Why is the area under a curve the integral?

I understand how derivatives work based on the definition, and the fact that my professor explained it step by step until the point where I can derive it myself.

However when it comes to the area under a curve for some reason when you break it up into an infinite amount of rectangles, magically it turns into the anti-derivative. Can someone explain why that is the definition of the integral and how Newton figured this out?

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There are 2 potential questions here. One is the question of why the definite Riemann integral gives the correct notion of "area under a curve" for a (nonnegative, Riemann integrable) function. The other, which seems to be what you're really asking, is the question of why an antiderivative evaluated at the endpoints of an interval and subtracted yields that definite integral. The latter question is answered by an understanding of the fundamental theorem of calculus. The purpose of this comment is just to help clarify your question. – Jonas Meyer Dec 23 '10 at 4:58
Its also possible that he is seeking the intuition for Calculus. Specifically integration – picakhu Dec 23 '10 at 5:19
@picakhu: I think that he or she is certainly seeking intuition, but the phrase "magically it turns into the anti-derivative" indicates that this is really about the fundamental theorem of calculus, and not just about the definition of the integral. – Jonas Meyer Dec 23 '10 at 5:25
I don't understand the difference between the two – qwertymk Dec 23 '10 at 5:27
@qwertymk: Well, now you have a fabulous explanation from Arturo Magidin, but I'll also mention that en.wikipedia.org/wiki/Riemann_integral is a good starting point for learning about how the integral is actually defined, while en.wikipedia.org/wiki/Fundamental_theorem_of_calculus is a good starting point for learning about the relationship with antiderivatives. Both include excellent graphics! – Jonas Meyer Dec 23 '10 at 5:32

First: the integral is defined to be the (net signed) area under the curve. The definition in terms of Riemann sums is precisely designed to accomplish this. The integral is a limit, a number. There is, a priori, no connection whatsoever with derivatives. (That is one of the things that makes the Fundamental Theorems of Calculus such a potentially surprising things).

Why does the limit of the Riemann sums actually give the area under the graph? The idea of approximating a shape whose area we don't know both from "above" and from "below" with areas we do know goes all the way back to the Greeks. Archimedes gave bounds for the value of $\pi$ by figuring out areas of inscribed and circumscribed polygons in a circle, knowing that the area of the circle would be somewhere between the two; the more sides to the polygons, the closer the inner and outer polygons are to the circle, the closer the areas are to the area of the circle.

The way Riemann tried to formalize this was with the "upper" and "lower" Riemann sums: assuming the function is relatively "nice", so that on each subinterval it has a maximum and a minimum, the "lower Riemann sum" is done by taking the largest "rectangle" that will lie completely under the graph by looking at the minimum value of the function on the interval and using that as height; and the "upper Riemann sum" is done by taking the smallest rectangle for which the graph will lie completely under it (by taking the maximum value of the function as the height). Certainly, the exact area under the graph on that interval will be somewhere between the two. If we let $\underline{S}(f,P)$ be the lower sum corresponding to some fixed partition $P$ of the interval, and $\overline{S}(f,P)$ be the upper sum, we will have that $$\underline{S}(f,P) \leq \int_a^b f(x)\,dx \leq \overline{S}(f,P).$$ (Remember that $\int_a^bf(x)\,dx$ is just the symbol we use to denote the exact (net signed) area under the graph of $f(x)$ between $a$ and $b$, whatever that quantity may be.)

Also, intuitively, the more intervals we take, the closer these two approximations (one from below and one from above) will be. This does not always work out if all we do is take "more" intervals. But one thing we can show is that if $P'$ is a refinement of $P$ (it includes all the dividing points that $P$ had, and possibly more points) then $$\underline{S}(f,P)\leq \underline{S}(f,P')\text{ and } \overline{S}(f,P')\leq \overline{S}(f,P)$$ so at least the approximations are heading in the right direction. To see why this happens, suppose you split one of the subintervals $[t_i,t_{i+1}]$ in two, $[t_i,t']$ and $[t',t_{i+1}]$. The minimum of $f$ on $[t_i,t']$ and on $[t',t_{i+1}]$ are each greater than or equal to the minimum over the whole of $[t_i,t_{i+1}]$, but it may be that the minimum in one of the two bits is actually strictly larger than the minimum over $[t_i,t_{i+1}]$. The areas we get after the split can be no smaller, but they can be larger than the ones we had before the split. Similarly for the upper sums.

