Unit 6Fundamental Theorem of Calculus, Part 1

Concept 4.71Built for May 2027 AP Exam

Fundamental Theorem of Calculus, Part 1

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The first part of the Fundamental Theorem says that differentiating an accumulation function gives back the integrand. If $g(x) = \int_a^x f(t)\,dt$ and $f$ is continuous, then $g'(x) = f(x)$ . The variable limit is what makes it work: the upper limit is the only thing changing, so the rate at which area accumulates is exactly the height of the curve at that limit. Two facts follow at once. The derivative of $g$ is read off the graph of $f$ with no computation, and the sign of $f$ decides where $g$ rises, falls, and turns. When the upper limit is a function of $x$ , the chain rule joins in. That version is where most points are lost.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A graph of $f$ with a companion function $g(x) = \int_a^x f(t)\,dt$ , then questions about where $g$ increases, where $g$ is concave, where $g$ has a critical point, or what $g'(c)$ equals. Sometimes $g(x) = \int_a^{x^2} f(t)\,dt$ and the question is $g'(2)$ .
What you’ll do: Replace the integral with its derivative in one line, then answer every part from the graph of $f$ . For a composite limit, multiply by the derivative of the upper limit.
Memory hook: Differentiate the integral, get the inside back. The integrand returns, evaluated at the upper limit.
Use it on the exam: For MCQs: When the integrand contains $t$ and the limit contains $x$ , the answer is $f$ at the limit. If the limit is $x^2$ , multiply by $2x$ ; forgetting that factor is the standard wrong choice. For FRQs: State the theorem by name on the line where you use it. “By the Fundamental Theorem of Calculus, $g'(x) = f(x)$ ” is a scored sentence; writing $g' = f$ with no justification is not.

Concept Review

The Area Function and Its Slope

Figure 1 — The shaded region is $g(x)$ , the area accumulated from $a$ to $x$ . Push the right edge a little further and the strip you add has height $f(x)$ , so the rate of growth of the area is the height of the curve.

Let $f$ be continuous on an interval containing $a$ , and define

$g(x) = \int_a^x f(t)\,dt$

Then $g'(x) = f(x)$ for every $x$ in that interval. Read this as a statement about rates. The area under the curve from $a$ to $x$ grows as $x$ moves, and the amount it grows per unit of $x$ is the height of the curve at $x$ , because the added sliver is a strip of width $dx$ and height $f(x)$ .

Object	What it is	How to read it
$g(a)$	the area from $a$ to $a$	always $0$
$g'(x)$	the rate the area grows	the height $f(x)$
$g''(x)$	the rate that rate changes	the slope $f'(x)$
$g(b) - g(a)$	the total accumulation	the value of the integral

The shaded region is $g(x)$, the area accumulated from $a$ to $x$. Push the right edge a little further and the strip you add has height $f(x)$, so the rate of growth of the area is the height of the curve.

The Chain Rule Version

When the upper limit is a function $u(x)$ rather than $x$ , the theorem still applies but picks up a factor:

$\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f\left(u(x)\right)u'(x)$

The reason is one sentence long. The integral grows $f(u)$ times as fast as $u$ grows, and $u$ grows $u'$ times as fast as $x$ , so the rates multiply. Lower limits can vary too, and then the two contributions subtract, because the lower limit removes area as it rises.

Reading $g$ From the Graph of $f$

Every property of $g$ is a property of $f$ one level down.

Question about $g$	Look at $f$ for	Answer
where is $g$ increasing?	where $f > 0$	those intervals
where does $g$ have a critical point?	where $f = 0$ or is undefined	those $x$ -values
where is $g$ concave up?	where $f$ is increasing	those intervals
where does $g$ have an inflection point?	where $f$ changes direction	those $x$ -values
what is the extreme value of $g$ ?	signed area, plus $g(a) = 0$	accumulate from $a$

The last row is the one that costs points, because the extrema of $g$ are not at the zeros of $f$ in any simple pattern. A zero of $f$ where $f$ switches from positive to negative is a local maximum of $g$ , and the value is the accumulated area up to that point. Nothing about $g$ can be read without accumulating.

Why Continuity Is in the Statement

The theorem needs $f$ continuous at the point where the derivative is taken. A jump in $f$ produces a corner in $g$ , and at that corner $g'$ does not exist, so the equation $g'(x) = f(x)$ fails there and only there. On exam graphs, which are built from segments and arcs, continuity always holds, so the equation is available everywhere. Recognizing why it holds is worth one sentence in a justification, and no more.

