From: 3blue1brown
Mathematics often prioritizes proving a fact with formulas over building an intuitive understanding [00:00:18]. This article aims to make it intuitively obvious that integrals are the inverse operation of derivatives [00:00:31].
The Car Example: Velocity to Distance
Imagine sitting in a car where you can only see the speedometer, not outside [00:00:55]. The car speeds up and slows down over 8 seconds [00:01:02]. The challenge is to figure out the total distance traveled or a distance function s(t) based solely on the speedometer readings [00:01:11].
Let’s say the velocity v(t) is modeled by the function t * (8 - t) in meters per second [00:01:38]. In a previous discussion, the opposite problem was addressed: knowing a distance function s(t) and figuring out the velocity function from it [00:01:48]. That problem showed how the derivative of a distance-time function yields a velocity-time function [00:02:00].
Now, with only velocity known, finding the distance function s(t) means asking: what function has t * (8 - t) as its derivative? [00:02:06] This process is known as finding the antiderivative of a function [00:02:19].
Connecting Distance to Area Under the Curve
While finding an antiderivative directly solves the problem, there’s a powerful intuitive connection to the area bounded by the velocity graph [00:02:27]. This intuition is key for understanding integral problems in various fields [00:02:35].
Constant Velocity Scenario
If the car moved at a constant velocity, calculating distance would be simple: multiply velocity (meters/second) by time (seconds) [00:02:42]. This product (distance) can be visualized as the area of a rectangle on a velocity-time graph [00:03:00]. On such a plot, the units of area (meters/second * seconds) naturally correspond to meters [00:03:08].
Approximating with Rectangles
The difficulty arises because velocity is constantly changing [00:03:22]. To tackle this, the velocity function can be approximated as if it were constant over many small time intervals [00:04:09].
- Divide the Time Axis: Chop the time axis (e.g., from 0 to 8 seconds) into many small intervals, each with a width
dt(e.g., 0.25 seconds) [00:04:28]. - Constant Velocity Approximation: For each interval, approximate the car’s velocity as constant. A convenient choice is the true velocity at the start of that interval (the height of the graph above the left side) [00:05:42].
- Calculate Interval Distance: The distance traveled in each small interval is approximately
v(t) * dt[00:06:17]. This can be visualized as the area of a thin rectangle [00:06:04]. - Summation: Sum the areas of all these rectangles [00:06:29].
The Integral: A Limit of Sums
The expression for this sum is written using a stretched ‘S’ symbol, representing an integral [00:06:36]: This notation implies two things [00:06:55]:
dtis a factor in each quantity being added [00:07:01].dtalso indicates the spacing between sampled time steps [00:07:05].
As dt becomes smaller and smaller, the number of rectangles increases, and the approximation becomes more precise [00:07:09]. The integral expresses the value that this sum approaches as dt approaches zero [00:07:33]. This limiting value is precisely the area bounded by the curve and the horizontal axis [00:07:39], which gives the exact distance traveled [00:07:54].
The concept of “area under a graph” is a general problem-solving tool, common in many scientific and mathematical problems that involve summing a large number of small things [00:09:02].
The Fundamental Theorem of Calculus
The relationship between integrals and derivatives is profound. Consider the area under the velocity curve from 0 to a variable T as a function s(T) [00:09:34]. This s(T) represents the distance traveled after T seconds [00:09:52].
What is the derivative of this distance function s(T)? [00:10:01] A tiny change in distance ds over a tiny change in time dt is, by definition, velocity [00:10:07].
Visually, a slight nudge dt to the input T adds a thin sliver of area ds [00:10:23]. The height of this sliver is v(T) (the height of the graph at that point) and its width is dt [00:10:32]. For small dt, ds is approximately v(T) * dt [00:10:39]. Therefore, the derivative of the area function, ds/dt, equals v(T), the value of the velocity function at that time [00:10:51].
This leads to a general argument: the derivative of any function that gives the area under a graph is equal to the function of the graph itself [00:11:06]. This is the essence of the Fundamental Theorem of Calculus.
Finding the Antiderivative
To find s(t) for v(t) = t * (8 - t) (or 8t - t^2), we need to find its antiderivative [00:11:18].
- For
8t: The derivative oft^2is2t[00:11:42]. Scaling it up, the derivative of4t^2is8t[00:11:45]. - For
-t^2: The derivative oft^3is3t^2[00:12:00]. Scaling it down, the derivative of(1/3)t^3ist^2[00:12:08]. Therefore, the derivative of-(1/3)t^3is-t^2[00:12:14].
So, an antiderivative of 8t - t^2 is 4t^2 - (1/3)t^3 [00:12:22].
The Constant of Integration and Definite Integrals
A crucial point is that adding any constant C to an antiderivative does not change its derivative (since the derivative of a constant is zero) [00:12:34]. Thus, there are infinitely many possible antiderivative functions, all of the form 4t^2 - (1/3)t^3 + C [00:12:54].
To find the specific distance traveled between two points, like 0 and 8 seconds, we use the bounds of the integral [00:13:08]. The integral from a lower bound (a) to an upper bound (b) of a function f(x) is evaluated by finding an antiderivative F(x) and computing F(b) - F(a) [00:15:13]. This is the Fundamental Theorem of Calculus [00:15:32].
For the car example, evaluating s(t) = 4t^2 - (1/3)t^3 at t = 8 gives the total distance traveled: 85.33 meters [00:14:07]. If integrating between t=1 and t=7, you would calculate s(7) - s(1) [00:14:23]. The constant C cancels out in this subtraction, confirming that any antiderivative can be used [00:14:45].
The remarkable aspect of the Fundamental Theorem of Calculus is that to compute an integral (which accounts for every input on a continuum), one only needs to evaluate the antiderivative at two inputs: the top and bottom bounds [00:15:38].
Negative Area (Signed Area)
What if the velocity function is negative, meaning the car moves backwards? [00:17:53] A tiny change in distance ds is still v(t) * dt, but v(t) would be negative, resulting in a negative ds [00:17:58].
On the graph, if a rectangle goes below the horizontal axis, its “area” represents distance traveled backward [00:18:16]. This backward distance is subtracted from the total [00:18:25]. Therefore, integrals measure the signed area between the graph and the horizontal axis [00:18:35].
Recap
- To find total distance from varying velocity, approximate the journey as constant velocity on many small intervals [00:16:26].
- The sum of distances (velocity * time) on these intervals corresponds to the sum of areas of thin rectangles [00:16:39].
- As the interval width approaches zero, this sum approaches the exact area under the velocity curve, which is the precise total distance traveled [00:16:46].
- Thinking of this area as a function of its right endpoint, its derivative is the original function (the height of the graph) [00:17:08].
- Thus, to find the area function, one needs to find an antiderivative of the original function [00:17:22].
- The constant of integration in antiderivatives is handled by subtracting the value of the antiderivative at the lower bound from its value at the upper bound [00:17:30].
- When a function’s graph dips below the horizontal axis, the area is counted as negative, representing “signed area” [00:18:37].
This understanding of integrals and antiderivatives forms the fundamental theorem of calculus, a powerful tool for solving a wide range of problems.