Yes. From a to a there is no width, so there is no area, and aa f(t)\,dt = 0 for every f.

Unit 6 · Topic 6.5 · about 30 minutes · 7 parts · 13 scored checks, 4 of them multiple choice

AP Calculus AB 6.5: Interpreting the Behavior of Accumulation Functions Involving Area

Q: Why is the variable inside the integral t and not x?

x is already the upper limit, the place where the total stops. Using a second letter for the variable of integration keeps the two jobs apart. Any letter other than x would work.

Q: Can g be negative and increasing at the same time?

Yes. If the area below the axis comes first, g is negative for a while, and as soon as f turns positive, g starts rising even though its values are still below 0. Increasing is about g'; negative is about g.

Q: What if the graph of f has a corner?

If f is continuous, g'(x) = f(x) still holds everywhere, so g is smooth. At a corner of f, g'' does not exist, but you can still read the concavity of g on each side from whether f is increasing or decreasing.

Q: How do I find values of g when f is only a table?

If you are told the graph is made of segments joining the points, the pieces are trapezoids and triangles, and the sum is exact. Otherwise the table only lets you estimate g, with a Riemann or trapezoidal sum.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 6: 15 to 20% of the multiple-choice section

What you'll be able to do

Represent an accumulation function as a definite integral, $g (x) = \int_{a}^{x} f (t) d t$ , and use a graph, a table, a formula or a description of $f$ to find values of $g$ and to describe where $g$ increases, decreases, has extreme values and bends.

Key vocabulary

Term	What it means	The exam trap
Accumulation function	A function defined by a definite integral with a variable upper limit, $g (x) = \int_{a}^{x} f (t) d t$ .	$g (x)$ and $f (x)$ are different numbers; say which one a question is asking about before you read the graph.
Definite integral	$\int_{a}^{b} f (t) d t$ , which for a graph made of lines and circles is the area between the graph and the $t$ -axis from $a$ to $b$ , counted with signs.	Before you add pieces of area, check which side of the axis each piece is on.
Graphical, numerical, analytical and verbal representations	A function given by a graph, a table of values, a formula, or a description in words.	Each one carries the same kind of facts about the accumulation function; work from the one you are given rather than looking for a formula.

Graph of a continuous function f for t from 0 to 8: a line segment from (0, -2) up to (2, 0) below the t-axis, a semicircle of radius 2 above the axis from (2, 0) through its top at (4, 2) to (6, 0), and a line segment from (6, 0) down to (8, -3) below the axis. The region above the axis and the two regions below it are shaded in different colors.

Part 1 Think first When is the rain barrel fullest?

Commit to a guess before you read on. Trying first, even when the guess is wrong, makes the lesson easier to remember.

Predict first

Water runs into a rain barrel from a gutter. At noon the flow is 0 gallons per hour. It climbs steadily to 6 gallons per hour at 2 p.m., then slows steadily to 0 at 5 p.m. At 5 p.m. someone opens the tap, and from then until 7 p.m. water leaves the barrel at a steadily growing rate, reaching 4 gallons per hour going out at 7 p.m. At what time between noon and 7 p.m. does the barrel hold the most water?

In short. An accumulation function $g (x) = \int_{a}^{x} f (t) d t$ keeps a running total of the signed area between the graph of $f$ and the axis from $a$ to $x$ , so any graph, table, formula or description of $f$ tells you about $g$ .

$g (a) = 0$ , and for $x > a$ each value $g (x)$ is the area above the axis minus the area below it, from $a$ to $x$ .
When $f$ is continuous, $g^{'} (x) = f (x)$ , so the sign of $f$ says where $g$ increases or decreases.
Whether $f$ is increasing or decreasing says where the graph of $g$ is concave up or concave down.

Part 2 Learn it Reading g from f

2a. A running total of signed area

Pick a starting input $a$ , and for each $x$ let

g (x) = \int_{a}^{x} f (t) d t .

The number $g (x)$ is the area between the graph of $f$ and the $t$ -axis from $a$ to $x$ , counted as positive where the graph is above the axis and negative where it is below. For now, take $x$ to the right of $a$ ; 6.6 shows what changes when $x$ is to the left. The letter $t$ is only the variable of integration; $x$ is where the total stops. Such a function is an accumulation function; in Think first it was the water gained since noon.

At $x = a$ no area has been collected yet, so $g (a) = 0$ .

