Integration and the fundamental theorem of calculus | Chapter 8, Essence of calculus

Summary

This video script explores the concept of integrals as the inverse of derivatives, using the analogy of a car's velocity and distance traveled over time. The presenter illustrates how integrals can be understood as the area under a velocity-time graph, which represents the distance covered. By approximating the area with rectangles and refining the approximation, the integral is revealed as a powerful tool for solving a variety of problems in math and science. The script delves into the fundamental theorem of calculus, highlighting the surprising simplicity of calculating an integral by evaluating an antiderivative at two points, despite it accounting for all inputs between those points.

Takeaways

• 🌟 Grothendieck's quote emphasizes the importance of understanding the intuitive side of mathematical facts beyond just proving them true.
• 🔍 The video focuses on integrals, aiming to make it clear that they are the inverse operation of derivatives.
• 🚗 The example of a car's motion with variable velocity is used to illustrate the concept of integrals and their relation to distance traveled.
• 📊 An integral is introduced as a way to find the area under a curve, which in the context of the car example, represents the distance traveled over time.
• ⏱ The concept of approximating the area under a curve by summing the areas of rectangles is explained, highlighting the process of refining this approximation.
• 📐 The integral is related to the antiderivative of a function, which is the function whose derivative is the given velocity function in the example.
• 📈 The fundamental theorem of calculus is alluded to, which connects derivatives and integrals, stating that the integral of a function can be found by evaluating its antiderivative at the bounds of integration.
• 📚 The script discusses the process of finding an antiderivative, which involves applying calculus rules such as the power rule to find the function that, when differentiated, gives the original function.
• 🔢 The importance of constants in antiderivatives is noted, as they can be added without affecting the derivative, and are adjusted based on the specific bounds of the integral.
• 📉 The script touches on the concept of negative area, explaining how it represents movement in the opposite direction (e.g., the car moving backward) and its significance in integral calculations.
• 🎨 The Art of Problem Solving is mentioned as a sponsor and a resource for fostering a love for creative math and exploring mathematical concepts beyond traditional education.

Q & A

What is the main concept the video aims to make almost obvious about integrals?

The main concept the video aims to make almost obvious is that integrals are an inverse of derivatives.

What is the example used in the video to illustrate the concept of integrals?

The example used in the video is a moving car scenario, where the car's velocity is measured over a period of 8 seconds, and the goal is to determine the distance traveled based on the velocity readings.

How does the video relate the concept of integrals to the area bounded by a velocity graph?

The video relates integrals to the area bounded by a velocity graph by showing that the area under the velocity curve between two points on the time axis represents the distance traveled by the car during that time interval.

What is the mathematical function used in the video to model the car's velocity over time?

The mathematical function used to model the car's velocity over time is v(t) = 8t - t^2 meters per second.

How does the video explain the process of finding the distance function s(t) from the velocity function?

The video explains that finding the distance function s(t) involves finding the antiderivative of the velocity function, which is the function whose derivative is the given velocity function.

What is the significance of the integral in the context of the video's explanation?

The integral is significant because it represents the sum of all the small areas (distances) under the velocity curve, which, as dt approaches 0, gives the exact distance traveled by the car.

How does the video discuss the concept of negative area in the context of integrals?

The video discusses the concept of negative area by explaining that if the velocity function is negative (indicating the car is moving backward), the area under the curve below the horizontal axis represents a negative distance, which would be subtracted from the total distance traveled.

What is the fundamental theorem of calculus mentioned in the video, and how does it relate to integrals?

The fundamental theorem of calculus states that the integral of a function can be computed by finding an antiderivative of the function and then evaluating it at the upper and lower bounds of the integration interval. This theorem connects the process of integration with the concept of derivatives.

How does the video illustrate the approximation of the car's motion with constant velocity on intervals?

The video illustrates this by dividing the time axis into small intervals and approximating the car's velocity as constant within each interval. The distance traveled on each interval is then calculated by multiplying the approximated constant velocity by the interval's width (dt).

What is the role of the赞助商 (sponsor) mentioned in the video, and how does it relate to the content?

The sponsor, The Art of Problem Solving, is mentioned as a company that provides resources for learning and exploring mathematics. The video creator recommends their books and courses for those interested in deepening their understanding of math beyond traditional education.

How does the video use the concept of 'signed area' in the context of integrals?

The video uses the concept of 'signed area' to explain that integrals measure the area between the graph and the horizontal axis, considering the sign (positive or negative) based on whether the graph is above or below the axis.

