Integration · Lesson 4 of 4 · ~6 min

Signed area

Predict

$f(x) = \sin x$ runs one full period on $[0, 2\pi]$: a hump above the axis, then a matching hump below. What is $\int_0^{2\pi} \sin x\,dx$?

Hold that thought. Drag bound b past $3.14$ and watch what the net ∫ readout does on the way to $2\pi$.

Try

Try itDrag bound b from $0$ to $6.28$, or press Sweep b → 2π. Past $b = 3.14$ the fill drops below the axis, dims, and picks up a NEG tag — and the net ∫ and total area readouts split.

Notice

Up to $b = 3.14$ the two readouts march together, reaching $2.0000$ side by side. Then they part ways. At $b = 4.71$ the net has fallen back to $1.0024$ while the total has climbed to $2.9976$. At $b = 6.28$ the net reads $0.0000$ and the total reads $4.0000$.

Explain

Each contribution $f(x)\,dx$ is a height times a width, and below the axis the height $\sin x$ is negative. The integral adds those negative slices in, so the running net falls — the definite integral computes net signed area:

\int_0^{2\pi} \sin x\,dx \;=\; \big[-\cos x\big]_0^{2\pi} \;=\; \cos 0 - \cos 2\pi \;=\; 0

Zero net does not mean nothing happened. To measure the region while ignoring sign, integrate the absolute value: $\int_0^{2\pi} |\sin x|\,dx = 4$, the total area stat. The AP exam runs both meanings through velocity: the net integral of $v$ is displacement; the integral of speed $|v|$ is distance traveled. The third stat, total minus net, is twice the region below the axis.

Apply

Checkpoint — prove it to yourself

At $b = 6.28$ the instrument shows net ∫ $= 0.0000$. What does that zero mean?

A particle moves with velocity $v(t) = \sin t$ for $0 \le t \le 2\pi$. What are its displacement and its distance traveled?

Transfer: $g(x) = x$ on $[-2, 2]$ — a line through the origin. What are the net signed area and the total area?

Key idea

One region, two measurements. The definite integral reports net signed area — below-axis slices count negative — while $\int |f|$ reports total area. Always ask which one the question wants.

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