Common Misconception Check

Myth 1: “ $g'(x) = f(x)$ means $g$ and $f$ are the same shape.” Truth: $g$ is one level up. Where $f$ is a straight line, $g$ is a parabola.
Myth 2: “You can antidifferentiate to find $g$ .” Truth: You usually cannot, and you never need to. The graph of $f$ plus the value $g(a) = 0$ determines everything asked.
Myth 3: “The chain rule factor is optional.” Truth: It is the whole difference between $g'(2) = f(2)$ and $g'(2) = 4f(4)$ when $g(x) = \int_0^{x^2} f$ .

Topic-Specific Study Method

Differentiate, Then Read the Picture

Borrowed idea: Name the theorem first. The derivative is the integrand, evaluated at the limit. Everything after that is graph reading on the integrand. Signs come first, because they decide where the accumulation turns. Then accumulate the areas in order, starting from the lower limit. Compare the running totals at both ends and wherever the integrand changes sign, since an interior turning point can beat both endpoints. The chain rule only adds a factor, and that factor is the derivative of whatever sits in the upper limit, so a squared or cubed limit never costs more than one extra multiplication.

Exam disguise	What you see on the page	What your brain must translate	Required execution
$g(x) = \int_0^x f$ with a graph of $f$	two functions, one picture	$g' = f$	answer every part from $f$
“Find $g'(3)$ ”	a single number wanted	the height of $f$ at $3$	read it off the graph
Upper limit $x^2$	a composite limit	chain rule factor $2x$	multiply after substituting
“Where is $g$ concave up”	a question about $g$	where $f$ increases	intervals of increase of $f$
“Absolute minimum of $g$ ”	an extreme value	accumulate signed area	start at $g(a) = 0$

① Write $g'(x) = f(x)$ first, before reading anything else. Every later part depends on that line being on the page.

② For sign questions, build a one-row sign chart of $f$ with the zeros marked. It converts a graph into an interval list you can read without looking back.

③ For extreme values, accumulate in order and keep a running total. Students who compute only the endpoint areas and compare them miss the interior turning point that the accumulation creates.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $g(x) = \int_0^x f(t)\,dt$ where $f$ is the function whose graph is a straight line from $(0,2)$ to $(3,-4)$ . Find $g'(1)$ , $g(3)$ , and the interval on which $g$ is decreasing.

Standard Execution:

Derivative. $g'(x) = f(x)$ , so $g'(1) = f(1)$ . The line through $(0,2)$ and $(3,-4)$ has slope $-2$ , so $f(1) = 0$ .

Accumulation. $g(3)$ is the signed area of the triangle from $0$ to $3$ . The line crosses the axis at $x = 1$ , giving $+1$ on $[0,1]$ and $-4$ on $[1,3]$ , so $g(3) = -3$ .

Decreasing. $g' = f < 0$ for $x > 1$ , so $g$ decreases on $(1,3)$ .

Check. $g$ rises to $g(1) = 1$ and then falls to $g(3) = -3$ , which matches the sign chart.

Exam Variation (Context / Parameter Shift): Let $F(x) = \int_2^{x^2} f(t)\,dt$ where $f(4) = 3$ and $f'(4) = -1$ . Find $F'(2)$ .

Solution: By the theorem with the chain rule, $F'(x) = f(x^2)\cdot 2x$ . At $x = 2$ this is $f(4)\cdot 4 = 12$ . The trap is answering $f(4) = 3$ , which drops the factor of $2x$ ; a second trap is answering $f(2)$ , which uses the limit itself instead of its square.

Common Traps & Failure Signals:

Variable mixed up (Signal: the answer contains $t$ ): the integrand is a dummy variable and must be replaced by the limit.
Chain factor dropped (Signal: a composite limit answered with a single height): multiply by $u'(x)$ .
Extreme read from zeros (Signal: a minimum claimed at a zero of $f$ without accumulating): the value of $g$ is an accumulated area.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Theorem Unnamed: Write the sentence.
Level Confusion: You answered questions about $g$ by looking at $g$ instead of at $f$ , which is the natural thing to do and the wrong one, because the graph on the page is always $f$ and its features are one level below the features being asked about.
Chain Omitted: You treated $\int_0^{x^3} f$ as if the limit were plain $x$ , so the answer came out short by the inner derivative, and the error survives every check that does not involve substituting a number.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Seven of the nine released exams since 2017 contain a part that differentiates an accumulation function or reads the behavior of one off the graph of its integrand
10-Year Total Count: 7 of 9 Exams · 9 FRQ Parts; Two of the seven ask two separate questions about the same accumulation function, and both count
FRQ Years Appeared: 2019 (Q3c), 2021 (Q4a, Q4b), 2022 (Q1d), 2023 (Q5c), 2024 (Q4b), 2025 (Q1D, Q4A), 2026 (Q6D) ---

