#readwise # Integration and The Fundamental Theorem of Calculus  ## Metadata - Author: [[3Blue1Brown]] - Full Title: Integration and The Fundamental Theorem of Calculus - URL: https://www.youtube.com/watch?v=rfG8ce4nNh0 ## Summary This text explains how to find the distance traveled by a car using its velocity function over time. It describes the relationship between velocity, distance, and area under a curve in calculus. By approximating the car’s velocity as constant over small time intervals, we can calculate the total distance. The fundamental theorem of calculus connects the concepts of differentiation and integration, showing that the area under the velocity curve gives the distance traveled. ## Document Notes 1. How does visualizing distance as the area under a velocity graph enhance our understanding of the relationship between integration and the fundamental theorem of calculus? 2. In what ways does the concept of negative area influence our interpretation of distance traveled when dealing with negative velocity functions? 3. How can approximating the area under a curve using rectangles illustrate the transition from a discrete sum to a continuous integral, and what implications does this have for understanding calculus in general? ## Highlights This guy, Grothendieck, is somewhat of a mathematical idol to me, and I just love this quote, don't you? Too often in math, we dive into showing that a certain fact is true with a long series of formulas before stepping back and making sure it feels reasonable, and preferably obvious, at least at an intuitive level. ([View Highlight](https://read.readwise.io/read/01jfmask5srgr9e8t31zghy2nx)) - Note: "One should never try to prove anything that is not almost obvious." – Alexander Grothendieck --- You might remember, in chapter 2 of this series, we were looking at the opposite situation, where you knew what a distance function was, $s(t)$, and you wanted to figure out the velocity function from that. There I showed how the derivative of a distance vs. The time function gives you a velocity vs. time function. So in our current situation, where all we know is velocity, it should make sense that finding a distance vs. time function is going to come down to asking what function has a derivative of $t(8-t)$. This is often described as finding the antiderivative of a function, and indeed, that's what we'll end up doing ([View Highlight](https://read.readwise.io/read/01jfmqaaz1ggrpap7gkk79tvd6)) - Tags: [[👻 ai highlighted]] --- the first video of this series already covered the basics of how this works, but now that we have more of a background with derivatives, we can take this idea to its completion. For a velocity example, think of this right endpoint as a variable, $T$. So we’re thinking of this integral of the velocity function between 0 and $T$, the area under this curve between those inputs, as a function where the upper bound is the variable. That area represents the distance the car has traveled after $T$ seconds, right? So in reality, this is a distance vs. time function, $s(T)$. Now ask yourself, what is the derivative of that function? On the one hand, a tiny change in distance over a tiny change in time is velocity; that is what velocity means. But there’s another way to see this, purely in terms of this graph and this area, which generalizes a lot better to other integral problems. A slight nudge of $dT$ to the input causes that area to increase, some little $ds$ represented by the area of this sliver. The height of that sliver is the height of the graph at that point, $v(T)$, and its width is $dT$. And for small enough $dT$, we can basically consider that sliver to be a rectangle, so this little bit of added area, $ds$, is approximately equal to $v(T) \times dT$. And because that’s an approximation that gets better and better for smaller $dT$, the derivative of that area function, $\frac{ds}{dT}$, at this point equals $v(T)$, the value of the velocity function at whatever time we started on. And that right there is a super general argument. The derivative of any function giving the area under a graph like this is equal to the function for the graph itself. ([View Highlight](https://read.readwise.io/read/01jfmqcd1p9hj4ksqph32vky7a)) ^l17wwx - Tags: [[👻 ai highlighted]] --- But there's a slight issue here. We could add any constant we want to this function, and its derivative is still $8t - t^2$. The derivative of a constant always goes to zero. If you were to graph $s(t)$, you could think of this in the sense that moving a graph of a distance function up and down does nothing to affect its slope at every input. So in reality, there are actually infinitely many different possible antiderivative functions, and every one of them looks like $4T^2 - \frac{1}{3} T^3 + C$, for some constant $C$. However, there is one piece of information we haven’t used yet that will let us zero in on which antiderivative to use: the lower bound of the integral. This integral has to be zero when we drag that right endpoint all the way to the left endpoint, right? The distance traveled by the car between 0 seconds and 0 seconds is… well, zero. As we found, the area as a function of capital $T$ is an antiderivative for the stuff inside. To choose what constant to add to this expression, you subtract off the value of that antiderivative function at the lower bound. If you think about it for a moment, that ensures that the integral from the lower bound to itself will indeed be zero. As it so happens, when you evaluate the function we have here at $T = 0$, you get zero. So in this specific case, you don’t need to subtract anything off. For example, the total distance traveled during the full 8 seconds is this expression evaluated at $T = 8$, which is $85.33 - 0$. So the answer as a whole is $85.33$. But a more typical example would be something like the integral between 1 and 7. That’s the area pictured here, and it represents the distance traveled between 1 second and 7 seconds. What you do is evaluate the antiderivative we found at the top bound, To evaluate the integral, you take the value at the top bound, $7$, and subtract off its value at the bottom bound, $1$. Notice, by the way, it doesn’t matter which antiderivative we choose here. If for some reason it had a constant added to it, like $5$, that constant would cancel out. More generally, any time you want to integrate some function, remember, you think of that as adding up values $f(x) \times dx$ for inputs in a certain range, and then asking what that sum approaches as $dx$ approaches $0$. The first step to evaluating that integral is to find an antiderivative, some other function, capital $F$, whose derivative is the thing inside the integral. Then the integral equals this antiderivative evaluated at the top bound minus its value at the bottom bound. And this fact right here that you’re staring at is the fundamental theorem of calculus ([View Highlight](https://read.readwise.io/read/01jfmqfqs9qhdcbchc1mh1e3vt)) ^hxrhvt --- There are actually infinitely many antiderivatives of a given function, since you can always just add some constant without affecting the derivative, so you account for that by subtracting off the value of whatever antiderivative function you choose at the bottom bound. ([View Highlight](https://read.readwise.io/read/01jfmc3nq8ym6633st02557evp)) ---