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I've known for some time that one of the fundamental theorems of calculus states:

$$ \int_{a}^{b}\ f'(x){\mathrm{d} x} = f(b)-f(a) $$

Despite using this formula, I've yet to see a proof or even a satisfactory explanation for why this relationship holds. Any ideas?

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    $\begingroup$ khanacademy.org/math/integral-calculus/… $\endgroup$
    – jimjim
    Nov 20, 2015 at 1:26
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    $\begingroup$ It only works with Euclidean metric, there are many hidden assumptions that I have not seen stated explicitly, this is special case of more generalized theorems, additional assumptions simplify it to the form known as FTC, If you are really interested in the Whys, then look up real and complex analysis. $\endgroup$
    – jimjim
    Nov 20, 2015 at 1:29
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    $\begingroup$ @Arjang, I'm aware of two versions of the fundamental theorem of calculus, and neither one requires a metric on the space you're integrating over. One version, Stokes' theorem, works on any orientable smooth manifold (en.wikipedia.org/wiki/…). The other version, the Lebesgue fundamental theorem of calculus, works on intervals, and uses the Lebesgue measure. $\endgroup$
    – Vectornaut
    Nov 20, 2015 at 6:24
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    $\begingroup$ @Arjang: There are no metrics involved here. $\mathrm{d} x$ is a "change in the real variable $x$" (one way to be more precise is a differential form), not any sort of measurement of distance. $\endgroup$
    – user14972
    Nov 20, 2015 at 6:35
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    $\begingroup$ @Arjang: That's an issue of setting up the integral, not the meaning of the integral once you've written it. If the "change in area" is not $f(x) \mathrm{d} x$, then integrating $f(x) \mathrm{d} x$ doesn't give you the area under the graph. For a graph $r = f(\theta)$ in polar coordinates, for example, the "change in area under the graph" is $\frac{1}{2} f(\theta)^2 \mathrm{d} \theta$. But if $\frac{1}{2} f(\theta)^2 = g'(\theta)$, then you still get $\int_0^{2 \pi} \frac{1}{2} f(\theta)^2 \mathrm{d} \theta = g(2\pi) - g(0)$. $\endgroup$
    – user14972
    Nov 20, 2015 at 7:11

6 Answers 6

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Intuitively, the fundamental theorem of calculus states that "the total change is the sum of all the little changes". $f'(x) \, dx$ is a tiny change in the value of $f$. You add up all these tiny changes to get the total change $f(b) - f(a)$.

In more detail, chop up the interval $[a,b]$ into tiny pieces: \begin{equation} a = x_0 < x_1 < \cdots < x_N = b. \end{equation} Note that the total change in the value of $f$ across the interval $[a,b]$ is the sum of the changes in the value of $f$ across all the tiny subintervals $[x_i,x_{i+1}]$: \begin{equation} f(b) - f(a) = \sum_{i=0}^{N-1} f(x_{i+1}) - f(x_i). \end{equation} (The total change is the sum of all the little changes.) But, $f(x_{i+1}) - f(x_i) \approx f'(x_i)(x_{i+1} - x_i)$. Thus, \begin{align} f(b) - f(a) & \approx \sum_{i=0}^{N-1} f'(x_i) \Delta x_i \\ & \approx \int_a^b f'(x) \, dx, \end{align} where $\Delta x_i = x_{i+1} - x_i$.

We can convert this intuitive argument into a rigorous proof. It helps a lot that we can use the mean value theorem to replace the approximation $f(x_{i+1}) - f(x_i) \approx f'(x_i) (x_{i+1} - x_i)$ with the exact equality $f(x_{i+1}) - f(x_i) = f'(c_i) (x_{i+1} - x_i)$ for some $c_i \in (x_i,x_{i+1})$. This gives us \begin{align} f(b) - f(a) & =\sum_{i=0}^{N-1} f'(c_i) \Delta x_i. \end{align} Given $\epsilon > 0$, it's possible to partition $[a,b]$ finely enough that that the Riemann sum $\sum_{i=0}^{N-1} f'(c_i) \Delta x_i$ is within $\epsilon$ of $\int_a^b f'(x) \, dx$. (This is one definition of Riemann integrability.) Since $\epsilon > 0$ is arbitrary, this implies that $f(b) - f(a) = \int_a^b f'(x) \, dx$.

