serves to specify the independent variable. – quasi May 21 '20 at 14:22
  • It may be worth pointing out that you have integrals such as $\oint_C(L~dx+M~dy)$ in Green's Theorem where the $dx$ and $dy$ aren't explicitly acting as the metaphorical closing parenthesis to the integral so there is precedence for notation allowing the $dx$ and similar to move around. – JMoravitz May 21 '20 at 14:30

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If you think of a definite integral for area as Leibniz did as an infinite sum of infinitely many infinitely thin rectangles each of which has base $dx$ then the position of $dx$ in the product "height times base" does not matter - but you do need it. The two expressions in the question are the same. See Why can't the second fundamental theorem of calculus be proved in just two lines? .

So $dx$ does more than simply indicate which variable you are integrating with respect to. That's really important in applications in physics.

When there is just one variable and you are more interested in theory than in applications you can in fact write $\int f$. Just remember that represents any of the infinitely many antiderivates (hence the "constant of integration").

Jam
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Ethan Bolker
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