Practice Examples

MCQ 1

The Plain Version

If $g(x) = \int_1^{x} \left(t^2 + 3\right)dt$ , then $g'(2) =$

MCQ 2

The Composite Limit

If $h(x) = \int_0^{x^3} f(t)\,dt$ and $f(8) = 5$ , then $h'(2) =$

MCQ 3

Reading Concavity

The graph of $f$ is decreasing on $(0,4)$ and increasing on $(4,7)$ . For $g(x) = \int_0^x f(t)\,dt$ , the graph of $g$ is concave up on

FRQ 4

An Accumulation Function on a Piecewise Graph

Let $f$ be the continuous function whose graph on $[0,6]$ consists of a line segment from $(0,4)$ to $(2,0)$ and a semicircle of radius $2$ centred at $(4,0)$ lying below the axis. Let $g(x) = \int_0^x f(t)\,dt$ .

(a) Find $g'(1)$ and $g'(4)$ . Give a reason for each answer.

(b) Find all values of $x$ in $(0,6)$ at which $g$ has a critical point, and classify it.

(c) Find the absolute minimum value of $g$ on $[0,6]$ . Justify your answer.

Cheatsheet

Core Reference

	Derivative	Note
$\int_a^{x} f(t)\,dt$	$f(x)$	the base case
$\int_a^{u(x)} f(t)\,dt$	$f(u)\,u'$	chain rule factor
$\int_{v(x)}^{u(x)} f(t)\,dt$	$f(u)\,u' - f(v)\,v'$	lower limit subtracts
$g(x) = \int_a^x f$	$g' = f$ , $g'' = f'$	two levels of graph reading

Sentences That Score

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Start Topic Practice Next Topic: 6.5 Fundamental Theorem of Calculus, Part 2

Unit 6Fundamental Theorem of Calculus, Part 1

Concept 4.71Built for May 2027 AP Exam

Fundamental Theorem of Calculus, Part 1

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A graph of $f$ with a companion function $g(x) = \int_a^x f(t)\,dt$ , then questions about where $g$ increases, where $g$ is concave, where $g$ has a critical point, or what $g'(c)$ equals. Sometimes $g(x) = \int_a^{x^2} f(t)\,dt$ and the question is $g'(2)$ .
What you’ll do: Replace the integral with its derivative in one line, then answer every part from the graph of $f$ . For a composite limit, multiply by the derivative of the upper limit.
Memory hook: Differentiate the integral, get the inside back. The integrand returns, evaluated at the upper limit.
Use it on the exam: For MCQs: When the integrand contains $t$ and the limit contains $x$ , the answer is $f$ at the limit. If the limit is $x^2$ , multiply by $2x$ ; forgetting that factor is the standard wrong choice. For FRQs: State the theorem by name on the line where you use it. “By the Fundamental Theorem of Calculus, $g'(x) = f(x)$ ” is a scored sentence; writing $g' = f$ with no justification is not.

Concept Review

The Area Function and Its Slope

Let $f$ be continuous on an interval containing $a$ , and define

$g(x) = \int_a^x f(t)\,dt$

Object	What it is	How to read it
$g(a)$	the area from $a$ to $a$	always $0$
$g'(x)$	the rate the area grows	the height $f(x)$
$g''(x)$	the rate that rate changes	the slope $f'(x)$
$g(b) - g(a)$	the total accumulation	the value of the integral

The Chain Rule Version

When the upper limit is a function $u(x)$ rather than $x$ , the theorem still applies but picks up a factor:

$\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f\left(u(x)\right)u'(x)$

Reading $g$ From the Graph of $f$

Every property of $g$ is a property of $f$ one level down.