The fundamental theorem of calculus is a perfect example of a theorem where: 1) the intuition is extremely clear; 2) the intuition can be converted directly into a rigorous proof.

Background knowledge: The approximation $f(x_{i+1}) - f(x_i) \approx f'(x_i) (x_{i+1} - x_i)$ is just a restatement of what I consider to be the most important idea in calculus: if $f$ is differentiable at $x$, then \begin{equation} f(x + \Delta x) \approx f(x) + f'(x) \Delta x. \end{equation} The approximation is good when $\Delta x$ is small. This approximation is essentially the definition of $f'(x)$: \begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}. \end{equation} If $\Delta x$ is a tiny nonzero number, then we have \begin{align} & f'(x) \approx \frac{f(x + \Delta x) - f(x)}{\Delta x} \\ \iff & f(x + \Delta x) \approx f(x) + f'(x) \Delta x. \end{align} Indeed, the whole point of $f'(x)$ is to give us a local linear approximation to $f$ at $x$, and the whole point of calculus is to study functions which are "locally linear" in the sense that a good linear approximation exists. The term "differentiable" could even be replaced with the more descriptive term "locally linear".

With this view of what calculus is, we see that calculus and linear algebra are connected at the most basic level. In order to define "locally linear" in the case where $f: \mathbb R^n \to \mathbb R^m$, we first have to invent linear transformations. In order to understand the local linear approximation to $f$ at $x$, which is a linear transformation, we have to invent linear algebra.

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  • $\begingroup$ If you're going to answer like this, then you should explain why $f'(x)dx$ is considered a tiny change in the value of $f$. Shouldn't a tiny change look like $f(x + \Delta x) - f(x)$? Or better yet: $f(x + dx) - f(x)$? Why is $f'(x)dx$ interpreted as a tiny change? $\endgroup$
    – layman
    Nov 20, 2015 at 1:35
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    $\begingroup$ I think I already know the answer anyway. Intuitively, $f'(x) = \dfrac{f(x + dx) - f(x)}{dx}$, so that $f'(x)dx = \dfrac{f(x + dx) - f(x)}{dx}dx = f(x + dx) - f(x) = $ a tiny change in the value of $f$. $\endgroup$
    – layman
    Nov 20, 2015 at 1:37
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    $\begingroup$ You need to add to this answer the more important idea! The convergence of the infinite sum when \Delta \to 0. This is IMHO key to understand why the equality holds, the sense of convergence. And I will add too the demonstration why the sum of partitions is equal to the whole difference. $\endgroup$
    – Masacroso
    Nov 20, 2015 at 3:06
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    $\begingroup$ +1 for invoking the Mean Value Theorem. My analysis professor taught me that it is the Most Important Theorem in Calculus. Like you said, all statements which use $f'$ to infer something about $f$ involve MVT. $\endgroup$ Nov 21, 2015 at 9:51
  • $\begingroup$ f(x) is decided to be a tiny change as that is mostly just assumed under calculus due to how calculus works, it is basically all just working with very small changes and intervals, adding up to a non-trivial change. $\endgroup$
    – Zoey
    Dec 25, 2021 at 2:46
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Others have said that the total change is the sum of the infinitely many infinitely small changes, and I agree. I will add another way of looking at it.

Think of $\displaystyle A = \int_a^x f(t) \, dt$, and imagine $x$ moving. Draw the picture, showing the $t$-axis, the graph of $f$, the vertical line at $t=a$ that forms the left boundary of the region whose area is the integral, and the vertical line at $t=x$ forming the right boundary, which is moving.

Now bring in what I like to call the "boundary rule":

[size of boundary] $\times$ [rate of motion of boundary] $=$ [rate of change of area]

The size of the boundary is $f(x)$, as you see from the picture described above.

The rate of motion of the boundary is the rate at which $x$ moves.

Therefore, the area $A$ is changing $f(x)$ times as fast as $x$ is changing; in other words: $$ \frac{dA}{dx} = f(x). $$ That is the fundamental theorem. It tells you that in order to find $A$ when you know $f(x)$, you need to find an anti-derivative of $f(x)$.