Outlines

00:00

📚 The Intuitive Approach to Integrals and Distance Traveled

This paragraph introduces the concept of integrals as the inverse of derivatives, using the analogy of a car's motion to explain how integrals can be understood intuitively. The narrator discusses the challenge of determining the distance traveled by a car based solely on its velocity readings over time. The idea of approximating the car's velocity as constant over small time intervals and summing these to estimate the total distance is introduced. This approach builds an intuition for integral problems and lays the groundwork for understanding the relationship between the area under a curve and the distance traveled.

05:02

📉 Approximating Distance with the Area Under a Curve

The second paragraph delves deeper into the approximation method by breaking down the time into small intervals and treating the car's velocity as constant within each interval. It explains how the distance traveled during each interval can be visualized as the area of a rectangle and how summing these areas approximates the total distance. The narrator introduces the integral notation and explains how as the time intervals become smaller, the approximation becomes more accurate, ultimately approaching the exact area under the velocity curve, which represents the true distance traveled.

10:07

🔍 The Fundamental Theorem of Calculus and Antiderivatives

This paragraph explains the connection between integrals and antiderivatives, focusing on the example of a car's velocity function. It describes how the derivative of the distance function gives the velocity, and conversely, how finding an antiderivative of the velocity function gives the distance function. The narrator discusses the process of finding an antiderivative for the given velocity function and emphasizes the importance of the constant of integration. The paragraph concludes by highlighting the fundamental theorem of calculus, which states that the integral of a function can be computed by evaluating its antiderivative at the bounds of integration.

15:09

📐 Understanding the Derivative as the Rate of Change of Area

The fourth paragraph explores the geometric interpretation of derivatives, relating the rate of change of area under a curve to the function representing the curve itself. It uses the example of a small change in time causing a small increase in the area under the velocity curve, which can be approximated as a rectangle. The narrator explains how this leads to the conclusion that the derivative of the area function is equal to the velocity at any given time, reinforcing the fundamental theorem of calculus and the concept that the antiderivative of a function encapsulates the cumulative effect of its values over an interval.

20:09

🎯 The Practical Application and Significance of Integrals

The final paragraph reflects on the practical significance of integrals, discussing how they can be used to solve real-world problems beyond just calculating distance traveled by a car. It emphasizes the power of integrals as a tool for summing small quantities to find a total and touches on the concept of negative area when the function dips below the horizontal axis. The narrator also acknowledges the support of The Art of Problem Solving and encourages viewers to explore mathematics further, highlighting the importance of fostering a love for creative problem-solving in mathematics.

Keywords

💡Integrals

Integrals are a fundamental concept in calculus, representing the process of finding the area under a curve between two points. In the context of the video, integrals are used to determine the distance traveled by a car when its velocity is given as a function of time. The video emphasizes that integrals are the inverse operation of derivatives, highlighting the relationship between the area under a curve (integral) and the rate of change (derivative).

💡Derivatives

Derivatives in calculus measure the rate at which a quantity changes with respect to another quantity. In the video, the concept of derivatives is introduced as the opposite of integrals, where knowing the distance function (s of t) allows one to find the velocity function by taking the derivative. The video script uses the example of a car's distance over time to illustrate how the derivative of distance with respect to time gives velocity.

💡Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that 'reverse' the process of differentiation. The video script explains that to find a distance-time function from a velocity-time function, one must determine an antiderivative of the velocity function. This is crucial for calculating the total distance traveled by approximating the area under the velocity curve.

💡Velocity Function

A velocity function is a mathematical representation of the speed of an object as it changes over time. In the video, the velocity function v(t) = t * (8 - t) is used to model the car's speed in meters per second at any given time t. The script discusses how this function can be integrated to find the distance traveled over a period of time.

💡Distance Function

A distance function, denoted as s(t) in the script, represents the total distance traveled by an object as a function of time. The video explains how to derive the distance function from the velocity function using integration, which is key to understanding the concept of integrals in the context of motion.

💡Area Under the Curve

The area under the curve is a visual representation of the integral, which corresponds to the accumulated value of a function over an interval. In the video, the area under the velocity curve is used to illustrate the total distance traveled by the car. The script makes a connection between the intuitive idea of area and the abstract concept of integration.

💡Approximation

Approximation in the context of the video refers to the method of breaking down the problem into smaller, more manageable parts, such as assuming constant velocity over small time intervals. The script describes how approximating the velocity function as a series of constant velocities allows for the calculation of the distance traveled over time, which becomes more accurate as the time intervals become smaller.