So, let's consider one particular sequence of partitions: divide the interval into 2 equal parts; then into 4; then into 8; then into 16; then into 32; and so on; then into $2^n$, etc. If $P_n$ is the partition that divides $[a,b]$ into $2^n$ equal parts, then $P_{n+1}$ is a refinement of $P_n$, and so we have: $$\underline{S}(f,P_1) \leq\cdots \leq \underline{S}(f,P_n)\leq\cdots \leq\int_a^b f(x)\,dx \leq\cdots \leq\overline{S}(f,P_n)\leq\cdots \leq \overline{S}(f,P_2)\leq\overline{S}(f,P_1).$$

Now, the sequence of numbers $\underline{S}(f,P_1) \leq \underline{S}(f,P_2)\leq\cdots \leq \underline{S}(f,P_n)\leq\cdots$ is increasing and bounded above (by the area). So the numbers have a supremum; call it $\underline{S}$. This number is no more than $\int_a^b f(x)\,dx$. And the numbers $\overline{S}(f,P_1) \geq \overline{S}(f,P_2)\geq\cdots \geq \overline{S}(f,P_n)\geq\cdots$ are decreasing and bounded below, so they have a minimum; call this $\overline{S}$; again, it is no less than $\int_a^bf(x)\,dx$. So we have: $$\lim_{n\to\infty}\underline{S}(f,P_n) = \underline{S} \leq \int_a^b f(x)\,dx \leq \overline{S} = \lim_{n\to\infty}\overline{S}(f,P_n).$$ What if we are lucky? What if actually we have $\underline{S}=\overline{S}$? Then it must be the case that this common value is the value of $\int_a^b f(x)\,dx$. It just doesn't have a choice! It's definitely trapped between the two, and if there is no space between them, then it's equal to them.

What Riemann proved was several things:

1. If $f$ is "nice enough", then you will necessarily get that $\underline{S}=\overline{S}$. In particular, continuous functions happen to be "nice enough", so it will definitely work for them (in fact, continuous functions turn out to be "very nice", not just "nice enough").

2. If $f$ is "nice enough", then you don't have to use the partitions we used above. You can use any sequence of partitions, so long as the "mesh size" (the size of the largest subinterval in the partition) gets smaller and smaller, and has limit of $0$ as $n\to\infty$; if it works for the partitions "divide-into-$2^n$-equal-intervals", then it works for any sequence of partitions whose mesh size goes to zero.

So, for example, we can take $P_n$ to be the partition that divides $[a,b]$ into $n$ equal parts, even though $P_{n+1}$ is not a refinement of $P_n$ in this case.

3. In fact, you don't have to do $\underline{S}(f,P)$ and $\overline{S}(f,P)$. For the partition $P$, just pick any rectangle that has as its height any value of the function in the subinterval (that is, pick an arbitrary $x_i^*$ in the subinterval $[t_i,t_{i+1}]$, and use $f(x_i^*)$ as the height). Call the resulting sum $S(f,P,x_1^*,\ldots,x_n^*)$. Then you have $$\underline{S}(f,P) \leq S(f,P,x_1^*,\ldots,x_n^*)\leq \overline{S}(f,P)$$ because $\underline{S}(f,P)$ is computed using the smallest possible values of $f$ throughout, and $\overline{S}(f,P)$ is computed using the largest possible values of $f$ throughout. But since we already know, from 1 and 2 above, that $\underline{S}(f,P)$ and $\overline{S}(f,P)$ have the same limit, then the sums $S(f,P,x_1^*,\ldots,x_n^*)$ also get squeezed and must have that same limit, which equals the integral.

In particular, we can always take the left endpoint (and get a "Left Hand Sum") or we can always take the right endpoint (and get a "Right Hand Sum"), and you will nevertheless get the same limit.

So in summary, you can pick any sequence of partitions, whichever happens to be convenient, so long as the mesh size goes to $0$, and you can pick any points on the subintervals (say, ones which make the calculations simpler) at each stage, and so long as the function is "nice enough" (for example, if it is continuous), everything will work out and the limit will be the number which must be the value of the area (because it was trapped between the lower and upper sums, and they both got squeezed together trapping the limit and the integral both between them).

Now, (1) and (2) above are the hardest part of what Riemann did. Don't be surprised if it sounds a bit magical at this point. But I hope that you agree that if the lower and upper sums for the special partitions have the same limits then that limit must be the area that lies under the graph.