The "boundary rule" also has some other nice consequences:

  • Imagine a growing sphere with changing radius $r$ and surface area $A$. The size of the boundary is $A$; the rate at which the boundary moves is the rate at which $r$ changes. Therefore the volume $V$ is changing $A$ times as fast as $r$ is changing. In other words $\dfrac{dV}{dr} = A$. That tells you the surface area is $4\pi r^2$ if you already knew that the volume was $\dfrac 4 3 \pi r^3$.

  • Imagine a cube whose side has length $x$, so the volume is $x^3$. It sits on the floor in the southwest corner of a room, so that its south, west, and bottom faces stay where they are and its north, east, and top faces move at the rate at which $x$ changes. Each of those $3$ faces has area $x^2$, so their total area is $3x^2$. The size of the moving boundary is $3x^2$ and the rate of motion of the boundary is the rate at which $x$ moves. In other words, this tells you that $\dfrac d {dx} x^3 = 3x^2$. And this generalizes to higher dimensions to explain why $\dfrac d{dx} x^n = nx^{n-1}$.

  • The north side of a rectangle has length $f$ and the east side has length $g$. The south and west sides are fixed and cannot move, so when $f$ and $g$ change, only the north and east sides move. The north side moves if the length of the east side changes, and the east side moves if the length of the north side changes. The rate of motion of the north side is the rate of change of the east side, so it is $g'$. The size of the north side is $f$. So the size of the boundary times the rate at which the boundary moves is $f \cdot g'$. And if they both move, the total rate of change of area is $f\cdot g' + f'\cdot g$. That must then be the rate of change of area, $(fg)'$. Hence we have the product rule.

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    $\begingroup$ That's true, although I was looking for a proof of the other fundamental theorem of calculus. $\endgroup$
    – tethernova
    Nov 20, 2015 at 1:58
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To combine asymptotic analysis with nonstandard analysis.

By the definition of derivative,

$$ f'(x) = \frac{f(x + \epsilon) - f(x)}{\epsilon} + o(1) $$

($o(1)$ means the error is infintiesimal)

If $H$ is a positive, infinite, nonstandard integer, then by the left endpoint rule, using the shorthand $\xi_i = a + i (b-a)/H$,

$$ \begin{align}\int_a^b f'(x) \, \mathrm{d}x &= \sum_{i=0}^{H-1} (\xi_{i+1} - \xi_i) f'\left(\xi_i \right) + o(1) \\&= \sum_{i=0}^{H-1} (\xi_{i+1} - \xi_i) \left(\frac{f(\xi_{i+1}) - f(\xi_i)}{\xi_{i+1} - \xi_i} + o(1)\right) + o(1) \\&= \sum_{i=0}^{H-1} \left(f(\xi_{i+1}) - f(\xi_i) + o\left(\frac{b-a}{H}\right) \right) + o(1) \\&= f(\xi_H) - f(\xi_0) + o(1) \\&= f(b) - f(a) \end{align}$$

where the very last step follows because both sides are standard, and so the infinitesimal difference must be zero.

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  • $\begingroup$ +1, but it would be good to have some brackets in the first sum so it was clear you're not summing that o(1) $\endgroup$ Nov 20, 2015 at 10:54
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There are really two FTCs. One is what you have written. The other is

$$\frac{d}{dx} \int_a^x f(y) dy = f(x)$$

for continuous $f$.

The latter is easier to understand. If you replace $x$ by $x+\Delta x$ for small positive $\Delta x$, then you add area which is "well-approximated" by a rectangle of height $f(x)$ and width $\Delta x$. You can intuitively justify this by just drawing a picture. In the rigorous proof you have to play with the errors to ensure the property above.

The FTC that you have written is a bit more difficult to understand. One way of looking at it is to consider the Riemann sum

$$\sum_{i=0}^{n-1} f' \left ( a + i \frac{b-a}{n} \right ) \frac{b-a}{n}.$$

On the one hand, this is a Riemann sum for $\int_a^b f'(x) dx$. On the other hand, this amounts to adding up approximations to the change in $f$ over $[a,b]$ by following the tangent line at $n$ points. Since the tangent line is the best possible linear approximation, you can hope that this approximation should be pretty good, at least if $n$ is large. And again, in the rigorous proof you have to play with error bounds to ensure that as $n \to \infty$ you actually get $f(b)-f(a)$.

