---
title: "AP Calculus 6.5 Accumulation Functions"
description: "Review behavior of accumulation functions in AP Calculus, including g(x) as an integral, signed area, increasing/decreasing, extrema, concavity, and graph interpretation."
canonical: "https://fiveable.me/ap-calc/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr"
type: "study-guide"
subject: "AP Calculus AB/BC"
unit: "Unit 6 – Integration and Accumulation of Change"
lastUpdated: "2026-06-09"
---

# AP Calculus 6.5 Accumulation Functions

## Summary

Review behavior of accumulation functions in AP Calculus, including g(x) as an integral, signed area, increasing/decreasing, extrema, concavity, and graph interpretation.

## Guide

An [accumulation function](/ap-calc/key-terms/accumulation-function "fv-autolink") $g(x)=\int_a^x f(t)\,dt$ stores the net signed area under $f$ from $a$ to $x$. Its behavior comes from the graph of $f$: $g$ increases where $f$ is positive, decreases where $f$ is negative, and changes concavity where $f$ changes from increasing to decreasing or back. For AP Calculus, read the [integrand](/ap-calc/key-terms/integrand "fv-autolink") graph before describing the accumulation function.

## Why This Matters for the AP Calculus Exam

This topic builds the skill of reading a [function](/ap-calc/unit-1/defining-continuity-at-point/study-guide/JbsR9iQfAzCznNOCG6JK "fv-autolink") defined by an integral without ever finding a formula for it. On both multiple-choice and free-response questions, you are often handed the graph of $f$ (frequently labeled as a derivative) and asked about the function it defines. You need to connect the sign of $f$ to whether $g$ is increasing or decreasing, the zeros of $f$ to the extrema of $g$, and the [slope](/ap-calc/key-terms/slope "fv-autolink") of $f$ to the concavity of $g$. Free-response questions in AP Calculus regularly use a graph made of semicircles and line segments so you can find exact areas with geometry, then interpret what those areas mean.

## Key Takeaways

- $g(x) = \int_a^x f(t)\,dt$ measures net signed area under $f$, so area below the x-axis counts as negative.
- By the [Fundamental Theorem of Calculus](/ap-calc/unit-6/fundamental-theorem-calculus-definite-integrals/study-guide/fEGd7E9gbOH8EtCf "fv-autolink"), $g'(x) = f(x)$ and $g''(x) = f'(x)$.
- $g$ increases where $f$ is positive and decreases where $f$ is negative; extrema of $g$ happen where $f$ changes sign.
- $g$ is [concave up](/ap-calc/key-terms/concave-up "fv-autolink") where $f$ is increasing and [concave down](/ap-calc/key-terms/concave-down "fv-autolink") where $f$ is decreasing; inflection points of $g$ occur where $f$ changes from increasing to decreasing or back.
- $g(a) = 0$ always, since the area from $a$ to $a$ is zero.
- When you only have a graph of $f$, use geometry (triangles, rectangles, semicircles) to compute exact accumulated area.

## The Fundamental Theorem of Calculus, Briefly

This is a quick summary. For the full setup, see the [6.4 guide](/ap-calc/unit-6/fundamental-theorem-calculus-accumulation-functions/study-guide/TyDEzN9M5wieA4Kw).

The Fundamental Theorem of Calculus connects an accumulation function to an [antiderivative](/ap-calc/key-terms/antiderivative "fv-autolink"). If

$$
g(x)=\int_{a}^{x}f(t)\,dt
$$

then

$$
g'(x) = f(x)
$$

The first equation says $g(x)$ is the accumulated area under $f(t)$ between the fixed lower bound $x = a$ and the variable upper bound $x$. Here $a$ is a constant and $x$ varies.

![Accumulation function diagram](https://storage.googleapis.com/static.prod.fiveable.me/images/Untitled.png-1708199653532-20503)

###### The accumulation function with boundaries $a$ and $x$ and the area under the curve $f(t)$ shaded and labeled as the value of $g$

The second equation says the derivative of $g(x)$ is $f(x)$. Equivalently, $g$ is an antiderivative of $f$. Putting these together: the accumulated area under $f$ equals the value of $g$ at that upper bound.