💡Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a theorem that links the concept of integration with differentiation. It states that the definite integral of a function can be computed by finding an antiderivative of the function and then evaluating it at the bounds of integration. The video script highlights this theorem as it explains how to calculate the distance traveled by evaluating the antiderivative at the upper and lower bounds of the time interval.

💡Signed Area

Signed area refers to the concept that when calculating the area under a curve, regions below the x-axis contribute a negative value to the total area. In the video, the script mentions the idea of negative area in the context of a car moving in reverse, indicating that the distance traveled backwards would subtract from the total distance.

💡The Art of Problem Solving

The Art of Problem Solving is an organization mentioned in the script that provides educational resources for students and educators. The video script acknowledges their support and recommends their books and courses for those interested in exploring mathematics beyond traditional schooling. The mention serves as an endorsement and a resource for further learning.

Highlights

Grothendieck's quote emphasizes the importance of intuitive understanding in mathematics.

The video aims to make the concept of integrals as inverse of derivatives almost obvious.

An example of a moving car is used to illustrate the concept of integrals.

The challenge of determining distance traveled based on speedometer readings is presented.

A velocity function v(t) = t * (8 - t) is introduced to model the car's speed over time.

The process of finding a distance function s(t) from a velocity function is explained.

The concept of antiderivatives is introduced as a method to find distance from velocity.

The relationship between the area under a velocity graph and distance traveled is discussed.

The idea of approximating variable velocity as constant over small time intervals is proposed.

The method of summing areas of rectangles to approximate distance is explained.

The symbol ∫ is introduced to represent the sum of areas as dt approaches 0.

The integral is described as a way to integrate values, leading to the area under a curve.

The integral of the velocity function is shown to represent the distance function s(t).

The fundamental theorem of calculus is introduced, relating derivatives and integrals.

The process of finding an antiderivative to evaluate an integral is detailed.

The concept of negative area and its significance in integrals is discussed.

The practical applications of integrals in various mathematical and scientific problems are hinted at.

The Art of Problem Solving is mentioned as a sponsor and resource for mathematical exploration.

Transcripts

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.

In this video, I want to talk about integrals,

and the thing that I want to become almost obvious is that they are an

inverse of derivatives.

Here we're just going to focus on one example,

which is a kind of dual to the example of a moving car that I talked about in chapter

2 of the series, introducing derivatives.

Then in the next video we're going to see how this same idea generalizes,

but to a couple other contexts.

Imagine you're sitting in a car, and you can't see out the window,

all you see is the speedometer.

At some point the car starts moving, speeds up,

and then slows back down to a stop, all over the course of 8 seconds.

The question is, is there a nice way to figure out how far you've

travelled during that time based only on your view of the speedometer?

Or better yet, can you find a distance function, s of t,

that tells you how far you've travelled after a given amount of time, t,

somewhere between 0 and 8 seconds?

Let's say you take note of the velocity at every second,

and make a plot over time that looks something like this.

And maybe you find that a nice function to model that velocity

over time in meters per second is v of t equals t times 8 minus t.

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 of t, and you wanted to figure out the velocity function from that.

There I showed how the derivative of a distance vs.

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 times 8 minus t.

This is often described as finding the antiderivative of a function, and indeed,

that's what we'll end up doing, and you could even pause right now and try that.

But first, I want to spend the bulk of this video showing how this question is related

to finding the area bounded by the velocity graph,

because that helps to build an intuition for a whole class of problems,

things called integral problems in math and science.

To start off, notice that this question would be a lot easier

if the car was just moving at a constant velocity, right?

In that case, you could just multiply the velocity in meters per second times the amount

of time that has passed in seconds, and that would give you the number of meters traveled.

And notice, you can visualize that product, that distance, as an area.

And if visualizing distance as area seems kind of weird, I'm right there with you.

It's just that on this plot, where the horizontal direction has units of seconds,

and the vertical direction has units of meters per second,

units of area just very naturally correspond to meters.

But what makes our situation hard is that velocity is not constant,

it's incessantly changing at every single instant.

It would even be a lot easier if it only ever changed at a handful of points,

maybe staying static for the first second, and then suddenly discontinuously

jumping to a constant 7 meters per second for the next second, and so on,

with discontinuous jumps to portions of constant velocity.

That would make it uncomfortable for the driver,

in fact it's actually physically impossible, but it would make your calculations

a lot more straightforward.

