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Accumulation Functions
and Area

Learn how to interpret the behavior of accumulation functions defined as integrals, connecting signed area to increasing, decreasing, and extreme values.

AP Calculus AB

What Is an Accumulation Function?

An accumulation function is a function defined by a definite integral with a variable upper limit. It "accumulates" the signed area under a curve \( f(t) \) from a fixed starting point \( a \) up to a variable endpoint \( x \).

The value \( F(x) \) represents the net signed area between the graph of \( f \) and the \( t \)-axis from \( t = a \) to \( t = x \). Area above the axis counts as positive, area below as negative.

Visualizing Signed Area

t f(t) Positive area Negative area a x

Why Does This Matter?

FTC Connection
The Fundamental Theorem of Calculus says \( F'(x) = f(x) \). The integrand is the derivative of the accumulation function.
Graph Reading
The AP exam frequently provides a graph of \( f \) and asks about the behavior of \( F \): increasing, decreasing, extrema, concavity.
Real-World
Accumulation functions model total distance, total change, net displacement: any quantity that builds up over time.

The Fundamental Theorem (Part 1)

This theorem is the bridge: to understand how \( F \) behaves (increasing, decreasing, concavity), we study \( f \) and \( f' \) directly.

\( f(x) > 0 \)
\( F'(x) > 0 \) so \( F \) is increasing
\( f(x) < 0 \)
\( F'(x) < 0 \) so \( F \) is decreasing
\( f(x) = 0 \)
\( F'(x) = 0 \): possible extremum of \( F \)

Connecting \( f \) to \( F \)

f(t) f > 0 f < 0 f > 0 F(x) max min

Follow the salt through a working mixing rig

A stirred transparent tank has one upper inlet and one lower outlet with a valve. The exact diagram links its 100 L salt balance to a computed approach to equilibrium.For net flow R(t) = 4t − t² gal/min and 10 gal initially, V(t) = 10 + 2t² − t³/3. Volume peaks at t = 4 minutes, when the net rate changes from positive to negative.

Signed net flow and inventory

For net flow R(t) = 4t − t² gal/min and 10 gal initially, V(t) = 10 + 2t² − t³/3. Volume peaks at t = 4 minutes, when the net rate changes from positive to negative.For net flow R(t) = 4t − t² gal/min and 10 gal initially, V(t) = 10 + 2t² − t³/3. Volume peaks at t = 4 minutes, when the net rate changes from positive to negative.

Key Distinction: \( f \) vs \( F \)

Property \( f(x) \) (the integrand) \( F(x) \) (accumulation)
Role Rate of change of \( F \) Net signed area from \( a \) to \( x \)
Sign of \( f \) Above/below axis Increasing/decreasing
Zero of \( f \) Crosses the \( t \)-axis Possible extremum of \( F \)
Sign of \( f' \) \( f \) increasing/decreasing Concavity of \( F \)
Zero of \( f' \) Extremum of \( f \) Inflection point of \( F \)

Concavity of \( F \)

Since \( F'(x) = f(x) \), it follows that \( F''(x) = f'(x) \). So the concavity of the accumulation function depends on whether the integrand is increasing or decreasing.

\( f'(x) > 0 \)
\( f \) is increasing, so \( F''(x) > 0 \) and \( F \) is concave up.
\( f'(x) < 0 \)
\( f \) is decreasing, so \( F''(x) < 0 \) and \( F \) is concave down.

How to Analyze \( F(x) \)

1
Identify the integrand \( f \). The graph or formula given is the function inside the integral, not \( F \) itself.
2
Find where \( f = 0 \). These are candidates for extrema of \( F \). Check if \( f \) changes sign.
3
Determine sign of \( f \). Where \( f > 0 \), \( F \) increases. Where \( f < 0 \), \( F \) decreases.
4
Find extrema of \( f \). Where \( f \) has a max or min, \( F \) has an inflection point (since \( F'' = f' \) changes sign).
5
Evaluate \( F(a) = 0 \). The accumulation function always starts at zero when \( x = a \).

Example 1: Reading the Graph

Let \( F(x) = \displaystyle\int_0^x f(t)\,dt \), where \( f \) is shown below. Find where \( F \) is increasing, decreasing, and has extrema.

t f(t) 2 4 6 8 t=4 t=6

Example 2: Evaluating \( F \) Numerically

Using the same graph of \( f \) from Example 1, suppose the area of the region above the axis on \( [0,4] \) is 10 and the area below the axis on \( [4,6] \) is 3. Find \( F(4) \) and \( F(6) \).

1
\( F(0) = \int_0^0 f(t)\,dt = 0 \). The accumulation always starts at zero.
2
\( F(4) = \int_0^4 f(t)\,dt = \) 10, the entire positive region.
3
\( F(6) = \int_0^6 f(t)\,dt = 10 + (-3) = \) 7. We add the negative area.

Example 3: Concavity and Inflection

t f(t) 2 4 6 max of f = inflection of F f increasing F concave up f decreasing F concave down

The Starting Value: \( F(a) = 0 \)

A common source of confusion: the accumulation function always has the value zero at the lower limit of integration, regardless of what \( f(a) \) equals.

\( F(a) \)
\( \displaystyle\int_a^a f(t)\,dt = 0 \). No interval, no area, so the value is always zero.
\( f(a) \)
This is the height of the integrand at \( t = a \). It equals \( F'(a) \), the initial slope of \( F \).

Absolute Extrema of \( F \)

To find the absolute maximum or minimum of \( F \) on a closed interval, compare the values of \( F \) at:

1
Endpoints of the interval: \( F(a) = 0 \) and \( F(b) \).
2
Critical points: where \( f(x) = 0 \) and \( f \) changes sign.
3
Compare values. Compute \( F \) at each candidate by adding/subtracting the geometric areas.

Common Mistakes

Confusing \( f \) and \( F \)
Students read the graph as \( F \) when the problem states it is \( f \). Always check: the graph shows the integrand, not the accumulation.
Forgetting signed area
Below-axis area is negative. Do not take absolute values unless the problem asks for total distance.
Thinking \( F(a) = f(a) \)
\( F(a) = 0 \) always. The starting value of the accumulation is zero, not the height of \( f \) at \( a \).

Summary Table

If you observe... On \( f \) Then \( F \)...
Positive values \( f(x) > 0 \) is increasing
Negative values \( f(x) < 0 \) is decreasing
Zero crossing (+ to -) \( f \) changes from + to - has a local maximum
Zero crossing (- to +) \( f \) changes from - to + has a local minimum
Increasing \( f' > 0 \) is concave up
Decreasing \( f' < 0 \) is concave down
Local extremum \( f' = 0 \), sign change has an inflection point

Question 1 of 5

Let \( F(x) = \displaystyle\int_0^x f(t)\,dt \). If \( f(3) = 0 \) and \( f \) changes from positive to negative at \( t = 3 \), what happens to \( F \) at \( x = 3 \)?

Question 2 of 5

If \( f \) is positive and increasing on \( (1, 5) \), what can you say about \( F(x) = \displaystyle\int_1^x f(t)\,dt \) on that interval?

Question 3 of 5

Given \( F(x) = \displaystyle\int_2^x f(t)\,dt \), what is the value of \( F(2) \)?

Question 4 of 5

The graph of \( f \) has a local maximum at \( t = 4 \). What does \( F(x) = \displaystyle\int_0^x f(t)\,dt \) have at \( x = 4 \)?

Question 5 of 5

Let \( F(x) = \displaystyle\int_0^x f(t)\,dt \). On \( [0,3] \) the area above the axis is 8, and on \( [3,7] \) the area below the axis is 5. What is \( F(7) \)?