Math · Calculus

Integral as Accumulation / Area

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Wrap-up

What you learned

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Read next: Derivative as Slope / Local Rate of ChangeConnect total change back to local rate

Key takeaway

The accumulation function is the signed area built from the start to the current bound.
The source value sets how the total is changing right now, not the total itself.
Negative source regions subtract area from whatever positive total was already built.

Common misconception

If the source height is negative, the accumulated total must already be negative too.

A negative source height only tells you that the running total is decreasing at that moment.

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Optimization / Maxima, Minima, and Constraints

Move one rectangle width under a fixed 24 meter perimeter and watch height, area, and local slope respond together so the square maximum emerges from the same bench.

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Limits and Continuity / Approaching a Value

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Reference and support

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Equation snapshotAccumulation rule · Accumulation slopeShow 3 equations

Accumulation rule
$A (x) = \int_{0}^{x} f (t) d t$
Defines A(x) as the signed area collected from 0 to the current bound x.
Accumulation slope
$A^{'} (x) = f (x)$
The local slope of the accumulation graph at x is exactly the current source height f(x).
Small-step accumulation
$Δ A \approx f (x) Δ x$
Over a small step in x, the change in the running total is approximately height times width.

Why it behaves this way

Explanation

This bench treats an integral as a running total, not a frozen shaded picture. As you move the upper bound, the signed area on the source graph and the matching point on the accumulation graph update together, so you can watch the total build, level off, or fall.

The key distinction is local height versus accumulated amount. The source value f(x) tells you whether the next thin slice adds or subtracts right now. The accumulation A(x) remembers all signed area from 0 to the current bound, so it can stay positive even after the source curve has dropped below the axis.

Key ideas

01A(x) is the signed area from 0 to the current upper bound, so moving the bound changes the running total rather than creating a separate static picture.

02When f(x) is above the axis, new slices add to A(x); when f(x) is below the axis, new slices subtract from A(x).

03Because A'(x) = f(x), the current source height gives the local slope of the accumulation graph, but that slope is not the same thing as the total accumulated so far.

Worked examples

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Frozen walkthrough

Step through the frozen example

Frozen walkthrough

Use the upper-bound slider, the signed-area shading, and the accumulation point together. First read the source height at the current bound, then read the total that has been built so far.

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Example 1 of 2

Frozen valuesUsing frozen parameters

At the current upper bound, what total signed area has accumulated from 0 to x?

x

Upper bound

1.2

1. Read the active bound on the source graph

The current upper bound is x = 1.2, and the source height there is f(x) = 0.52.

2. Read the matching point on the accumulation graph

At that same x-value, the accumulation graph shows A(x) = 1.01.

3. Connect the local slope to the source height

Because A'(x) = f(x), the accumulation graph has local slope 0.52 at this point.

Current running total

A (x) = 1.01

The signed area collected from 0 up to this bound is still net positive.

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Starter track

Step 6 of 6

Functions and Change

Integral as Accumulation / Area appears later in this track, so it is cleaner to start from the beginning first.

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