When the graph of $f$ is made of lines and circles, geometry gives every area: a triangle is $\frac{1}{2} b h$ , and a half circle of radius $r$ is $\frac{1}{2} π r^{2}$ . A formula can hide a circle, too: $y = \sqrt{r^{2} - t^{2}}$ is the top half of a circle of radius $r$ centered at the origin.

For example, let the graph of $f$ run level from $(0, 3)$ to $(2, 3)$ and then straight down to $(5, - 3)$ , crossing the axis at $t = 3.5$ , and let $g (x) = \int_{0}^{x} f (t) d t$ . The rectangle from 0 to 2 gives $g (2) = 6$ . The triangle above the axis from 2 to 3.5 adds 2.25, so $g (3.5) = 8.25$ . The triangle below the axis from 3.5 to 5 subtracts 2.25, so $g (5) = 6$ .

Quick check · Calculate

No calculator. The graph of $h$ is made of two line segments: one from $(0, 3)$ to $(3, 0)$ and one from $(3, 0)$ to $(6, - 2)$ . Let $k (x) = \int_{0}^{x} h (t) d t$ . What is $k (6)$ ?

Your answer

2b. The sign of f says where g rises and falls

By the Fundamental Theorem of Calculus from the last topic, when $f$ is continuous, $g^{'} (x) = f (x)$ . So a graph of $f$ is a graph of $g^{'}$ , and what Unit 5 says about a derivative applies to $g$ :

where $f (x) > 0$ , $g$ is increasing, and where $f (x) < 0$ , $g$ is decreasing;
where $f$ changes from positive to negative, $g$ has a relative maximum, and where $f$ changes from negative to positive, $g$ has a relative minimum.

The height of $f$ says how fast $g$ is growing, not how large $g$ is. In the example from the last section, $f$ is at its highest on $[0, 2]$ , yet $g$ keeps growing past $t = 2$ , all the way to $t = 3.5$ , where $f$ crosses from positive to negative. That is where $g$ has its relative maximum, $g (3.5) = 8.25$ .

For an absolute maximum or minimum on a closed interval, use the Candidates Test: compare $g$ at those points with $g$ at both endpoints.

Quick check · Decide and justify

No calculator. Let $g (x) = \int_{0}^{x} (t - 1) (t - 4) d t$ for $0 \leq x \leq 6$ . A classmate says $g$ has a relative maximum at $x = 1$ . Is the classmate right?

Which reason supports your answer?

2c. Whether f rises or falls says how g bends

Concavity comes from the derivative. The graph of $g$ is concave up where $g^{'} = f$ is increasing and concave down where $f$ is decreasing. So for concavity, read whether the graph of $f$ is going up or down, not which side of the axis it is on. A point of inflection of $g$ sits where $f$ changes from increasing to decreasing, or the other way, which on a graph of $f$ is a peak or a valley.

A table works the same way. Suppose the graph of $f$ is made of segments joining these points:

Values of f, joined by line segments
$t$	0	1	2	3	4
$f (t)$	1	3	4	3	0

Every value before $t = 4$ is positive, so $g (x) = \int_{0}^{x} f (t) d t$ increases on $(0, 4)$ . The values climb to 2 and then fall, so the graph of $g$ is concave up on $(0, 2)$ , concave down on $(2, 4)$ , and has a point of inflection at $x = 2$ .

Words carry the same facts. If a pump's rate is positive but falling, the total pumped still goes up, just more slowly each minute: increasing and concave down.

Quick check · Match

No calculator. Let $g (x) = \int_{a}^{x} f (t) d t$ , where $f$ is continuous. Tap a fact about $f$ , then the fact about $g$ it gives.

Common mistakes

The mistake	Why it is wrong	Do this instead
Reading $g (x)$ as the height of the graph of $f$ at $x$ .	$g (x)$ is the signed area from $a$ to $x$ . The height $f (x)$ is how fast that total is changing at $x$ .	Before you answer, ask whether the question is about $g$ , an area, or about $g^{'} = f$ , a height.
Putting the maximum of $g$ where $f$ is highest.	Where $f$ is highest, $g$ is growing fastest, and it keeps growing for as long as $f$ stays positive.	Look for where $f$ changes from positive to negative, then compare with the endpoints.
Reading the concavity of $g$ from the sign of $f$ .	Concavity depends on whether $g^{'} = f$ is increasing or decreasing, not on whether it is positive.	Use the sign of $f$ for where $g$ rises and falls, and the direction of $f$ for how $g$ bends.

Part 3 Watch it done The smallest value of an accumulation function

Watch each step, then finish the last one yourself.

