← Unit 8

The Integral as Accumulation

One idea runs under half of this unit: a definite integral adds up a rate to give a total change. Net change, motion, and average value are all this one engine — here's how it works, then you'll drill it.

1 · Rate in, total out

Here's a question you already know how to answer with arithmetic: if water flows into a tank at a steady 5 liters per minute for 3 minutes, how much went in? You multiplied — 5 × 3 = 15 liters. Rate times time equals total.

But what if the flow rate changes every instant — fast at first, then trickling, then surging again? You can't just multiply, because there's no single rate to multiply by. This is the exact problem the integral was built to solve: it's what "rate times time" becomes when the rate won't hold still. Slice the time into tiny pieces so small the rate is nearly constant across each one, multiply rate by the tiny time on each slice, and add up all those tiny totals. That sum — taken to the limit — is the integral. So the integral of a rate isn't some abstract symbol; it's the grown-up version of rate × time.

f′(t)
The rate — how fast the quantity is changing at the instant t. A flow rate, a velocity, a rate of growth. This is what you're given.
f′(t) dt
Rate times a sliver of time dt — the tiny bit of total change accumulated during that one instant. This is "rate × time" on a slice so small the rate barely moves.
∫ₐᵇ
The sum of every sliver from a to b. Add up all those tiny changes and you get the net change in the quantity over the whole interval.
= f(b) − f(a)
And here's the payoff: that whole sum equals just the ending amount minus the starting amount. That's the Fundamental Theorem — you don't have to add up infinitely many slivers by hand, you just subtract two values.

This is why the idea is so powerful: you never need a formula for the quantity itself, only for its rate. You don't track the tank's water level moment by moment — you take the flow rate, integrate it, and out comes the total added. A capacitor charging, a population growing, a car's odometer climbing: in every case, total accumulated = integral of the rate. Everything else on this page is this one idea wearing a different costume.

2 · Motion: where the rate is velocity

Now give that rate a specific job. Velocity is the rate your position changes — so by the idea above, position is just the accumulation of velocity. Integrate velocity over some time and you get how much position changed. Simple enough. But motion hides a fork in the road that costs students points on the exam constantly, and it's worth slowing all the way down for.

The fork is this: which "how far" are you asking about? There are two, and they're different numbers. Imagine walking 3 steps forward, then 3 steps back. How far did you travel? Six steps — your feet did real work. But where did you end up? Right where you started — net zero. Both answers are correct; they're answering different questions. Velocity handles them with one small distinction.

Displacement
Integrate velocity exactly as it is: ∫v dt. When the particle moves backward, velocity is negative, so it subtracts from the total. You get the net result — where it ended up relative to the start. (In the walk: zero.)
Total distance
Integrate the speed instead: ∫|v| dt. The absolute value flips backward motion to positive, so every bit of travel counts no matter the direction. You get every inch covered. (In the walk: six steps.)

They're equal only when the particle never reverses. The moment it turns around, they split — and the visual below lets you watch exactly that happen. Sweep the slider: the two numbers start identical and stay locked while the particle moves forward, then pull apart the instant it reverses.

Displacement ∫v dt
0.00
 
Total distance ∫|v| dt
0.00
 
they're still equal — keep sweeping
Sweep time  t = 0.00
v(t) = (t−1)(t−3): forward, then backward (red), then forward again.
Where students slip: they compute ∫v dt and call it the distance. By t = 4 above, that gives 4/3 — but the particle actually travelled 4, three times farther. The tell is always the same: if velocity changes sign anywhere on the interval, displacement and distance differ. For distance, find where v = 0 (the turning points), split the integral there, and add the pieces as positive values. Skipping that split is the single most common error in this whole unit.

3 · The accumulation function

So far we've integrated over a fixed interval to get one number. But what if you let the finish line move? Fix the start at a, and let the upper limit slide. Now the integral isn't a single value — it's a function of where you stop. That's the accumulation function: a running total that grows (or shrinks) as you sweep to the right.

The classic picture is a bank balance, or water in a tank. The rate is money flowing in and out each day; the accumulation function is your balance over time. On days money flows in (positive rate), the balance climbs. On days more flows out than in (negative rate), the balance drops. The rate drives the total — and that relationship is the whole point.

Watch the two graphs together below. The top is the rate f; the bottom is the accumulation F being drawn as you sweep. Three things to notice — they're the entire concept:

Sweep  x = 0.00
Accumulated total F(x)
0.00
= area under f from 0 to x
rate > 0
F rises. When the top curve is above the axis, you're adding area, so the running total climbs.
rate < 0
F falls. When the top curve dips below the axis, you're subtracting area, so the total drops back down.
rate = 0
F turns. Right where the rate crosses zero, F reaches a peak or a valley — because that's where it switches from adding to subtracting (or back). The rate hitting zero is F's slope hitting zero.
That last point is the Fundamental Theorem staring back at you: F′(x) = f(x). The rate is the slope of the accumulated total. Where the rate is big and positive, F climbs steeply; where the rate is zero, F is momentarily flat. The two graphs aren't just related — one is literally the slope-story of the other.

4 · Average value: the total, spread flat

Think about how you average test scores: add them up, divide by how many there are. But a function has infinitely many values across an interval — you can't "add them up and divide by how many." So what could the average of a function even mean? Here's the move that rescues it: instead of averaging points, ask about area.

Picture the area under a wavy curve. Now ask: what single flat height would trap that exact same area over the same interval? If you could press the wave down into a level rectangle without spilling any area, the height of that rectangle is the average value. It's the "fair" constant — the steady level that accounts for the same total as all the ups and downs combined. (Same logic as average speed: the one steady speed that would've covered your trip's distance in the same time.)

Deriving the formula
1
Call the average height c. We want the rectangle of height c across the interval [a,b] to have the same area as the curve.
2
The rectangle's area is height × width: c · (b − a).
3
The curve's area is the integral: ∫ₐᵇ f(x) dx.
4
Set them equal — that's the whole condition:
5
Solve for c by dividing both sides by (b − a). That is the average-value formula — not memorized, but built.
∫ₐᵇ f dx
The accumulated total — the whole area under the curve. By itself this is a total, not an average.
1/(b−a)
Dividing by the width spreads that total evenly across the interval. Dividing a total by how much you spread it over is exactly what turns a total into an average — same as distance ÷ time giving average speed.
The thing to remember: an integral gives you a total, never an average on its own. The division by b − a is the entire difference between "how much area" and "what average height." Forgetting to divide is forgetting to average.

See it directly: the area under the curve and the dashed rectangle at the average height enclose exactly the same area. The wave has been pressed flat into a level of equal size — that height is favg.

One guarantee worth knowing: the Mean Value Theorem for Integrals says a continuous curve must actually hit its average height at some point on the interval — the wave genuinely crosses the flat line, it isn't just an abstract average.

Three ideas, one engine. Net change integrates a rate to get a total. Motion is that with velocity as the rate — plus the displacement-vs-distance fork. Average value takes that total and divides by the width. Once you see the integral as "add up a rate, slice by slice," all three stop being separate formulas to memorize and become the same idea asked three ways.

5 · Drill it

Six questions across net change, the displacement-vs-distance trap, accumulation, and average value. Each shows a worked explanation after you answer.

Score: 0 / 0
Got it? The same "add up a rate" idea becomes geometric next: shading the region between two curves in Area Between Curves, or head back to Unit 8.