Thanks to that work of Riemann, then (at least for continuous functions) we can define $\int_a^b f(x)\,dx$ to be the limit of, say, the left hand sums of the partitions we get by dividing $[a,b]$ into $n$ equal parts, because these partitions have mesh size going to $0$, we can pick any points we like (say, the left end points), and we know the limit is going to be that common value of $\underline{S}$ and $\overline{S}$, which has to be the area. So that, under this definition, $\int_a^b f(x)\,dx$ really is the net signed area under the graph of $f(x)$. It just doesn't have a choice but to be that, when $f$ is "nice enough".

Second, the area does not turn into "the" antiderivative. What happens is that it turns out (perhaps somewhat magically) that the area can be computed using an antiderivative. I'll go into some more details below.

As to how Newton figured this out, his teacher, Isaac Barrow, was the one who discovered there was a connection between derivatives and tangents; some of the basic ideas were his. They came from studying some simple functions and some simple formulas for tangents he had discovered.

For example, the tangents to the parabola $y=x^2$ were interesting (there was generally geometric interest in tangents and in "squaring" regions, also known as finding the "quadrature" of a region, that is, finding a way to construct a square or rectangle that had the same area as the region you were considering), and led to associating the parabola $y=x^2$ to lines of the form $y=2x$. It does not take too much experimentation to realize that if you look at the area under $y=2x$ from 0 to a, you end up with $a^2$, establishing a connection. Barrow did this with arguments with infinitesimals (which were a bit fuzzy and not set on entirely correct and solid logical foundation until well into the 20th century), which were generally laborious, and only for some curves. When Newton extended Barrow's methods to more general curves and tangents, he also extended the discovery of the connection with areas, and was able to prove what is essentially the Fundamental Theorem of Calculus.

Now, here is one way to approach the connection. We want to figure out the value of, say, $$\int_0^a f(x)\,dx$$ for some $a$. This can be done using limits and Riemann sums (Newton and Leibniz had similar methods, though not set up quite as precisely as Riemann sums are). But here is an absolutely crazy suggestion: suppose you can find a "master function" $\mathcal{M}$, which, when given any point $b$ between $0$ and $a$, will give you the value of $\int_0^b f(x)\,dx$. If you have such a master function, then you can use it to find the value of the integral you want just by taking $\mathcal{M}(a)$!

In fact, this is the approach Barrow had taken: his great insight was that instead of trying to find the quadrature a particular area, he was trying to solve the problem of squaring several different (but related) areas at the same time. So he was looking for, for instance, a "master function" for the region was like a triangle except that the top was a parabola instead of a line (like the area from $0$ to $a$ under $y=x^2$), and so on.

On its face, this is a completely ludicrous suggestion. It's like telling someone who is trying to know how to get from building A to building B that if he only memorizes the map for the entire city first, then he can use that knowledge to figure out how to get form A to B. If we are having trouble finding the integral $\int_0^a f(x)\,dx$, then the "master function" seems to require us to find not just that area, but also all areas in between! It's like telling someone who is having trouble walking that he should just run very slowly when he wants to walk.

But, again, the interesting thing is that even though we may not be able to say what the "master function" is, we can say how it changes as b changes (remember, $\mathcal{M}(b) = \int_0^b f(x)\,dx$ is a number that depends on $b$, so $\mathcal{M}$ is a function of $b$). Because figuring out how functions change is easier than computing their values (just think about derivatives, and how we can easily figure out the rate of change of $\sin(x)$, but we have a hard time actually computing specific values of $\sin(x)$ that are not among some very simple ones). (This is also something Barrow already knew, as did Newton).

For "nice functions" (if $f$ is continuous on an interval that contains $0$ and $a$), we can do it using limits and some theorems about "nice" functions: Using limits, we have: \begin{align} \lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}}{h} &= \lim_{h\to 0} \frac{1}{h}\left(\int_0^{b+h}f(x)\,dx - \int_0^bf(x)\,dx\right)\\\ &= \lim_{h\to 0}\frac{1}{h}\int_b^{b+h}f(x)\,dx. \end{align} Since we are assuming that $f$ is continuous on $[0,a]$, it is continuous on the interval with endpoints $b$ and $b+h$ (I say it this way because $h$ could be negative). So it has a maximum and a minimum (continuous function on a finite closed interval). Say the maximum is $M(h)$ and the minimum is $m(h)$. Then $m(h) \leq f(x) \leq M(h)$ for all $x$ in the interval, so we know, since the integral is the area, that $$hm(h) \leq \int_b^{b+h}f(x)\,dx \leq hM(h).$$ That means that $$m(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq M(h)\text{ if h\gt 0}$$ and $$M(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq m(h)\text{ if h\lt 0.}$$

