Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I've been a calculus student now for about two years, and I've gone as high as differential equations, but I am still a bit puzzled by the fact that the area under the curve of some function is defined the way it is.

Let me clarify. so let's say you take the anti derivative of $$ \int (x^2+4x) \ dx = \frac{x^3}{3}+2x^2 $$ I don't have a clear intuition of why the area is exponentially greater than the original function. if any one could explain why that is, I would be very grateful. I know that this must sound like a stupid question, especially form someone who's been in math for quite some time, but I just don't have a good grasp on that concept.

share|cite|improve this question
Just think of it in the case of $f(x)=1$. Then the area under $f(x)$ from $0$ to $t$ is $t$. – Thomas Andrews Oct 26 '11 at 13:57
Is that some new sense of "exponentially greater" that I'm not aware of? – Henning Makholm Oct 26 '11 at 14:10
Yes, the usage of "exponentially greater" is very imprecise here. – Thomas Andrews Oct 26 '11 at 14:20
ya sorry about that... I'm sometimes a bit careless with my words. what i meant by that is that you are adding an to the current exponent, thus it grows quite a bit faster because of the new exponent. – AlexW.H.B. Oct 26 '11 at 14:24
Indeed, when I first saw the question, I assumed you were integrating a function like $e^{(e^x)}$... – MartianInvader Oct 26 '11 at 19:00
up vote 7 down vote accepted

The function $\frac{x^3}{3}+2x^2$ does not grow exponentially faster than $x^2+4x$. The function $\frac{x^3}{3}+2x^2$ is a polynomial, and it grows roughly like $(x^2+4x)^{3/2}$. That is not dramatically faster than $x^2+4x$. The following is an example of faster relative growth that is more familiar.

Look at the area of the region under the curve $y=x$, above the $x$-axis, from $x=0$ to $x=w$. If you draw the "curve" $y=x$, you will see that the region is a triangle with base $w$ and height $w$, so the region has area $w^2/2$. (The same result can be obtained by integration, but that's overkill.)

So the function grows like $w$, and the area grows like $w^2$, more precisely like a constant times $w^2$.

The most "extreme" example is a constant function like $f(x)=1$. The area from $0$ to $w$ under the curve is $w$. This grows much faster than $f$, which does not grow at all, but the rate of growth of the area would not be considered fast. And the fact that the area of a rectangle of constant height grows linearly with the base is unsurprising.

In general, imagine a curve $y=f(x)$, where $f(x)$ is positive, and, for simplicity, increasing. Let $W$ be the region below $y=f(x)$, above the $x$-axis, from $x=0$ to $x=w$. Then the region $W$ is contained in the rectangle with base the interval from $0$ to $w$, and height $f(w)$. So the area of $W$ is $\le wf(w)$. Thus, if $f(w)$ grows "fast," then the rate of growth of the area is not much faster than the rate of growth of $f(w)$.

share|cite|improve this answer
Thank you very much for this explanation. I am starting to grasp the concept. – AlexW.H.B. Oct 26 '11 at 14:21

Simplify still further the situation and assume that you look for the area under the curve of equation $y=ax$ between the abscissæ $0$ and $x\geqslant0$. This is the area of a triangle with a right angle and sides $x$ and $ax$ hence the result is $\frac12ax^2$, which is of the order of the length of the interval times the maximal value of $f$ on the interval. If the height of the curve is of the order of its maximum for a positive fraction of the abscissæ one is considering, the heuristics that the area is of the order of the length of the interval times the greatest value of the function on the interval is valid.

On the contrary the area under the curve of equation $y=\mathrm e^{ax}$ between the abscissæ $0$ and $x\geqslant0$ is $a^{-1}(\mathrm e^{ax}-1)$ which is equivalent to the height $\mathrm e^{ax}$ of the higher point. This is because, when seen from afar, the curve is roughly constant and not very high for most of the time but it climbs steeply at the very last moment to reach its maximum. Hence the area under the curve is of the order of the greatest value of the function on the interval.

share|cite|improve this answer
thank you so much for your explanation. and your time. This is a big help. – AlexW.H.B. Oct 26 '11 at 14:14

The most simplest explanation is that you add one dimension. The way the Riemann integral is defined gives a hint that you're summing the infinitesimal squares of infinitesimal rectangles. Each rectangle has a height equal to $f(x)$ and there are as much rectangles as $[0,x]$ (since you're asking for the intuition I hope I can allow myself to be informal).

So, roughly speaking you obtain the anti-derivative which if $F(x)\approx x f(x)$. Here $\approx$ is very approximate and good only if $f(x)$ is constant. In all other cases, you'll have e.g. additional coefficient like with power functions $\displaystyle{x^n\to\frac{x^{n+1}}{n+1}}$.

The more complex your function the more dangerous is this intuition, e.g. the exponent function $e^x$ counts the square under its graph from $-\infty$ to $x$ without any additional power of $x$, i.e. $$ \int\limits_{-\infty}^x e^t\,dt = e^x. $$

share|cite|improve this answer
But then why does "adding a dimension" not change $e^x$ at all? – user7530 Oct 26 '11 at 14:00
@user7530 Because mathematics it the science where intuitively obvious things can be dangerous and false. There all the fun begins. The intuition I've described works with only the class of functions somehow close to the constant. – Ilya Oct 26 '11 at 14:02
thank you so much for your explanation. and your time. This is a big help. I think math becomes much easier if you can conceptualize what you are doing. – AlexW.H.B. Oct 26 '11 at 14:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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