### Reading the Behavior of g from f

An integrally defined function has the same kinds of features as any other function: intervals of increase and decrease, extrema, concavity, and inflection points. Many AP questions hand you a graph or formula for $f$ and ask you to describe $g$.

#### Quick Reference Chart

| If g(x) is… | then g'(x) | and g''(x) |
|---|---|---|
| Increasing | + | --- |
| Decreasing | - | --- |
| Relative Maximum | 0 and changes from + to - | is - |
| Relative Minimum | 0 and changes from - to + | is + |
| Concave Up | --- | is + |
| Concave Down | --- | is - |
| Inflection Point | --- | Changes Sign |

Increasing and decreasing come from the [first derivative](/ap-calc/key-terms/first-derivative "fv-autolink"); concavity comes from the [second derivative](/ap-calc/unit-3/calculating-higher-order-derivatives/study-guide/Mh7ZnLES3ycIzEEKWl95 "fv-autolink"). A relative minimum or maximum cannot also be an inflection point.

The key move with [accumulation](/ap-calc/unit-8/using-accumulation-functions-definite-integrals-applied-contexts/study-guide/nUlJKvXqRcsfLnVMd5fG "fv-autolink") problems is that you are working backward. Instead of differentiating the given graph or formula, treat it as the already-found derivative of your target function. For example, if

$$
G(x) =\int_a^x 6n^2\,dn
$$

then $6x^2$ is $G'(x)$, so use it to find [critical points](/ap-calc/key-terms/critical-points "fv-autolink") or differentiate it again for concavity.

---

## How to Use This on the AP Calculus Exam

Almost all questions on this topic come as a graph, a table, or an equation, and all three show up regularly. The graph type is the most common, so that is the focus here.

### Free Response

Graph questions usually define $G(x) =\int_a^x f(t)\,dt$ and give you the graph of $f(t)$. The graph you see is not your target function. It is the derivative of your target function. Here is an official College Board free-response question from the [2022 exam](https://apcentral.collegeboard.org/media/pdf/ap22-frq-calculus-ab.pdf).

![2022 AP Calculus AB FRQ Question 3](https://storage.googleapis.com/static.prod.fiveable.me/images/Untitled_1.png-1708199653540-40981)

###### Question 3 from the 2022 AP Calculus AB FRQ Exam. Courtesy of College Board

Let $f$ be a [differentiable function](/ap-calc/key-terms/differentiable-function "fv-autolink") with $f(4)=3$. On the interval $0 \leq x \leq 7$, the graph of $f'$, the derivative of $f$, consists of a semicircle and two line segments, as shown in the figure above.

**(a)** Find $f(0)$ and $f(5)$.

**(b)** Find the x-coordinates of all [points of inflection](/ap-calc/key-terms/points-of-inflection "fv-autolink") on the graph of $f$ for $0<x<7$. Justify your answer.

**(c)** Let $g$ be the function defined by $g(x)=f(x)-x$. On what intervals, if any, is $g$ decreasing for $0 \leq x \leq 7$? Show the analysis that leads to your answer.

**(d)** For the function $g$ defined in part (c), find the [absolute minimum](/ap-calc/key-terms/absolute-minimum "fv-autolink") value on the interval $0 \leq x \leq 7$. Justify your answer.

#### Part (a)

Part (a) asks for $f(0)$ and $f(5)$. Since the graph is $f'(x)$, the original function satisfies

$$
f(x) = f(\text{start}) + \int_{\text{start}}^{x} f'(t)\,dt
$$

By the Fundamental Theorem of Calculus, the integral of $f'$ equals the net signed area between the curve and the x-axis, which is the change in $f$ over those bounds. Set up two equations.