As $h\to 0$, the interval gets smaller, the difference between the minimum and maximum value gets smaller. One can prove that both $M$ and $m$ are continuous functions, and that $m(h)\to f(b)$ as $h\to 0$, and likewise that $M(h)\to f(b)$ as $h\to 0$. So we can use the Squeeze Theorem to conclude that since the limit of $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ is squeezed between two functions that both have the same limit as $h\to 0$, then $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ also has a limit as $h\to 0$ and is in fact that same quantity, namely $f(b)$. That is $$\frac{d}{db}\mathcal{M}(b) = \lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}(b)}{h} = \lim_{h\to 0}\frac{1}{h}\int_{b}^{b+h}f(x)\,dx = f(b).$$

That is: when $f$ is continuous, the "Master function" for areas turns out to have a rate of change equal to $f$. This is not that crazy, if you think about it: how is the area under $y=f(x)$ from $x=0$ to $x=b$ changing? Well, it's changing by whatever $f$ is.

This means that, whatever the "Master function" turns out to be, it will be an antiderivative of $f(x)$.

We also know, because we are very good with derivatives, that if $\mathcal{F}(x)$ and $\mathcal{G}(x)$ are two functions, and $\mathcal{F}'(x) = \mathcal{G}'(x)$ for all $x$, then $\mathcal{F}$ and $\mathcal{G}$ differ by a constant: there exists a constant $k$ such that $\mathcal{F}(x) = \mathcal{G}(x)+k$ for all $x$.

So, we know that the "Master function" is an antiderivative. If, by some sheer stroke of luck, we happen to find any antiderivative $F(x)$ for $f(x)$, then we know that the only possible difference between $\mathcal{M}(b)$ and $F(b)$ is a constant. What constant? Well, luckily we know one value of $\mathcal{M}(b)$: we know that $\mathcal{M}(0) = \int_0^0f(x)\,dx$ should be $0$. So, $M(0) = 0 = F(0)-F(0)$, which means the constant has to be $-F(0)$. That is, we must have $M(b) = F(b)-F(0)$ for all $b$.

So, if we find any antiderivative $F$ of $f$, then $\mathcal{M}(b) = F(b)-F(0)$ is in fact the "Master function" we were looking for, the one that gives all the integrals between $0$ and $a$, including $0$ and including $a$. So that we have that two very different processes (computing areas using limits of Riemann sums, and derivatives) are connected: if $f(x)$ is continuous, and $F(x)$ is any antiderivative for $f(x)$ on $[0,a]$, then $$\int_0^a f(x)\,dx = \mathcal{M}(a) = F(a)-F(0).$$

But the integral did not "magically turn" into an antiderivative. It's that the "Master function" which can be used to keep track of all integrals of $f(x)$ has rate of change equal to $f$, which gives us a "back door" to computing integrals.

Newton was able to prove this because he had the guide of Barrow's insight that this was happening for the functions he worked with. Barrow's insight was achieved because he had the brilliant idea of trying to come up with a "Master function" instead of trying to rectify lots of different areas one at a time, and he noticed the connection because he had already worked with tangents/derivatives for those functions. Leibniz likewise had access to Barrow's ideas, so the connection between the two was also known to him.

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Just a tiny note: "finding a way to construct a rectangle that had the same area as the region you were considering" would be what they called in those days "quadrature". Otherwise, this answer is a work of Art, as expected. :) – J. M. Dec 23 '10 at 5:49
It was intended to be a (horrible) pun, but sure, I won't do it again. ;) – J. M. Dec 23 '10 at 6:11
@J.M.: Oh, I got the pun all right (hence the winking smiley, and I did see your smiley). Just wanted to nip any possibility of the nickname getting picked up in the bud. – Arturo Magidin Dec 23 '10 at 6:12
That was awesome! – MyUserIsThis Feb 24 '13 at 20:07
i read it...i got lost pretty fast – Muhammad Umer Mar 17 '14 at 0:41

One way you can perhaps "justify"/give an intuitive reason is to consider the following figure:

$A(x)$ is the area under the curve from $0$ to $x$, the brown region.

$A(x+dx)$ is the area under the curve from $0$ to $x + dx$, the brown + gray.

