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Clown school

Who cares about area?

Finding the shaded area under a function seems like a very specific problem, but it turns out to be surprisingly useful. It's a fundamental technique in math, physics, and beyond.

A Slow Start

Suppose a car is driving at 60mph for 3 hours. How far does it travel in that time?

It's an easy question, right? Just multiply:

60 mileshour×3 hours=180 miles60\ \frac{\text{miles}}{\cancel{\text{hour}}} \times 3\ \cancel{\text{hours}} = 180\ \text{miles}

Multiplying makes the units cancel, and we're left with miles.

This problem can be represented with a rectangle of width 3 (for the 3 hours of driving) and height 60 (for the speed of the car). The area of the rectangle is 603=180 miles60 \cdot 3 = 180\ \text{miles}.

01230mph20mph40mph60mph
Car driving at 60 mph
6003010204050

Of course, doing this doesn't seem all that helpful until you consider a harder problem...

Changing Gears

Suppose a car is driving for 3 hours. During the first hour, it drives at 20mph. During the second hour, it drives at 40mph. During the third hour, it drives at 60mph. How far does the car travel in total?

Now our problem has three separate phases, which can be represented on a graph:

01230mph20mph40mph60mph
Car driving at 20 mph
6003010204050

We can calculate the distance the car travels during each of the three phases:20 mileshour×1 hour=20 miles40 mileshour×1 hour=40 miles60 mileshour×1 hour=60 miles\begin{align*} 20\ \frac{\text{miles}}{\text{hour}} \times 1\ \text{hour} &= 20\ \text{miles} \\ 40\ \frac{\text{miles}}{\text{hour}} \times 1\ \text{hour} &= 40\ \text{miles} \\ 60\ \frac{\text{miles}}{\text{hour}} \times 1\ \text{hour} &= 60\ \text{miles} \end{align*} All together, the car travels a total distance of 20 miles+40 miles+60 miles=120 miles20\text{ miles} + 40\text{ miles} + 60\text{ miles} = 120\text{ miles}.

Notice that what we actually just calculated is the shaded area of the graph.

It turns out that this is true in general: The area under a velocity curve equals the distance. (Even if that curve is, well, curvy.)

Smooth Sailing

Let's try something harder.

Suppose a car begins at rest, and accelerates at a rate of 20 mileshour220\ \frac{\text{miles}}{\text{hour}^2}.

That means that the velocity is increasing constantly, like this:

01230mph20mph40mph60mph
Car driving at 0 mph
6003010204050
What is the total distance travelled after 3 hours?

Well, it's the area under the curve of course!

This is a triangle, and we know that the area of a triangle is 12×Base×Height\frac{1}{2} \times \text{Base} \times \text{Height}. In this case, 12×3 hours×60 mileshour=90 miles\frac{1}{2} \times 3\text{ hours} \times 60\ \frac{\text{miles}}{\text{hour}} = 90\text{ miles}You might convince yourself that this is reasonable because the average speed of the car is 30mph, and it drives for 3 hours.

So... why is area useful?

If you've been following along with calculus so far, you probably have an intuition for how taking the derivative of a function takes you from position to velocity to acceleration.

Position
Velocity
Acceleration
Side note
Acceleration is followed by jerk, and then—amusingly—snap, crackle, and pop. Which I find absolutely magical.

In the above problems we saw that finding the area under a function moved us in the opposition direction, from velocity to position. And in fact, this works for reversing the entire lineup. This is a very useful technique!

Why is finding the area the opposite of taking a derivative?

This seems a bit confusing at first. Taking the area under the function—a process which we will call "integrating"—somehow does the opposite of taking the derivative. But why?

One way to think about it is that the derivative was like slope for curves, and slope is ΔyΔx\frac{\Delta y}{\Delta x}. Derivatives are like curvy division. The integral, on the other hand, is like rectangular area for curves, and the area is Δx×Δy\Delta x \times \Delta y. Integrals are like curvy multiplication.

Since derivatives are like division, and integrals are like multiplication, perhaps it's not so surprising that they are opposites.