Let $g (x) = \int_{0}^{x} (2 - t) d t$ for $0 \leq x \leq 5$ . Find the absolute minimum value of $g$ on $[0, 5]$ .

Step 1 · Find the derivative
The integrand $2 - t$ is continuous, so $g^{'} (x) = 2 - x$ . It is 0 only at $x = 2$ .
Step 2 · Read the sign
$g^{'} (x) > 0$ on $(0, 2)$ and $g^{'} (x) < 0$ on $(2, 5)$ . So $g$ rises and then falls, and its only critical point, $x = 2$ , is a relative maximum.
Why this step?
Why must the absolute minimum be at an endpoint?
On a closed interval, a continuous function takes its absolute minimum at a critical point or at an endpoint. The only critical point here is a maximum, so the minimum has to be at $x = 0$ or at $x = 5$ .
Step 3 · Find the areas
$g (0) = 0$ . From 0 to 2 the line $y = 2 - t$ is above the axis, a triangle of area $\frac{1}{2} (2) (2) = 2$ . From 2 to 5 it is below the axis, a triangle of area $\frac{1}{2} (3) (3) = 4.5$ .
Your step · Calculate

No calculator. Finish it. For $g (x) = \int_{0}^{x} (2 - t) d t$ , $g (0) = 0$ , the region from 0 to 2 has area 2 above the axis, and the region from 2 to 5 has area 4.5 below the axis. What is the absolute minimum value of $g$ on $[0, 5]$ ?

Your answer

Part 4 Try it A circle and a sketch

Try it · Multiple choice

No calculator. Let $g (x) = \int_{- 3}^{x} \sqrt{9 - t^{2}} d t$ for $- 3 \leq x \leq 3$ . What is $g (0)$ ?

Try it · Sketch the graph

No calculator. Let $f (t) = 4 - t$ , and let $g (x) = \int_{0}^{x} f (t) d t$ for $0 \leq x \leq 4$ . Sketch the shape of the graph of $g$ : drag the points so it rises or falls, and bends, the way $g$ does. The first point is fixed at $g (0) = 0$ . The exact heights are up to you.

Drag a point up or down, or tap it and use the arrow keys. Only the shape is checked.

Part 5 Explain it From the graph of f to the behavior of g

Free response is written by hand on the exam, so answer on paper first. Nothing you type here is sent anywhere. Then open the model answer and score yourself one point at a time.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. The graph of the continuous function $f$ at the top of this lesson is made of a line segment from $(0, - 2)$ to $(2, 0)$ , a semicircle above the $t$ -axis from $(2, 0)$ to $(6, 0)$ , and a line segment from $(6, 0)$ to $(8, - 3)$ . Let $g (x) = \int_{0}^{x} f (t) d t$ for $0 \leq x \leq 8$ .

(a) Find $g (6)$ and $g (8)$ .

(b) Find the $x$ -coordinate of each point at which $g$ has a relative maximum on the interval $0 < x < 8$ . Justify your answer.

(d) Water flows into and out of a tank, and $f (t)$ is the rate of flow in liters per minute, $t$ minutes after noon, with positive values meaning water flows in. Using correct units, interpret the meaning of $g (8)$ in the context of the problem.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $g (6) = \int_{0}^{2} f (t) d t + \int_{2}^{6} f (t) d t = - \frac{1}{2} (2) (2) + \frac{1}{2} π (2)^{2} = - 2 + 2 π$ .

$g (8) = g (6) + \int_{6}^{8} f (t) d t = (2 π - 2) - \frac{1}{2} (2) (3) = 2 π - 5$ .

(b) $g^{'} (x) = f (x)$ . $f$ changes from positive to negative only at $x = 6$ , so $g$ has a relative maximum at $x = 6$ . (At $x = 2$ , $f$ changes from negative to positive, which gives a relative minimum.)

(d) $g (8) = 2 π - 5$ , about 1.28 liters, is the net change in the water in the tank from noon to 12:08 p.m.: the tank holds $2 π - 5$ liters more water at $t = 8$ than at $t = 0$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Finds $g (6) = 2 π - 2$ : the triangle below the axis subtracts 2 and the semicircle adds $2 π$ .
Finds $g (8) = 2 π - 5$ , subtracting the triangle of area 3 on $[6, 8]$ .
Answers $x = 6$ only, because $g^{'} = f$ changes from positive to negative there.
Answers $(4, 8)$ , or $(4, 6)$ and $(6, 8)$ , because $g^{'} = f$ is decreasing there.
Says the tank holds $2 π - 5$ liters (about 1.28 liters) more water at $t = 8$ than at $t = 0$ , with units.

