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Is there some single, persuading visualization that can be used to convince students of the intuitive truth of the fundamental theorem of calculus, in the form $$ \int_a^b f(t) \, dt = F(b) - F(a) \;? $$ I'm not seeking a "proof-without-words," but rather an image that, when supplemented by an appropriate verbal explanation, is quite convincing. The Wikipedia image doesn't quite do it for me.

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    $\begingroup$ This image in the same article. $\endgroup$ – Paracosmiste Oct 13 '18 at 15:33
  • $\begingroup$ @BPP: Yes, that is essentially James Cook's first figure illustrating FTC I. $\endgroup$ – Joseph O'Rourke Oct 13 '18 at 15:34
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    $\begingroup$ My intuition for the second fundamental theorem of calculus is "the total change is the sum of all the little changes". $f'(x) dx$ is a tiny change in the value of $f$. We sum up all these tiny changes to get the total change $f(b) - f(a)$. I think it would be possible to illustrate this idea with a picture. $\endgroup$ – littleO Oct 14 '18 at 11:01
  • $\begingroup$ Incidentally, I admit I am quite puzzled by the distinction FTC I and FTC II. $\endgroup$ – Benoît Kloeckner Oct 17 '18 at 13:48
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I'd make this a comment, but I'm not certain I know how. Maybe this is the picture to lead students to see FTC I (I think of your theorem as FTC II, but I know books vary on this terminology)

enter image description here

Hopefully this leads students to see $\frac{dA}{dx} = f(x)$ where $A(x) = \int_a^x f(t) \, dt$. I'm not sure what the picture is for FTC II.

EDIT: I thought about FTC II a bit and here is my best stab at it currently: If $F$ is an antiderivative of $f$ then $\frac{dF}{dx} = f$. Apply the Mean Value Theorem on $[x_{i-1}, x_i]$ to select $x_i^* \in [x_{i-1}, x_i]$ for which $$ \frac{F(x_i)-F(x_{i-1})}{x_i-x_{i-1}} = f(x_i^*) $$ Then if we set $x_i-x_{i-1} = \Delta x$ we find $$ F(x_i)-F(x_{i-1}) = f(x_i^*) \Delta x $$ Therefore, the signed-area $A_i$ bounded by $y=f(x)$ on $[x_{i-1},x_i]$ is well approximated by: $$ A_i = f(x_i^*) \Delta x = F(x_i)-F(x_{i-1}) $$ Notice the signed area under $y=f(x)$ on $[a,b]$ we defined to be $\int_a^b f(x)dx$ and it is well approximated by $A_1+A_2+ \cdots + A_n$. Thus, \begin{align} \int_a^b f(x)dx &= A_1+A_2+ \cdots + A_n \\ &= F(x_1)-F(x_o)+F(x_2)-F(x_1)+ \cdots + F(x_{n})-F(x_{n-1}) \end{align} But, we set-up the partition of $[a,b]$ with $x_o=a$ and $x_n=b$ hence, $$ \int_a^b f(x)dx = F(x_n)-F(x_o) = F(b)-F(a). $$ My picture for the argument above at the moment is: enter image description here

I guess a proof by picture alone must somehow picture both $f$ and $F$ and somehow connect the area under $y=f(x)$ to the secant line for $y=F(x)$. I'm not sure I have a good generic picture for that. Certainly I can do it for simple graphs such as $y = constant$ or even $y = x$, but I'm not sure that is what we're after here. See pages 231-232 of my calculus notes

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  • $\begingroup$ James: I've replaced your triangle with Deltas (`\Delta'), which is the more common notation. If you don't like it, please rollback my edit. $\endgroup$ – Xander Henderson Oct 13 '18 at 16:08
  • $\begingroup$ @XanderHenderson no complaint here. Thanks for the adjustment. $\endgroup$ – James S. Cook Oct 14 '18 at 0:06
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This is about the Riemann sums but the logic is there. enter image description here

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    $\begingroup$ I never get the $4$-long block when I need it. This is interesting. $\endgroup$ – James S. Cook Oct 16 '18 at 0:43
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Here’s a picture that illustrates the comment of littleO (with enough time staring at it). It uses the graph of $F$, while the pictures in the answer of James S. Cook use the graph of $f$.

FTCII

I’m using FTC II in the form $$ \int_{x=a}^b \frac{dy}{dx}dx =y|_{x=b} - y|_{x=a} $$ with $y=F(x)$, $\frac{dy}{dx}=f(x)$.

(It becomes even more obvious if one writes $dy$ for $\frac{dy}{dx}dx$ in the integral. This is mathematically correct, but it seems that many people feel uncomfortable doing this.)

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  • $\begingroup$ So, the continuous summation of the slope (rise/run) gives the net rise. Each little rise is seen to be the slope times the little run, but the slope of $F$ is $f$ hence this gives the integral. I feel like I'm still missing something from the picture, I need to stare longer... $\endgroup$ – James S. Cook Oct 18 '18 at 5:11
  • $\begingroup$ @JamesS.Cook. I would say "the continuous summation of the little rises gives the net rise". The integral is not summing the slope $dy/dx$ but the rise $dy/dx\cdot dx$ (which is just $dy$). $\endgroup$ – Michael Bächtold Oct 18 '18 at 6:31
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Try the Derek Owens YouTube playlist for the FTC.

He starts with a few worked examples, then goes into the intuition for why the FTC is true. As far as visuals go, this is quite good.

It is not a single, iconic image, but it is still very good, nonetheless, at getting across the basic idea.

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This may not be the iconic image for the introduction of the fundamental theorem of calculus, but I think it is the iconic image of Calculus Reform.

The problem associated to the image is to use FTC to determine the sign of $\displaystyle{\int_0^2 f''(x)\, dx}$, $\displaystyle{\int_0^2 f'(x)\, dx}$, and $\displaystyle{\int_0^2 f (x)\, dx}$

NewCenturyIconiImage

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  • $\begingroup$ Great problem!!! $\endgroup$ – Steven Gubkin Oct 14 '18 at 14:51

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