Integration and the Fundamental Theorem of Calculus

Start here

This is Chapter 8 of the Essence of Calculus series. So far every chapter has been about derivatives: take a function, ask how fast it changes. This chapter runs the machine backwards. We start with a rate and ask for the total. That question is integration, and the punchline is that it is secretly the same tool you already have, used in reverse.

The car problem

You are in a car. The speedometer is the only thing you can see, and it tells you your velocity at every instant. Say it reads v(t) over an eight second trip. Question: how far did you travel?

If the speed were constant, this is grade school. Distance is speed times time. Drive 60 for 2 hours, you go 120. Done.

But the needle keeps moving. The speed at second 3 is different from the speed at second 7. There is no single number to multiply by 8. The thing that makes this hard is exactly the thing calculus was built to chew through: a quantity that changes continuously.

Here is the move. Plot velocity against time and look at the area under that curve. That area is the total distance. The rest of this chapter is about why that is true and how to compute it.

Slice it into rectangles

The trick is the one from Chapter 1: when something curvy is hard, chop it into tiny straight pieces.

Pick a tiny chunk of time, call its width dt. Over a window that short the velocity barely changes, so pretend it is constant at v(t). The distance covered in that sliver is just speed times time:

distance in one sliver ≈ v(t) · dt

That is the area of a skinny rectangle: height v(t), width dt. Now tile the whole trip with these rectangles and add their areas:

 v(t)
   ^            ___
   |        ___|   |__
   |     __|   |   |  |___
   |  __|  |   |   |  |   |__
   | |  |  |   |   |  |   |  |
   | |  |  |   |   |  |   |  |
   +-+--+--+---+---+--+---+--+----> t
   0    dt   dt  dt   ...     8
   |<--- total distance = sum of areas --->|

Add up every v(t)·dt and you get an approximation of the distance. Make dt smaller and the rectangles hug the curve tighter, so the approximation gets better. In the limit, as dt shrinks toward zero, the staircase becomes the smooth area and the sum becomes an integral:

total distance = ∫  v(t) dt     from t = 0 to t = 8

The elongated S is a stretched "sum". The dt is the leftover whisper of the rectangle width, kept in the notation to remind you what you summed.

The twist: an area that grows

Computing that limit by hand sounds horrible. We do not. Instead we ask a sneaky question.

Forget the fixed endpoints for a second. Let the right edge slide. Define A(x) as the area under v(t) from the start up to some movable point x. As you push x to the right, the area grows. So A(x) is itself a function: feed it a right edge, get back an area.

Now nudge x by a tiny dx. How much new area appears? You glued on one more skinny rectangle: height v(x), width dx. So:

dA  ≈  v(x) · dx

Divide both sides by dx:

dA
---  =  v(x)
dx

Read that slowly. The derivative of the area function is the original height. The area piles up exactly as fast as the curve is tall, which is the whole reason a tall curve sweeps out area quickly and a flat one crawls.

So to find the area function A(x), you do not sum anything. You hunt for a function whose derivative is v(x). That function is called an antiderivative, and you already know how to find them from Chapters 3 through 7.

The fundamental theorem

Suppose you find any antiderivative F, meaning F'(x) = f(x). Then the area under f from a to b is just the change in F across that span:

∫  f(x) dx  =  F(b) - F(a)        from a to b

That is the Fundamental Theorem of Calculus. Why subtract? F(b) is the accumulated area from the start out to b, F(a) is the accumulated area out to a. Take the difference and the overlap from the start to a cancels, leaving exactly the slab between a and b.

No limits of giant sums in sight. You reverse a derivative, plug in two numbers, subtract. The infinite sum was real, but the theorem lets you skip it.

Worked example

Say the velocity is v(t) = t², and you want the distance traveled between t = 0 and t = 8.

Step one, find an antiderivative. You want F with F'(t) = t². The power rule run backwards (bump the exponent up by one, divide by the new exponent) gives:

F(t) = t³ / 3        because   F'(t) = 3t²/3 = t²   ✓

Step two, evaluate at the bounds and subtract:

∫ t² dt  =  F(8) - F(0)  =  8³/3 - 0³/3  =  512/3  ≈ 170.67

So the car covered about 170.67 units of distance. The whole curvy-velocity problem collapsed into one cube divided by three, minus zero. That is the leverage the fundamental theorem buys you.

The same idea as Chapter 1

Go back to the very first chapter, where we found the area of a circle by slicing it into thin concentric rings and unrolling them into a triangle. That was this exact move in disguise: chop a hard region into many tiny pieces, approximate each piece as something simple (a thin rectangle), add them up, and take the limit as the pieces vanish.

A circle's rings, a car's velocity slivers, the area under any curve. Different stories, one engine. Integration is the art of adding up an infinity of tiny things, and the fundamental theorem is the shortcut that makes it painless.

Where this lands in AI

Integrals are everywhere once probability enters the room, because a continuous probability is an area.

• Expectations and probabilities. The expected value of a continuous variable is an integral: E[x] = ∫ x · p(x) dx. The probability of landing in a range is the area under the density over that range.
• Normalizing constants. A distribution has to integrate to 1. The denominator that forces that, the partition function in energy based models or the evidence in Bayes' rule, is an integral. It is also the hard part of most of those methods.
• Continuous losses. Many objectives are integrals over a data distribution rather than a clean finite sum.

The catch: most of these integrals have no neat antiderivative. So in practice we approximate them by sampling, drawing many points and averaging, which is the core of Monte Carlo methods. The fundamental theorem tells you what the exact answer means; sampling tells you how to estimate it when the algebra runs out.

Quick gotchas

Area below the axis counts as negative. If v(t) dips under zero (driving in reverse), that chunk subtracts from the total. The integral gives signed area, not raw area. For total distance you would integrate the absolute speed.

The dt is not decoration. It tells you which variable you are summing over and carries the units of the width. Drop it and the notation stops meaning anything.

An antiderivative is not unique. t³/3 and t³/3 + 7 both differentiate to t². The constant cancels in F(b) - F(a), which is why definite integrals do not care about it. Indefinite integrals carry a + C to remember the whole family.

What you walked away with

• The area under a rate curve is the accumulated total: area under velocity is distance.
• Slice into rectangles of width dt, sum v(t)·dt, shrink dt to zero, and the sum becomes the integral ∫ v(t) dt.
• The area function A(x) has derivative f(x), because each new sliver adds f(x)·dx. Integration and differentiation are inverses.
• Fundamental theorem: find an antiderivative F, then the area from a to b is F(b) - F(a). No infinite sum required.
• In AI this powers expectations, normalizing constants, and continuous losses, with Monte Carlo sampling standing in when the integral is intractable.

Next up, Chapter 9: we put the pieces together to find the average value of a continuously changing function, and watch integrals and derivatives shake hands one more time. See you there.