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  • $\begingroup$ It's maybe worth noting that the FTC as written by OP is easily equivalent to the one you commented if we assume $f'$ is continuous. So the only difficulty is in extending to the case where $f'$ is only Riemann integrable. It's curious and nice that it works, but for intuition purposes, I'm happy with assuming $C^1$ functions. $\endgroup$
    – Pedro
    Nov 21, 2015 at 9:27
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    $\begingroup$ @Pedro I somewhat disagree. I would actually argue that the second FTC, or rather its conclusion, is stronger. In the $C^1$ case for instance, the second FTC tells you $\int_a^x f'(y) dy = f(x)-f(a)$, then you recover the first FTC by differentiating both sides. By contrast, it is not obvious to me how to directly conclude the second FTC from the first FTC, even in the $C^1$ case. Similarly, in the Lebesgue context, the second FTC requires a stronger hypothesis about $f$ than the first one. I do agree that for purposes of intuition, understanding just the second FTC is sufficient, however. $\endgroup$
    – Ian
    Nov 24, 2015 at 20:01
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By definition, we know (other definitions exist, but if the functions are smooth, these definitions are equivalent):

$$f'(x) = \lim_{h\to 0} {f(x+h) - f(x) \over h}$$

and

$$ \int_a^b g(x)\, dx = \lim_{h\to 0} \sum_{i=0}^{(a-b)/ h-1} h g(a + ih)$$

Combining,

$$ \begin{align} \int_a^b f'(x)\, dx &= \lim_{h\to 0} \sum_{i=0}^{(a-b)/ h-1} h f'(a + ih)\\ &= \lim_{h\to 0} \sum_{i=0}^{(a-b)/ h-1} h \times {f(a + ih + h) - f(a + ih) \over h}\\ &= \lim_{h\to 0} \sum_{i=0}^{(a-b)/ h-1} {f(a + (i+1)h) - f(a + ih)}\\ &= f(b) - f(a) \end{align} $$

Literally all the other terms just cancel out.

PS: Yes, I recognize my answer is virtually the same as Hurkyl's, but with minor change of details.

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    $\begingroup$ Mulling over the details, I think going from the first to the second line of "combining" is actually a rather nontrivial step. The overall idea is fine, of course, just that the detail of justifying that step is tricky to get right. $\endgroup$
    – user14972
    Nov 21, 2015 at 17:31
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The fundamental theorem of calculus has a nice physics analogy. Suppose that $v(t)$ is the velocity function of a particle. To compute the displacement between times $t=a$ and $t=b$, you have to work out the area under the graph of the velocity function. Symbolically, $$ s(b)-s(a)=\int_{a}^{b}v(t) \, dt \, , $$ where $s(t)$ is the particle's displacement function.

How might we obtain this result? We know from the formula $$ \text{displacement} = \text{velocity} \cdot \text{time} $$ that to compute the displacement over some time interval we have to compute the average velocity over that interval, and then multiply by the time taken. To approximate the average velocity, we could partition the interval $[a,b]$ into $n$ subintervals of width $\delta t$. Then, we could take 'samples' of the velocity function at regular intervals to estimate the average: Samples In the above graph, for instance, I've divided the velocity function into $10$ intervals, and estimated the mean value by summing the $10$ rightmost $y$-values and then dividing by $10$. In general, the average velocity can be estimated as $$ \text{average velocity}\approx\frac{\sum_{i=1}^{n}f(a+i\delta t)}{n} \, , $$ with this approximation turning into exactly equality when we take the limit of the above quotient as $n$ tends to infinity. The time taken is obviously $b-a$, meaning that $$ \text{displacement} = \lim_{n \to \infty}\frac{b-a}{n}\sum_{i=1}^{n}f(a+i\delta t) \, . $$ Keeping in mind that $$ \delta t=\frac{\text{interval width}}{\text{no. of partitions}}=\frac{b-a}{n} \, , $$ we obtain $$ \text{displacement} = \lim_{n \to \infty}\sum_{i=1}^{n}f(a+i\delta t) \cdot \delta t $$ But this is virtually the definition of $$ \int_{a}^{b}v(t) \, dt \, , $$ meaning that $$ \int_{a}^{b}v(t) \, dt = s(b) - s(a) \, . $$

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