Question about $g$	Look at $f$ for	Answer
where is $g$ increasing?	where $f > 0$	those intervals
where does $g$ have a critical point?	where $f = 0$ or is undefined	those $x$ -values
where is $g$ concave up?	where $f$ is increasing	those intervals
where does $g$ have an inflection point?	where $f$ changes direction	those $x$ -values
what is the extreme value of $g$ ?	signed area, plus $g(a) = 0$	accumulate from $a$

Why Continuity Is in the Statement

Common Misconception Check

Myth 1: “ $g'(x) = f(x)$ means $g$ and $f$ are the same shape.” Truth: $g$ is one level up. Where $f$ is a straight line, $g$ is a parabola.
Myth 2: “You can antidifferentiate to find $g$ .” Truth: You usually cannot, and you never need to. The graph of $f$ plus the value $g(a) = 0$ determines everything asked.
Myth 3: “The chain rule factor is optional.” Truth: It is the whole difference between $g'(2) = f(2)$ and $g'(2) = 4f(4)$ when $g(x) = \int_0^{x^2} f$ .

Topic-Specific Study Method

Differentiate, Then Read the Picture

Exam disguise	What you see on the page	What your brain must translate	Required execution
$g(x) = \int_0^x f$ with a graph of $f$	two functions, one picture	$g' = f$	answer every part from $f$
“Find $g'(3)$ ”	a single number wanted	the height of $f$ at $3$	read it off the graph
Upper limit $x^2$	a composite limit	chain rule factor $2x$	multiply after substituting
“Where is $g$ concave up”	a question about $g$	where $f$ increases	intervals of increase of $f$
“Absolute minimum of $g$ ”	an extreme value	accumulate signed area	start at $g(a) = 0$

① Write $g'(x) = f(x)$ first, before reading anything else. Every later part depends on that line being on the page.

② For sign questions, build a one-row sign chart of $f$ with the zeros marked. It converts a graph into an interval list you can read without looking back.

③ For extreme values, accumulate in order and keep a running total. Students who compute only the endpoint areas and compare them miss the interior turning point that the accumulation creates.

Training & Error Diagnosis

Standard Execution:

Derivative. $g'(x) = f(x)$ , so $g'(1) = f(1)$ . The line through $(0,2)$ and $(3,-4)$ has slope $-2$ , so $f(1) = 0$ .

Accumulation. $g(3)$ is the signed area of the triangle from $0$ to $3$ . The line crosses the axis at $x = 1$ , giving $+1$ on $[0,1]$ and $-4$ on $[1,3]$ , so $g(3) = -3$ .

Decreasing. $g' = f < 0$ for $x > 1$ , so $g$ decreases on $(1,3)$ .

Check. $g$ rises to $g(1) = 1$ and then falls to $g(3) = -3$ , which matches the sign chart.

Exam Variation (Context / Parameter Shift): Let $F(x) = \int_2^{x^2} f(t)\,dt$ where $f(4) = 3$ and $f'(4) = -1$ . Find $F'(2)$ .

Solution: By the theorem with the chain rule, $F'(x) = f(x^2)\cdot 2x$ . At $x = 2$ this is $f(4)\cdot 4 = 12$ . The trap is answering $f(4) = 3$ , which drops the factor of $2x$ ; a second trap is answering $f(2)$ , which uses the limit itself instead of its square.

Common Traps & Failure Signals:

Variable mixed up (Signal: the answer contains $t$ ): the integrand is a dummy variable and must be replaced by the limit.
Chain factor dropped (Signal: a composite limit answered with a single height): multiply by $u'(x)$ .
Extreme read from zeros (Signal: a minimum claimed at a zero of $f$ without accumulating): the value of $g$ is an accumulated area.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Theorem Unnamed: Write the sentence.
Level Confusion: You answered questions about $g$ by looking at $g$ instead of at $f$ , which is the natural thing to do and the wrong one, because the graph on the page is always $f$ and its features are one level below the features being asked about.
Chain Omitted: You treated $\int_0^{x^3} f$ as if the limit were plain $x$ , so the answer came out short by the inner derivative, and the error survives every check that does not involve substituting a number.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Seven of the nine released exams since 2017 contain a part that differentiates an accumulation function or reads the behavior of one off the graph of its integrand
10-Year Total Count: 7 of 9 Exams · 9 FRQ Parts; Two of the seven ask two separate questions about the same accumulation function, and both count
FRQ Years Appeared: 2019 (Q3c), 2021 (Q4a, Q4b), 2022 (Q1d), 2023 (Q5c), 2024 (Q4b), 2025 (Q1D, Q4A), 2026 (Q6D) ---

Practice Examples

MCQ 1

The Plain Version

If $g(x) = \int_1^{x} \left(t^2 + 3\right)dt$ , then $g'(2) =$

MCQ 2

The Composite Limit

If $h(x) = \int_0^{x^3} f(t)\,dt$ and $f(8) = 5$ , then $h'(2) =$

MCQ 3

Reading Concavity

The graph of $f$ is decreasing on $(0,4)$ and increasing on $(4,7)$ . For $g(x) = \int_0^x f(t)\,dt$ , the graph of $g$ is concave up on

FRQ 4

An Accumulation Function on a Piecewise Graph

(a) Find $g'(1)$ and $g'(4)$ . Give a reason for each answer.