You could just compute the distance traveled on each interval by multiplying the constant

velocity on that interval by the change in time, and then just add all of those up.

So what we're going to do is approximate the velocity function as if it

was constant on a bunch of intervals, and then, as is common in calculus,

we'll see how refining that approximation leads us to something more precise.

Here, let's make this a little more concrete by throwing in some numbers.

Chop up the time axis between 0 and 8 seconds into many small intervals,

each with some little width dt, something like 0.25 seconds.

Consider one of those intervals, like the one between t equals 1 and 1.25.

In reality, the car speeds up from 7 m per second to about 8.4 m per

second during that time, and you could find those numbers just by

plugging in t equals 1 and t equals 1.25 to the equation for velocity.

What we want to do is approximate the car's motion

as if its velocity was constant on that interval.

Again, the reason for doing that is we don't really know

how to handle situations other than constant velocity ones.

You could choose this constant to be anything between 7 and 8.4.

It actually doesn't matter.

All that matters is that our sequence of approximations,

whatever they are, gets better and better as dt gets smaller and smaller.

That treating this car's journey as a bunch of discontinuous jumps

in speed between portions of constant velocity becomes a less-wrong

reflection of reality as we decrease the time between those jumps.

So for convenience, on an interval like this, let's just approximate the

speed with whatever the true car's velocity is at the start of that interval,

the height of the graph above the left side, which in this case is 7.

In this example interval, according to our approximation,

the car moves 7 m per second times 0.25 seconds.

That's 1.75 meters, and it's nicely visualized as the area of this thin rectangle.

In truth, that's a little under the real distance traveled, but not by much.

The same goes for every other interval.

The approximated distance is v of t times dt, it's just that you'd be plugging in a

different value for t at each one of these, giving a different height for each rectangle.

I'm going to write out an expression for the sum of

the areas of all those rectangles in kind of a funny way.

Take this symbol here, which looks like a stretched s for sum,

and put a 0 at its bottom and an 8 at its top,

to indicate that we'll be ranging over time steps between 0 and 8 seconds.

And as I said, the amount we're adding up at each time step is v of t times dt.

Two things are implicit in this notation.

First of all, that value dt plays two separate roles.

Not only is it a factor in each quantity we're adding up,

it also indicates the spacing between each sampled time step.

So when you make dt smaller and smaller, even though it decreases the area of

each rectangle, it increases the total number of rectangles whose areas we're adding up,

because if they're thinner, it takes more of them to fill that space.

And second, the reason we don't use the usual sigma notation to indicate a sum is that

this expression is technically not any particular sum for any particular choice of dt.

It's meant to express whatever that sum approaches as dt approaches 0.

And as you can see, what that approaches is the

area bounded by this curve and the horizontal axis.

Remember, smaller choices of dt indicate closer approximations for the original question,

how far does the car actually go?

So this limiting value for the sum, the area under this curve,

gives us the precise answer to the question in full unapproximated precision.

Now tell me that's not surprising.

We had this pretty complicated idea of approximations that

can involve adding up a huge number of very tiny things.

And yet, the value that those approximations approach can be described so simply,

it's just the area underneath this curve.

This expression is called an integral of v of t,

since it brings all of its values together, it integrates them.

Now at this point, you could say, how does this help?

You've just reframed one hard question, finding how far the car has traveled,

into an equally hard problem, finding the area between this graph and the horizontal axis.

And you'd be right.

If the velocity-distance duo was the only thing we cared about, most of this video,

with all the area under a curve nonsense, would be a waste of time.

We could just skip straight ahead to finding an antiderivative.

But finding the area between a function's graph and the horizontal axis

is somewhat of a common language for many disparate problems that can be

broken down and approximated as the sum of a large number of small things.

You'll see more in the next video, but for now I'll just say in

the abstract that understanding how to interpret and how to

compute the area under a graph is a very general problem-solving tool.

In fact, 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, capital 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 travelled after T seconds, right?

So in reality, this is a distance vs.

time function, s of 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 of 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 of t times dt.

And because that's an approximation that gets better and better for smaller dt,

the derivative of that area function, ds, dt, at this point equals vt,

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.

So, if our velocity function is t times 8-t, what should s be?

What function of t has a derivative of t times 8-t?

It's easier to see if we expand this out, writing it as 8t minus t squared,

and then we can just take each part one at a time.

What function has a derivative of 8t?