Now for for really small $dx$, we can consider the gray region to be a rectangle of side width $dx$ and height $f(x)$.

Thus $\dfrac{A(x+dx) - A(x)}{dx} = f(x)$.

Thus as $dx \to 0$, we see that $A'(x) = f(x)$.

It is kind of intuitive to define area by approximating by very thin rectangles. The above gives an intuition as to why the derivative of the area gives the curve.

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+1.. For graphics and explanation. – night owl Apr 17 '11 at 5:29
how does this make F(x) = A(x) – Muhammad Umer Mar 24 '14 at 20:23
@MuhammadUmer: Because of fundamental theorem of calculus. See this:en.wikipedia.org/wiki/… (and I suggest you read the whole page). – Aryabhata Mar 25 '14 at 0:23
thanks, i've read. Why does the proof of 1st part, For a given f(t), define the function F(x) as.., start with assumption that F(x) or integral = area..because that is what it's made equivalent to when mean value theorem is used which is area. – Muhammad Umer Mar 25 '14 at 1:56
@MuhammadUmer: I don't understand what you are trying to say. The second part applies here. – Aryabhata Mar 25 '14 at 22:13

I strongly recommend that you take a look at the first chapter of Gilbert Strang's Calculus textbook: http://ocw.mit.edu/resources/res-18-001-calculus-online-textbook-spring-2005/textbook/MITRES_18_001_strang_1.pdf. This chapter provides an insightful introduction to integration that likely takes an approach that is very different from your professor's.

A typical explanation of integration is as follows:

We want to know the area under a curve. We can approximate the area under a curve by summing the area of lots of rectangles, as shown above. It is clear that with hundereds 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 animation may help with the intuition,

We define the integral to be the limit described and depicted above:

$\int _{ a }^{ b }{ f(x)dx=\lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ f({ x }_{ i })\Delta x } } }$

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Question is asking about why of what you showed and stopped there with last part. Somehow infinite triangles equals the difference in antiderivatives??? – Muhammad Umer Mar 16 '14 at 22:58

heres a very simple explanation of how an integral works. Forget about things needed for the proof. This is just for the concept

take the function f'(x) and its antidervative, f(x)

We can find the area under a graph by taking the average of all the y values and multiplying by Delta X, creating a rectangle of equal area.

 (1)Area = Yaverage of f'(x)* DeltaX
average y for f'(x) is average slope of f(x)


finding average slope of f(x) is easy. Simply go to the original function and do the average slope calculation

 (2)average slope =(y1-y2)/(deltaX)


Now substitute formula 2 into 1:

  area of f'(x) = (y1-y2)/delaX  * deltaX
area of f'(x) = (y1-y2)


y1 = f(x1)

 area of f'(x)=f(x1)-f(x2)


Took me a while to figure this out. IMO the mean sum of tiny squares hinders understanding. mean value theorem is good to find average slope. Its convoluted and imprecise though when people say there is a value c that does this or that. Irrelevant to finding the average slope. Just supply the formula for finding average slope. Best of luck

Such a simple concept to be so over complicated. I hope they change the way it is taught in canadian schools

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I remember thinking about this problem in high school. Here is the intuitive explanation that popped into my head one morning before I got out of bed. I am assuming the question you are asking is why the antiderivative evaluated at the endpoints gives you the area under the curve.

Let $f(x)$ be a function on an interval $[a,b]$. We want to calculate $\int_a^b f(x)dx$, i.e. the area under the curve of $f$ from $a$ to $b$. Intuitively, $\int_a^b f(x)dx$ is the infinite sum of all rectangles with height $f(x)$ and width $dx$, and we are summing over all $x$ in the interval $[a,b]$. Let's take for granted that this expression $\int_a^b f(x)dx$ really is referring to the area under the curve.

Let $y = F(x)$ be an antiderivative of $f$. So $\frac{dy}{dx} = f$.