(b) Find all values of $x$ in $(0,6)$ at which $g$ has a critical point, and classify it.

(c) Find the absolute minimum value of $g$ on $[0,6]$ . Justify your answer.

Cheatsheet

Core Reference

	Derivative	Note
$\int_a^{x} f(t)\,dt$	$f(x)$	the base case
$\int_a^{u(x)} f(t)\,dt$	$f(u)\,u'$	chain rule factor
$\int_{v(x)}^{u(x)} f(t)\,dt$	$f(u)\,u' - f(v)\,v'$	lower limit subtracts
$g(x) = \int_a^x f$	$g' = f$ , $g'' = f'$	two levels of graph reading

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 6.5 Fundamental Theorem of Calculus, Part 2

Unit 6Fundamental Theorem of Calculus, Part 1

Concept 4.71Built for May 2027 AP Exam

Fundamental Theorem of Calculus, Part 1

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A graph of $f$ with a companion function $g(x) = \int_a^x f(t)\,dt$ , then questions about where $g$ increases, where $g$ is concave, where $g$ has a critical point, or what $g'(c)$ equals. Sometimes $g(x) = \int_a^{x^2} f(t)\,dt$ and the question is $g'(2)$ .
What you’ll do: Replace the integral with its derivative in one line, then answer every part from the graph of $f$ . For a composite limit, multiply by the derivative of the upper limit.
Memory hook: Differentiate the integral, get the inside back. The integrand returns, evaluated at the upper limit.
Use it on the exam: For MCQs: When the integrand contains $t$ and the limit contains $x$ , the answer is $f$ at the limit. If the limit is $x^2$ , multiply by $2x$ ; forgetting that factor is the standard wrong choice. For FRQs: State the theorem by name on the line where you use it. “By the Fundamental Theorem of Calculus, $g'(x) = f(x)$ ” is a scored sentence; writing $g' = f$ with no justification is not.

Concept Review

The Area Function and Its Slope

Let $f$ be continuous on an interval containing $a$ , and define

$g(x) = \int_a^x f(t)\,dt$

Object	What it is	How to read it
$g(a)$	the area from $a$ to $a$	always $0$
$g'(x)$	the rate the area grows	the height $f(x)$
$g''(x)$	the rate that rate changes	the slope $f'(x)$
$g(b) - g(a)$	the total accumulation	the value of the integral

The Chain Rule Version

When the upper limit is a function $u(x)$ rather than $x$ , the theorem still applies but picks up a factor:

$\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f\left(u(x)\right)u'(x)$

Reading $g$ From the Graph of $f$

Every property of $g$ is a property of $f$ one level down.

Question about $g$	Look at $f$ for	Answer
where is $g$ increasing?	where $f > 0$	those intervals
where does $g$ have a critical point?	where $f = 0$ or is undefined	those $x$ -values
where is $g$ concave up?	where $f$ is increasing	those intervals
where does $g$ have an inflection point?	where $f$ changes direction	those $x$ -values
what is the extreme value of $g$ ?	signed area, plus $g(a) = 0$	accumulate from $a$

Why Continuity Is in the Statement

Common Misconception Check

Myth 1: “ $g'(x) = f(x)$ means $g$ and $f$ are the same shape.” Truth: $g$ is one level up. Where $f$ is a straight line, $g$ is a parabola.
Myth 2: “You can antidifferentiate to find $g$ .” Truth: You usually cannot, and you never need to. The graph of $f$ plus the value $g(a) = 0$ determines everything asked.
Myth 3: “The chain rule factor is optional.” Truth: It is the whole difference between $g'(2) = f(2)$ and $g'(2) = 4f(4)$ when $g(x) = \int_0^{x^2} f$ .

Topic-Specific Study Method

Differentiate, Then Read the Picture

Exam disguise	What you see on the page	What your brain must translate	Required execution
$g(x) = \int_0^x f$ with a graph of $f$	two functions, one picture	$g' = f$	answer every part from $f$
“Find $g'(3)$ ”	a single number wanted	the height of $f$ at $3$	read it off the graph
Upper limit $x^2$	a composite limit	chain rule factor $2x$	multiply after substituting
“Where is $g$ concave up”	a question about $g$	where $f$ increases	intervals of increase of $f$
“Absolute minimum of $g$ ”	an extreme value	accumulate signed area	start at $g(a) = 0$

① Write $g'(x) = f(x)$ first, before reading anything else. Every later part depends on that line being on the page.