We know that the derivative of t squared is 2t,

so if we just scale that up by a factor of 4, we can see that the derivative

of 4t squared is 8t.

And for that second part, what kind of function do

you think might have negative t squared as a derivative?

Using the power rule again, we know that the derivative of a cubic term,

t cubed, gives us a square term, 3t squared.

So if we just scale that down by a third, the

derivative of 1 third t cubed is exactly t squared.

And then making that negative, we'd see that negative

1 third t cubed has a derivative of negative t squared.

Therefore, the antiderivative of our function,

8t minus t squared, is 4t squared minus 1 third t cubed.

But there's a slight issue here.

We could add any constant we want to this function,

and its derivative is still 8t minus t squared.

The derivative of a constant always goes to zero.

And if you were to graph s of 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's actually infinitely many different

possible antiderivative functions, and every one of them looks

like 4t squared minus 1 third t cubed plus c, for some constant c.

But 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 travelled by the car between 0 seconds and 0 seconds is… well, zero.

So as we found, the area as a function of capital

T is an antiderivative for the stuff inside.

And 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 equals zero,

you get zero.

So in this specific case, you don't need to subtract anything off.

For example, the total distance travelled during the full 8 seconds

is this expression evaluated at t equals 8, which is 85.33 minus 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 travelled between 1 second and 7 seconds.

What you do is evaluate the antiderivative we found 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 chose 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, and remember,

you think of that as adding up values f of x times dx for inputs in a certain range,

and then asking what is that sum approach 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.

And I want you to appreciate something kind of crazy about this fact.

The integral, the limiting value for the sum of all these thin rectangles,

takes into account every single input on the continuum,

from the lower bound to the upper bound.

That's why we use the word integrate, it brings them all together.

And yet, to actually compute it using an antiderivative,

you only look at two inputs, the top bound and the bottom bound.

It almost feels like cheating.

Finding the antiderivative implicitly accounts for all the

information needed to add up the values between those two bounds.

That's just crazy to me.

This idea is deep, and there's a lot packed into this whole concept,

so let's recap everything that just happened, shall we?

We wanted to figure out how far a car goes just by looking at the speedometer.

And what makes that hard is that velocity is always changing.

If you approximate velocity to be constant on multiple different intervals,

you could figure out how far the car goes on each interval with multiplication,

and then add all of those up.

Better and better approximations for the original problem correspond to

collections of rectangles whose aggregate area is closer and closer to

being the area under this curve between the start time and the end time.

So that area under the curve is actually the precise distance

traveled for the true nowhere constant velocity function.

If you think of that area as a function itself,

with a variable right endpoint, you can deduce that the derivative

of that area function must equal the height of the graph at every point.

And that's really the key right there.

It means that to find a function giving this area,

you ask, what function has v of t as a derivative?

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.

By the way, one important thing to bring up before we leave is the idea of negative area.

What if the velocity function was negative at some point, meaning the car goes backwards?

It's still true that a tiny distance traveled ds on a little time interval is

about equal to the velocity at that time multiplied by the tiny change in time.

It's just that the number you'd plug in for velocity would be negative,

so the tiny change in distance is negative.

In terms of our thin rectangles, if a rectangle goes below the horizontal axis,

like this, its area represents a bit of distance traveled backwards,

so if what you want in the end is to find a distance between the car's

start point and its end point, this is something you'll want to subtract.

And that's generally true of integrals.

Whenever a graph dips below the horizontal axis,

the area between that portion of the graph and the horizontal axis is counted as negative.

What you'll commonly hear is that integrals don't measure area per se,

they measure the signed area between the graph and the horizontal axis.

Next up, I'm going to bring up more context where this idea

of an integral and area under curves comes up,

along with some other intuitions for this fundamental theorem of calculus.

Maybe you remember, chapter 2 of this series introducing the derivative

was sponsored by The Art of Problem Solving, so I think there's

something elegant to the fact that this video, which is kind of a duel to that one,

was also supported in part by The Art of Problem Solving.

I really can't imagine a better sponsor for this channel,

because it's a company whose books and courses I recommend to people anyway.

They were highly influential to me when I was a student developing a love for

creative math, so if you're a parent looking to foster your own child's love

for the subject, or if you're a student who wants to see what math has to offer

beyond rote schoolwork, I cannot recommend The Art of Problem Solving enough.

Whether that's their newest development to build the right intuitions in

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topics and contest preparation, going to aops.com slash 3blue1brown,

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I consider these videos a success not when they teach people a particular bit of math,

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