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Is it okay to have a function $f(x)$ defined by $$f(x)=\int_0^x{x^2-x\,dx}\,?$$ If so, what would $f'(x)$ be?

I've seen many questions like these on my math competitions (they give functions defined by integrals and both the limits of the integral and the integrand contain the same variable, as in the example above) and it seems like they do have a/n (numerical) answer, and they accept the answer as correct, although people are able to dispute why it shouldn't be okay to have such function.

Parcly Taxel
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NestorV S
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    just another nitpicky remark, $\int f(x) - g(x),dx$ should be denoted as $\int (f(x)-g(x)),dx$ simply for the sake of clarity. Most people will understand your intent, however it's best to be as clear as possible. – Dando18 Jul 15 '17 at 06:04
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    @Dando18: I never understood that particular nitpicking point on clarity. What could it possibly mean other than that that writing parentheses would "clarify" it? – user541686 Jul 15 '17 at 08:29
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    @Mehrdad: Playing devil's advocate... we spend a lot of time training our spatial recognition to automatically parse arithmetic expressions, and we spend a lot of time training ourselves to read parentheses as delimiters. We spend comparatively little time training ourselves to read $\int \ldots \mathrm{d}x$ as delimiters. $\int f(x) - g(x) , dx$ is simply harder to read than $\int(f(x) - g(x)) , dx$, since it conflicts with our training. That the intent is unambiguous doesn't change that fact. –  Jul 15 '17 at 09:08
  • @Hurkyl, you're right, but IMO we should just stop using the integral in this strange operation-parenthisation double role then. I always write $\int\limits_l^r\mathrm{d}x\ (f(x)-g(x))$, which nicely groups the integration domain with the variable it quantifies over – a huge benefit when dealing with nested integrals. – leftaroundabout Jul 15 '17 at 09:19
  • @Hurkyl: The thing is I personally find it easier to read, so it's actually harder for me to read it with extra parentheses. I seem to be in the minority though, everyone seems to say the opposite haha. – user541686 Jul 15 '17 at 09:40
  • @Mehrdad: My personal experience (i.e. no longer playing devil's advocate) is that Dando's example has enough clutter that the parentheses don't really affect readability either way, but my eye really wants to group $\int x^2 - x , dx$ as $(\int x^2) - (x , dx)$ before settling down into the correct reading. Doubly so if typeset as $\int x^2 - xdx$ is written. –  Jul 15 '17 at 09:51
  • @Hurkyl: Oh, yeah, you definitely need the spacing. Without it I want to poke my eyes out... – user541686 Jul 15 '17 at 09:57
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    @Mehrdad I know this context is single variable, but just as an aside I've often seen $\int_C \vec{f},d\vec{r}$ written as $\int_C u(x,y)dx + v(x,y)dy$, which in some cases can be confusing without proper parenthesis. – Dando18 Jul 15 '17 at 13:16
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    I think I'm the only one that has more problem with the use of $x$ as a dummy variable than no parentheses. – Shashi Jul 15 '17 at 15:46
  • It is of course formally correct to use $x$ as the bound variable, since it never refers to the original $x$, but it is still not far short of a hanging offence. While I agree that the integrand is better bracketed, I personally would incline to square brackets: $ \int_0^x \left[ t^2 - t \right] dt $. – PJTraill Jul 15 '17 at 18:33
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    Hopefully people will make more note of this than the parentheses, which are honestly less of a problem, and I will point out that we can have multiple variables in an integral. The doubling up on variables becomes an extreme problem when we start to get expressions such as: $$\int_{a(x)}^{b(x)}f(x,t)~\mathrm dt$$ If you slacked on the notation here, you'd end up with: $$\int_{a(x)}^{b(x)}f(x,x)~\mathrm dx$$ And as you can see, we've lost the original clarity. – Simply Beautiful Art Jul 15 '17 at 23:18
  • Examples of integrals that come in this form include the formula for Leibniz integral rule, and particularly, everything in multivariable calculus. As the name implies, you will eventually see expressions such as $$\iint_{x^2+y^2\le1}xy~\mathrm dx~\mathrm dy$$ And there isn't much room for misplacing your $x$'s and your $y$'s, much less naming them ambiguously. – Simply Beautiful Art Jul 15 '17 at 23:20
  • For everyone interested in the parenthesis issue, see Is it required to use brackets inside an integral? – Simply Beautiful Art Jul 15 '17 at 23:23

6 Answers6

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Writing $\int_0^x f(x)\,dx$ is the same as $\int_0^x f(t)\,dt$. However, the former is much more ambiguous and should be avoided for sake of clarity.