Choose any finite sequence of points $a = x_0 \leq \cdots \leq x_t =b$. We have the 'net change' $F(b) - F(a)$, and on the other hand we have the 'small changes' $F(x_1) - F(a), F(x_2) - F(x_1)$ etc. But if you add all the small changes:

$$[F(x_1) - F(a)] + [F(x_2) - F(x_1)] + \cdots + [F(b) - F(x_{t-1})]$$ you get the net change, $F(b) - F(a)$. The sum of the small changes is the net change. Now imagine infinitely many points there. Each small change, which I might have written as $\Delta y$, is now written as $dy$ (this is the notational tradition in calculus). And again, the net change is the (now infinite) sum of the (infinitely small) small changes: $$F(b) - F(a) = \int_a^b dy$$ But $$\int_a^b dy = \int_a^b \frac{dy}{dx}dx = \int_a^b f(x)dx$$

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There are of course many logical difficulties with this explanation, in particular the problem of summing up infinitely (uncountably many) terms, as well as the general problem of declaring "since this works for finitely many things, it must also be true for infinitely many things." If you study real analysis, you will learn what is "really" going on. – D_S May 26 at 5:41
But I still think this is a good intuitive way to think about what the fundamental theorem of calculus is saying. When I think FTOC, I think "The net change is the sum of the small changes." – D_S May 26 at 5:42

The main concept which make relation between integral and derivative is mean value theorem: $$F(b)-F(a)=(b-a)f'(c), c\in (a,b)$$ by using this theorem we can answer the desired question.

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For all smooth functions the Taylor formula applies: $$f(x+\varepsilon)=f(x)+\varepsilon f'(x)$$ including $A(x)$ as the area function (for another function). So: $$A(x+\varepsilon)-A(x)=\varepsilon A'(x)$$ $$\varepsilon f(x)+\frac{1}{2}\varepsilon.\varepsilon f'(x)=\varepsilon A'(x)$$ where LHS1 is the rectangle area and LHS2 is the triangle area. LHS2 is now neglected: $$\varepsilon f(x)=\varepsilon A'(x)$$ $$f(x)=A'(x)$$ $$\int f(x)=A(x)$$

NB Please see the link for an explanation of why neglecting nilsquare infinitesimals is entirely in accordance with limit theory. This note was prompted by statements to the effect that proofs using older techniques should not be allowed due to an alleged lack of 'rigour'.

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This is not correct. It should be $f(x+\varepsilon)=f(x)+\varepsilon f'(\xi)$ where $\xi$ is a number between $x$ and $x+\varepsilon$. – BigbearZzz May 26 at 2:41
You realize that $f'(x)$ is defined by this equation? – selfawareuser May 26 at 2:43
Do you realize that you are making no sense? Unless you are talking about the non-standard real number $\varepsilon$ but still, that would still be inaccurate. – BigbearZzz May 26 at 2:45
Feel free to explain yourself. – selfawareuser May 26 at 2:46
To whom? To you or to the OP? If you can't understand that your answer are just at best heuristic or at worst blatantly wrong, then I guess there's no helping you. – BigbearZzz May 26 at 2:47

Idea behind definite integral (area under the curve) comes directly from derivative definition. Let's bring latter definition:

$$f(x) = \lim_{\Delta x\to 0}{\frac{\Delta F(x)}{\Delta x}} \implies f(x) = \frac{dF(x)}{dx}$$

Multiply both sides by $dx$:

$$f(x)dx = dF(x)$$

From the above, area of infinitely narrow ($dx$ wide) stripe $f(x)dx$ is equal to $dF(x)$.

The task is to find the area of all such stripes in range of $x \in \{A, B\}$.

Given: $dF(x) = F(x + dx) - F(x) \\$ \left\{ \begin{aligned} f(x )dx = F(x + dx) - F(x ) \\ f(x + dx)dx = F(x + 2 dx) - F(x + dx) \\ f(x + 2 dx)dx = F(x + 3 dx) - F(x + 2 dx) \\ \cdots \\ f(x + N dx)dx = F(x + (N+1) dx) - F(x + N dx) \end{aligned} \right.

Sum up above group of equations:

$$f(x)dx + f(x + dx)dx + f(x + 2 dx)dx + \cdots + f(x + N dx)dx \\ = F(x + dx) - F(x ) + F(x + 2 dx) - F(x + dx) + F(x + 3 dx) - F(x + 2 dx) + \cdots + F(x + (N+1) dx) - F(x + N dx)$$

As a result, we end up with definite integral (of single variable function), which in fact corresponds to the area, we've been looking for:

$$\sum_{n=0}^N f(x + n dx)dx \\= F(x + (N+1) dx) - F(x) \\= F(B) - F(A)$$

Sum of all stripes in given range is equal to the difference between its antiderivative on both ends.

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This is just a bunch of formulas without any explanation. I don't say it is wrong, but I don't see how it could help OP either... – mickep Dec 1 '15 at 14:18