# Iconic image to explain the fundamental theorem of calculus?

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.

• This image in the same article. – user5402 Oct 13 '18 at 15:33
• @BPP: Yes, that is essentially James Cook's first figure illustrating FTC I. – Joseph O'Rourke Oct 13 '18 at 15:34
• 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. – littleO Oct 14 '18 at 11:01
• Incidentally, I admit I am quite puzzled by the distinction FTC I and FTC II. – Benoît Kloeckner Oct 17 '18 at 13:48

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)

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:

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

• 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. – Xander Henderson Oct 13 '18 at 16:08
• @XanderHenderson no complaint here. Thanks for the adjustment. – James S. Cook Oct 14 '18 at 0:06

This is about the Riemann sums but the logic is there.

• I never get the $4$-long block when I need it. This is interesting. – James S. Cook Oct 16 '18 at 0:43

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 $$f$$.

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.)

• 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... – James S. Cook Oct 18 '18 at 5:11
• @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 $dy$). – Michael Bächtold Oct 18 '18 at 6:31

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.

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 $$\displaystyle{\int_0^2 f (x)\, dx}$$

• Great problem!!! – Steven Gubkin Oct 14 '18 at 14:51