The first is

$$
f(0) + \int_0^4 f'(x)\,dx = f(4)
$$

which rearranges to

$$
f(0) = f(4)- \int_0^4 f'(x)\,dx

$$

>💡 Watch your bounds. The lower bound is where you start and the upper bound is where you end. Arrange every setup as f(low bound) + (integral from low to high) = f(high bound).

Here, $f(4) = 3$, and $\int_0^4 f'(x)\,dx$ is the area under $f'(x)$ from $x = 0$ to $x = 4$, which is the semicircle region. Its area is

$$
\frac{\pi}{2}\,r^2 =\frac{\pi}{2}\cdot 2^2  = 2\pi
$$

Since this region is below the x-axis, the net signed area is negative. So

$$
f(0) = 3- (-2\pi)=3+2\pi

$$

For $f(5)$, set up

$$
f(4) + \int_4^5 f'(x)\,dx = f(5)
$$

The region from $x=4$ to $x=5$ lies under the first line segment, which has slope $\frac{2-0}{6-4} = 1$ using the points $(4,0)$ and $(6,2)$. At $x = 5$, the height is $1$, so the triangle has base $1$ and height $1$, giving area $\frac{1}{2}$. Then

$$
f(5) = 3 + \frac{1}{2}=3.5
$$

Write down your integral setups and clearly label your final answers as $f(0)$ and $f(5)$.

#### Part (b)

This asks for the x-coordinates of points of inflection of $f$. Since you are given $f'$, the inflection points of $f$ occur where the slope of $f'$ changes sign (that is, where $f''$ changes sign). On the graph, this happens at $x = 2$, where the slope of $f'$ changes from negative to positive, and at $x = 6$, where it changes from positive to negative.

>💡 There is a corner in $f'$ at $x = 4$, but it is not an inflection point of $f$. The slope of $f'$ goes from very steep and positive to a smaller positive constant without changing sign, so $f''$ does not change sign there.

#### Part (c)

The new function is

$$
g(x) =f(x) -x
$$

so

$$
g'(x) = f'(x) -1
$$

A function decreases when its derivative is negative, so $g$ is decreasing when

$$
f'(x)-1 < 0
$$

or

$$
f'(x) < 1
$$

From the graph, this holds on $0<x<5$.

#### Part (d)

This asks for the absolute minimum of $g(x)$. Since $g'(x) = f'(x) - 1$, a [critical point](/ap-calc/key-terms/critical-point "fv-autolink") happens where $f'(x) = 1$, which occurs at $x = 5$. The candidates are the [endpoints](/ap-calc/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ "fv-autolink") $x=0$ and $x=7$ plus the critical point $x = 5$.

| x | f(x) | g(x) = f(x) - x |
|---|---|---|
| 0 | 3 + 2pi | 3 + 2pi |
| 5 | 3.5 | 3.5 - 5 = -1.5 |
| 7 | 7.5 | 6.5 - 7 = -0.5 |

The values $f(0)$ and $f(5)$ come from part (a). The value $f(7)$ comes from $f(5) + \int_5^7 f'(x)\,dx = 3.5 + 3 = 6.5$, using geometry to find the area.

![Region from x=5 to x=7 split into a triangle and rectangle](https://storage.googleapis.com/static.prod.fiveable.me/images/Untitled_2.png-1708199653541-6920)

###### Integral between $x =5$ and $x =7$ broken into a triangle and a rectangle. Courtesy of College Board and Julianna Fontanilla

So the absolute minimum of $g(x)$ is $-1.5$, which occurs at $x=5$.

### Problem Solving Tips

1. 💭 **Anchor the relationship.** If $g(x)=\int_{a}^{x}f(t)\,dt$, then $g'(x) = f(x)$ and $g''(x) = f'(x)$. Write this somewhere on your paper.
2. 📈 **Use the graph.** Mark zeros, maxima, and minima of $f$, then relate each back to a feature of $g$.
3. 🧠 **Check every candidate.** Evaluate $g$ at critical points where $f$ changes sign, and don't forget the endpoints.
4. ✅ **Track which function you need.** It is easy to mix up $g$, $g'$, and $g''$. Confirm you are analyzing the right derivative before you answer.

