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

Why do you get a function's (changing) slope when you take its derivative and why do you get the area under the function when you take its integral? What is the easiest reasoning behind this?

share|cite|improve this question
Why they work is merely a matter of proofs and definitions. The intuitive reasoning behind it (which what I think you're asking), is geometric. I'm not good at drawing so hopefully someone will give you an answer soon. – Git Gud Feb 24 '13 at 12:30
Because derivative is defined in terms of slopes of tiny line segments and integral is defined in terms of areas of tiny rectangles. – Karolis Juodelė Feb 24 '13 at 12:30
Check this out for integrals :… – user10444 Feb 24 '13 at 12:31
Read Derivative and Integral on Wikipedia. I am sure you'll find the answers. – Heitor Fontana Feb 24 '13 at 12:32

I strongly recommend that you take a look at the first chapter of Gilbert Strang's Calculus. In it, Strang gives an intuitively insightful introduction to differentiation and integration.

But for a brief and fairly standard explanation:


Consider the above image. The derivative essentially takes the limit as $h \to 0$. It is clear that for small $h$ you get a good approximation for the tangent line assuming $f$ is somewhat smooth. In the limit, this approximation becomes the tangent line. This concept is defined mathematically as:

$$f'(x_0)=\lim\limits_{h\to0}\frac{f(x_0)+h - f(x_0)}{h}.$$

Graphically, this is:

enter image description here


Again, consider the above image. With integration we take the limit as the number of rectangles approaches infinity (and hence each rectangles width approaches zero). It is clear that with hundreds or thousands of rectangles, the sum of the area of each rectangle is very nearly the area under the curve. In the limit, we get that the sum is exactly equal to the area. This limit is written mathematically as:

$$\lim\limits_{n\to\infty} \sum\limits_{i=1}^n f(x_i)\Delta x_i$$

where $\Delta x_i$ is the width of the $i$'th partition. Graphically, this process looks something like:

enter image description here

share|cite|improve this answer

It only works under euclidean metric; change the metric, and they are not that simply related.

Or, just take the Riemann sum of the differences; all the terms cancel out but the first and last.

Or, just differentiate the Riemann sums and all the mid terms cancel each other out but first and last.

Of course, all this has to be still true with more general integration and differentiation methods, but seeing the simple arithmetic that relates the two takes away the mystery.

If the graph was drawn on surface of sphere, the relationship between integration and differentiation would be much more different as one being the inverse of other.

share|cite|improve this answer
-1. Given that the question is fairly elementary, an answer that talks about generalisations to non-Euclidean situations will probably not help very much. Also, the answer has several typos, which I corrected. – bubba Feb 24 '13 at 13:08
Given that the question asks why, without the generalisation looking only at the specific does not explain the why, but the how. Tx – Arjang Feb 24 '13 at 13:35
I came here because I googled why it works (not how), and this is the only answer that appropriately handles the question. +1 – ivanmp Sep 27 '15 at 14:13

For insight, probably the best is to always consider the quantities like $dx$ as an infinitesimal quantity, i.e. such a positive real-like 'number' which is smaller than all $\frac1n\ (n\in\Bbb N)$. This context can be made precise, using nonstandard analysis. Using this language, real functions can be extended to the nonstandard reals (this contains the reals, together with infinitesimals at each point). Let us introduce the notation $$a\approx b\ \text{ iff }\ a-b\ \text{is infinitesimal}.$$

For a function $f$, we define $df(x):=f(x+dx)-f(x)$, where $dx$ denotes an arbitrary infinitesimal. We have that $f$ is continuous at $x$ iff $df(x)$ is always infinitesimal (for any infinitesimal choice of $dx$), i.e. iff $f(x+dx)\approx f(x)$.

For an arbitrary $h>0$ we have that $\displaystyle\frac{f(x+h)-f(x)}h$ is just the slope of the line going through $(x,f(x))$ and $(x+h,f(x+h))$. In ordinary analysis, we can take its limit as $h\to 0$, then the sector will converge to the tangent at $x$. In the nonstandard approach, it is just defined as $$f'(x):\approx\frac{df(x)}{dx}$$ Intuitively, if the distance of $x$ and $x+dx$ is infinitesimal, then the difference (of the slopes) of the sector with $h=dx$ to the tangent is also infinitesimal.

For the integral, $f(x)\cdot dx$ is already the (signed) area of a rectangle of height $f(x)$ and infinitesimal width $dx$. If $f(x)$ is real, then $f(x)\cdot dx$ is again infinitesimal, and we can 'sum' these up for all reals $x\in [a,b]$ -- these are so much that their sum is usually already not infinitesimal. $$\int_a^b f :=\sum_{x\in [a,b]}f(x)\cdot dx$$ This only wanted to give just an insight for the original and now called nonstandard approach. For more details, see the wikipedia article.

share|cite|improve this answer

An intuitive reason that the derivative of the integral is the original function (for "nice" functions) is this:

Let $f$ be the function and $F$ be its integral, so $F(x) = \int_a^x f(t) dt$, where $a$ is the initial integration point. It does not matter what $a$ is - it will soon exit, stage right.

The derivative of $F$ at $x$ is approximately $\frac{F(x+h)-F(x)}{h}=\frac1{h}\int_x^{x+h} f(t) dt$ (told you $a$ would go away, now all ok, say hey!). Since $f$ is "nice", over a small interval, such as $[x, x+h]$, it is nearly constant, so each $f(t)$ in $[x, x+h]$ is $f(x)$ with an error that gets smaller and goes to zero as $h \to 0$ (that is what "nice" means).

This $\int_x^{x+h} f(t) dt \approx \int_x^{x+h} f(x) dt = h f(x)$ so $F'(x) \approx \frac{F(x+h)-F(x)}{h} =\frac1{h}\int_x^{x+h} f(t) dt \approx \frac1{h} (h f(x)) = f(x) $.

I explicitly deny any originality in this post, aside from my usage of non-essential phrasing.

share|cite|improve this answer

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.