Part 6 Check yourself A function given by a table

All three case questions use the same table. Answer them from it.

The graph of a continuous function $f$ is made of line segments joining the points in the table. Let $g (x) = \int_{0}^{x} f (t) d t$ for $0 \leq x \leq 6$ .

Values of f, joined by line segments
$t$	0	1	2	3	4	5	6
$f (t)$	$- 2$	0	3	4	1	0	$- 1$

Case question 1 · Calculate

No calculator. What is $g (4)$ ?

Your answer

Case question 2 · Multiple choice

No calculator. Which open interval is the full set of $x$ in $(0, 6)$ where the graph of $g$ is concave down?

Case question 3 · Multiple choice

No calculator. At which value of $x$ does $g$ attain its absolute maximum on the interval $[0, 6]$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds where the graph of $g (x) = \int_{0}^{x} (t^{2} - 4 t) d t$ is concave up on $(0, 5)$ . Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. Tap a blank, then tap an answer.

Let $g (x) = \int_{3}^{x} f (t) d t$ . Then $g (3) =$ . On any interval where the graph of $f$ is below the $t$ -axis, $g$ is .

Part 7 Keep it Bring back earlier ideas

Two questions from earlier topics, so they stay fresh.

Bring it back · Calculate

No calculator. Water flows into a tank at $r (t)$ liters per hour, where $t$ is in hours. Measurements give $r (0) = 4$ , $r (2) = 7$ , $r (5) = 3$ and $r (6) = 1$ . Using a right Riemann sum with the three subintervals these values mark, estimate $\int_{0}^{6} r (t) d t$ , in liters.

Your answer

Bring it back · Multiple choice

No calculator. If $h (x) = \sin (x^{2})$ , what is $h^{'} (x)$ ?

In three lines

$g (x) = \int_{a}^{x} f (t) d t$ is the signed area from $a$ to $x$ , so $g (a) = 0$ and geometry gives its values.
$g^{'} = f$ , so the sign of $f$ gives where $g$ rises, falls and has relative extrema.
Whether $f$ increases or decreases gives where the graph of $g$ is concave up or down.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a question gives you $f$ and asks about $g (x) = \int_{a}^{x} f (t) d t$ , write $g^{'} (x) = f (x)$ first, then justify every claim about $g$ with $g^{'}$ or $f$ : " $g$ has a relative maximum at $x = c$ because $g^{'} = f$ changes from positive to negative at $c$ ." A justification that names only "the graph" or "it" does not say which function you mean.

Questions students ask

Why is the variable inside the integral

t

and not

x

$x$ is already the upper limit, the place where the total stops. Using a second letter for the variable of integration keeps the two jobs apart. Any letter other than $x$ would work.

g (a)

always 0?

Yes. From $a$ to $a$ there is no width, so there is no area, and $\int_{a}^{a} f (t) d t = 0$ for every $f$ .

Can

g

be negative and increasing at the same time?

Yes. If the area below the axis comes first, $g$ is negative for a while, and as soon as $f$ turns positive, $g$ starts rising even though its values are still below 0. Increasing is about $g^{'}$ ; negative is about $g$ .

What if the graph of

f

has a corner?

If $f$ is continuous, $g^{'} (x) = f (x)$ still holds everywhere, so $g$ is smooth. At a corner of $f$ , $g^{''}$ does not exist, but you can still read the concavity of $g$ on each side from whether $f$ is increasing or decreasing.

How do I find values of

g

when

f

is only a table?

If you are told the graph is made of segments joining the points, the pieces are trapezoids and triangles, and the sum is exact. Otherwise the table only lets you estimate $g$ , with a Riemann or trapezoidal sum.

Course alignment, for teachers

AP Calculus AB topic 6.5, Unit 6: Integration and Accumulation of Change.

Skill 2.D: Identify how mathematical characteristics or properties of functions are related in different representations.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first When is the rain barrel fullest?

Part 2 Learn it Reading g from f

2a. A running total of signed area

2b. The sign of f says where g rises and falls

2c. Whether f rises or falls says how g bends

Part 3 Watch it done The smallest value of an accumulation function

Part 4 Try it A circle and a sketch

Part 5 Explain it From the graph of f to the behavior of g

Part 6 Check yourself A function given by a table

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic