15
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ This image in the same article. $\endgroup$
    – user5402
    Oct 13, 2018 at 15:33
  • $\begingroup$ @BPP: Yes, that is essentially James Cook's first figure illustrating FTC I. $\endgroup$ Oct 13, 2018 at 15:34
  • 1
    $\begingroup$ 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, 2018 at 11:01
  • 1
    $\begingroup$ Incidentally, I admit I am quite puzzled by the distinction FTC I and FTC II. $\endgroup$ Oct 17, 2018 at 13:48

6 Answers 6

16
$\begingroup$

This is about the Riemann sums but the logic is there. enter image description here

$\endgroup$
0
9
$\begingroup$

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

$\endgroup$
0
3
$\begingroup$

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.)

$\endgroup$
2
  • $\begingroup$ 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... $\endgroup$ Oct 18, 2018 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 $dy$). $\endgroup$ Oct 18, 2018 at 6:31
2
$\begingroup$

Below is an image that shows the net change in $F$ over each interval as a (finite) "differential," which is the same in this case as a term in a Riemann sum for the integral. The net change is given by the change in $y$ for the black, dashed secant line and is equal to the change in $y$ along the purple, solid tangent line obtained via the Mean Value Theorem. The sum of the net changes over all intervals gives $F(b)-F(a)$ and is equal to the sum of the differentials, which is a Riemann sum. The appropriate limit is the integral of $F'=f$.

enter image description here

Here is an animation of the same idea, with different labeling:

enter image description here

One is accustomed to resort to area as an intuitive basis, since most or all textbooks do. I am. But position and velocity make a good basis as well, taking $\Delta F$ to be the change in position and $\Delta x$ the change in time. It is from that point of view I find the above figures intuitively convincing, rather than trying to think in terms of area being the height of the graph. It reflects a typical application of integration, to break a motion from $F(a)$ to $F(b)$ into indefinitely many parts, apply some formula, and integrate/sum to obtain that can be used to solve problems. Such applications were a monthly if not weekly feature of the year-long mathematical methods physics course I took, and the above shows the fundamental first step.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.