## Common Misconceptions

- **The given graph is the function itself.** When a problem gives you the graph of $f$ and defines $g(x) = \int_a^x f(t)\,dt$, the graph is $g'$, not $g$. Read features of $g$ off of $f$.
- **Area is always positive.** Accumulation is net signed area. Regions below the x-axis subtract from the total, so a semicircle below the axis contributes a negative value.
- **Extrema of g come from zeros of f', not f.** The extrema of $g$ happen where $f$ (which equals $g'$) changes sign, not where $f$ has a maximum or minimum.
- **Every [corner](/ap-calc/key-terms/corner "fv-autolink") is an inflection point.** A corner or sharp change in $f$ only signals an inflection point of $g$ if the slope of $f$ actually changes sign there.
- **Forgetting endpoints in an absolute extremum search.** On a [closed interval](/ap-calc/key-terms/closed-interval "fv-autolink"), compare $g$ at the endpoints as well as at the interior critical points.
- **Mixing up g, g', and g''.** Increasing and decreasing of $g$ come from the sign of $f$; concavity of $g$ comes from whether $f$ is rising or falling.

## Related AP Calculus Guides

- [Unit 6 Overview: Integration and Accumulation of Change](/ap-calc/unit-6/review/study-guide/GcRakhgwYqAcIAg0y7WB)
- [6.11 Integrating Using Integration by Parts](/ap-calc/unit-6/integrating-using-integration-by-parts/study-guide/O4P3LchNoZnWElf8zETV)
- [6.1 Integration and Accumulation of Change](/ap-calc/unit-6/integration-accumulation-change/study-guide/NWRV9MaRJO4Eno32l5Xp)
- [6.12 Integrating Using Linear Partial Fractions](/ap-calc/unit-6/using-linear-partial-fractions/study-guide/VNjHMatlmaZoFyt7yLPH)
- [6.3 Riemann Sums, Summation Notation, and Definite Integral Notation](/ap-calc/unit-6/riemann-sums-summation-notation-definite-integral-notation/study-guide/RnM03H2k6l3ewvxX)
- [6.4 The Fundamental Theorem of Calculus and Accumulation Functions](/ap-calc/unit-6/fundamental-theorem-calculus-accumulation-functions/study-guide/TyDEzN9M5wieA4Kw)

## Vocabulary

- **accumulation function**: Functions that represent the accumulated total of a quantity over an interval, typically defined as g(x) = ∫[a to x] f(t) dt.
- **definite integral**: The integral of a function over a specific interval [a, b], representing the net signed area between the curve and the x-axis.

## FAQs

### How do accumulation functions involving area behave?

An accumulation function such as g(x) = integral from a to x of f(t) dt stores net signed area under f. Because g prime equals f, the sign and behavior of f tell you where g increases, decreases, has extrema, and changes concavity.

### How do you tell where an accumulation function is increasing or decreasing?

If g(x) = integral from a to x of f(t) dt, then g prime equals f. So g increases where f is positive and decreases where f is negative.

### Where do extrema occur for an accumulation function?

Extrema of g occur where g prime changes sign. Since g prime equals f, look for x-values where f crosses the x-axis and changes from positive to negative or negative to positive.

### How do you determine concavity of an accumulation function?

Concavity of g comes from g double prime. Since g prime equals f, g double prime equals f prime. That means g is concave up where f is increasing and concave down where f is decreasing.

### Why does signed area matter for accumulation functions?

Accumulation functions measure net signed area. Area above the x-axis adds to the total, while area below the x-axis subtracts from the total, so you must track sign when evaluating g values.

### How is Topic 6.5 tested on the AP Calculus exam?

AP Calculus questions often give a graph, table, or formula for f and define g as an integral of f. You may need to find g values using area, identify increasing/decreasing intervals, determine extrema, or explain concavity and inflection points.

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