User:IssaRice/Fundamental theorem of calculus

There's a typical picture of FTC1 that you see in places like Pugh's analysis book or 3Blue1Brown's video on FTC. This explanation makes sense, but I want point out a few different ways of thinking about the picture. One is, like the 3b1b video says, to look at the incremental area. You get $\Delta A(x)\approx f(x)\Delta x$. So then you divide by $\Delta x$ and take the limit as $\Delta x\to 0$ and get $A'(x)=\lim _{\Delta x\to 0}{\frac {\Delta A(x)}{\Delta x}}=f(x)$.

Another way of looking at this that I saw in one of John Stillwell's books [i think it was mathematics and its history, in the calculus chapter, in the section talking about leibniz] is that $\int f(x)\,dx$ is the sum of infinitely many quantities $f(x)\,dx$. So the incremental thing you add is $f(x)\,dx$, i.e. $d\int f(x)\,dx=f(x)\,dx$. Now if you divide by $dx$ you get ${\frac {d\int f(x)\,dx}{dx}}=f(x)$, which again is FTC1.

Finally, let's look at $A(x)=\int _{a}^{x}f(t)\,dt$. What is the rate of change of this area function? As x changes, A(x) changes a bit. The rate of change is jut the height of the graph, since the area increases "one vertical line at a time". Or you can think of it as, "the rate at which area, approximated as a rectangle, changes, as the width of the rectangle changes, is the height of that rectangle".

Now, on to FTC2. It bothers me that people are so enthusiastic about showing you visualizations of FTC1, but then they don't even mention how to visualize FTC2. Like, they probably don't have a visualization, so then they are like "well let's just be quiet about it here and nobody will ask". Even 3b1b does this, and i am like >:( wtf. So anyway, I think FTC2 can be visualized using the same picture as FTC1, but you just have to sort of use your brain in a different way? like you have to think about it in a different way, but it's the same picture!

So, let's be clear about what we are trying to show. If $F$ is some nice function, and is an antiderivative of $f$ (i.e., $F'=f$), then we have $\int _{a}^{b}f(x)\,dx=F(b)-F(a)$. So how do we visualize this? well the trick is, let's say you start with a function F. how the heck do you visualize it? well one way is to let F be a curve. then F' becomes the instantaneous slope. but it's hard to visualize that slope changing over time... so an alternative is, you let F be area, and then f just becomes the height (i.e., the curve). that allows us to visualize both functions nicely, just like in FTC1. but now, if we fix some point a, then let the area from a to x be F(x), we have a problem. because now F(a) is an area of 0, so we can't represent an arbitrary function F. so the trick is... for the moment, let's make the problem easier on ourselves by requiring that F(a)=0. Then, as long as F is "smooth" enough, we can represent F as an area. [PROOF?]

Now, what is the height of this graph? why, by FTC1 it must be that the height is F', which is f. so $F(x)-0=\int _{a}^{x}f$ as we wanted.

Now what if F is some function such that F(a) is not 0? then we can define G(x) := F(x) - F(a), so that G(a)=0. Apply the argument/visualization above, to get $G(x)=\int _{a}^{x}G'$. But $G'=F'$ so $F(x)-F(a)=\int _{a}^{x}F'$, and we are done.