② For sign questions, build a one-row sign chart of $f$ with the zeros marked. It converts a graph into an interval list you can read without looking back.

③ For extreme values, accumulate in order and keep a running total. Students who compute only the endpoint areas and compare them miss the interior turning point that the accumulation creates.

Training & Error Diagnosis

Standard Execution:

Derivative. $g'(x) = f(x)$ , so $g'(1) = f(1)$ . The line through $(0,2)$ and $(3,-4)$ has slope $-2$ , so $f(1) = 0$ .

Accumulation. $g(3)$ is the signed area of the triangle from $0$ to $3$ . The line crosses the axis at $x = 1$ , giving $+1$ on $[0,1]$ and $-4$ on $[1,3]$ , so $g(3) = -3$ .

Decreasing. $g' = f < 0$ for $x > 1$ , so $g$ decreases on $(1,3)$ .

Check. $g$ rises to $g(1) = 1$ and then falls to $g(3) = -3$ , which matches the sign chart.

Exam Variation (Context / Parameter Shift): Let $F(x) = \int_2^{x^2} f(t)\,dt$ where $f(4) = 3$ and $f'(4) = -1$ . Find $F'(2)$ .

Solution: By the theorem with the chain rule, $F'(x) = f(x^2)\cdot 2x$ . At $x = 2$ this is $f(4)\cdot 4 = 12$ . The trap is answering $f(4) = 3$ , which drops the factor of $2x$ ; a second trap is answering $f(2)$ , which uses the limit itself instead of its square.

Common Traps & Failure Signals:

Variable mixed up (Signal: the answer contains $t$ ): the integrand is a dummy variable and must be replaced by the limit.
Chain factor dropped (Signal: a composite limit answered with a single height): multiply by $u'(x)$ .
Extreme read from zeros (Signal: a minimum claimed at a zero of $f$ without accumulating): the value of $g$ is an accumulated area.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Theorem Unnamed: Write the sentence.
Level Confusion: You answered questions about $g$ by looking at $g$ instead of at $f$ , which is the natural thing to do and the wrong one, because the graph on the page is always $f$ and its features are one level below the features being asked about.
Chain Omitted: You treated $\int_0^{x^3} f$ as if the limit were plain $x$ , so the answer came out short by the inner derivative, and the error survives every check that does not involve substituting a number.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Seven of the nine released exams since 2017 contain a part that differentiates an accumulation function or reads the behavior of one off the graph of its integrand
10-Year Total Count: 7 of 9 Exams · 9 FRQ Parts; Two of the seven ask two separate questions about the same accumulation function, and both count
FRQ Years Appeared: 2019 (Q3c), 2021 (Q4a, Q4b), 2022 (Q1d), 2023 (Q5c), 2024 (Q4b), 2025 (Q1D, Q4A), 2026 (Q6D) ---

Practice Examples

MCQ 1

The Plain Version

If $g(x) = \int_1^{x} \left(t^2 + 3\right)dt$ , then $g'(2) =$

MCQ 2

The Composite Limit

If $h(x) = \int_0^{x^3} f(t)\,dt$ and $f(8) = 5$ , then $h'(2) =$

MCQ 3

Reading Concavity

The graph of $f$ is decreasing on $(0,4)$ and increasing on $(4,7)$ . For $g(x) = \int_0^x f(t)\,dt$ , the graph of $g$ is concave up on

FRQ 4

An Accumulation Function on a Piecewise Graph

(a) Find $g'(1)$ and $g'(4)$ . Give a reason for each answer.

(b) Find all values of $x$ in $(0,6)$ at which $g$ has a critical point, and classify it.

(c) Find the absolute minimum value of $g$ on $[0,6]$ . Justify your answer.

Cheatsheet

Core Reference

	Derivative	Note
$\int_a^{x} f(t)\,dt$	$f(x)$	the base case
$\int_a^{u(x)} f(t)\,dt$	$f(u)\,u'$	chain rule factor
$\int_{v(x)}^{u(x)} f(t)\,dt$	$f(u)\,u' - f(v)\,v'$	lower limit subtracts
$g(x) = \int_a^x f$	$g' = f$ , $g'' = f'$	two levels of graph reading

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