This means the derivative would be,

$$ \frac{d}{dx}\left[ \int_0^x (x^2-x)\,dx \right] = \frac{d}{dx}\left[ \int_0^x (t^2-t)\,dt \right] = x^2 - x $$

Dando18
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    $\int_0^xf(x)~\mathrm dx=\int_0^xf(t)~\mathrm dt$ is quite a bad habit, and it will become especially ambiguous when your integrals contain multiple variables. – Simply Beautiful Art Jul 15 '17 at 23:12
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Let $\phi$ be a continuous, real-valued function defined in some neighborhood of $0$. Writing $$ f(x) = \int_{0}^{x} \phi(x)\, dx $$ is bad syntax: it uses the single letter $x$ for both

  • The input value of the function $f$;

  • The dummy variable of integration.

Setting $x = 2$, for example, invites the reader to ponder the meaning of $$ f(2) = \int_{0}^{2} \phi(2)\, d2, $$ which is almost surely not intended. (The problem is "$d2$".)

Strictly speaking, "no", it is not okay to have a function $f$ defined by $$ f(x) = \int_{0}^{x} (x^{2} - x)\, dx. $$

By contrast, equations such as $$ g(x) = \int_{0}^{x} \phi(t)\, dt,\qquad h(x) = \int_{0}^{x} (x - t)\phi(t)\, dt, $$ do properly define functions. Note carefully that in the defintion of $h$, the input value $x$ appears in the integrand but is not the variable of integration.

The fundamental theorem of calculus gives $$ g'(x) = \phi(x),\qquad h'(x) = \int_{0}^{x} \phi(t)\, dt = g(x). $$ The second is an edifying exercise; to give a proof, it may help to write $$ h(x) = \int_{0}^{x} (x\phi(t) - t\phi(t))\, dt = x\int_{0}^{x} \phi(t)\, dt - \int_{0}^{x} t\phi(t))\, dt = xg(x) - \int_{0}^{x} t\phi(t)\, dt. $$

Andrew D. Hwang
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It is just a basic property,

$$I=\int_{0}^{t}f(x)dx$$

Differentiating

$$I'=f(t)$$

$$f(x)=\int_{0}^{x}x^2-xdx$$

Note that t is a variable

Differentiating w.r.t x

$$f'(x)=x^2-x$$

Reasoning is

$$f(x)=[\frac{x^3}{3}-\frac{x^2}{2}]_{0}^{x}$$

$$f(x)=\frac{x^3}{3}-\frac{x^2}{2}$$

$$f'(x)=x^2-x$$

Crazy
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A function like this is OK, but can be simplified due to the fundamental theorem of calculus. If $G(x)$ is the antiderivative of $g(x)$: $$f(x)=\int_0^xg(x)\,dx=[G(x)]_0^x=G(x)-G(0)$$ $$f'(x)=(G(x))'-(G(0))'=g(x)$$ Similar derivations can be made for domains of integration other than $[0,x]$.

Parcly Taxel
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  • Especially if the upper bound is, say, $\sqrt{x}$, then $f'(x) = (G(\sqrt{x}))' - (G(0))' = g(\sqrt{x})\cdot\frac{1}{2\sqrt{x}}$. – ComFreek Jul 15 '17 at 09:11
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The variable of integral $x $ and the upper bound of integral are different. They arent the same. Since we have by integration property that $\int _a^b f (x)dx=\int _a^bf (t)dt $ we can write it that way though some people find it wrong. Also a case is possible where both the limit of integral and the variable are same like for example. $\int_0 ^x x^2f (t)dt $ then both the x are the same as the variable of integral is t and not x and we can pull it out from the integral operator. As for the part of $f'(x) $ we have a well known theorem ie Newton Leibnitz theorem or Fundamental theorem of calculus which says that if $f (x)= \int _{h(x)}^{g (x)}k (t)dt $ then $f'(x)=k (g (x)).g'(x)-k (h (x).h'(x) $

Archis Welankar
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I'm going to say "no, it is not okay". As you will eventually find out, you can have integrals such as

$$\int_0^x3x^2+y~\mathrm dy$$

Which should evaluate to $3x^3+\frac12x^2$ (feel free to work it out yourself)

Such integrals appear very often in multivariable calculus, and the above integral was probably part of a larger problem:

$$\int_0^1\int_0^x3x^2+y~\mathrm dy~\mathrm dx$$

Particularly, surface integrals tend to be integrals involving multiple variables, often related to one another. The above example comes from the following integral:

$$\iint_{0<y<x<1}(3x^2+y)~\mathrm dA$$

So when you've come to problems like these, it will become extremely unclear which variables refer to what and what things mean if you don't name your variables properly. Note that unlike most of the other answers, in the first integral of my answer, the $x$ in the upper bound and the $x$ inside the integral are one and the same!

So overcome your bad habits soon and use your variables correctly, as any assumptions you make about how you can name your variables will one day bite you back!

Simply Beautiful Art
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