Unit 6Fundamental Theorem of Calculus, Part 2

Concept 4.150Built for May 2027 AP Exam

Fundamental Theorem of Calculus, Part 2

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The second part of the Fundamental Theorem turns a definite integral into two subtractions. Find any antiderivative $F$ of $f$ , evaluate it at the upper limit, evaluate it at the lower limit, and subtract. The constant of integration cancels, which is why any antiderivative works. The theorem also runs backwards with unusual force on the exam: when the integrand is already a derivative, the integral collapses to the difference of the original function at the two endpoints, so a table of values answers the question without a single antiderivative being computed. The one structural warning is that the antiderivative must be valid across the whole interval. A vertical asymptote inside the interval voids the computation.

On the AP Exam

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

What you’ll see: “Evaluate $\int_0^{2}\left(2f'(x) + 4\right)dx$ ” with a table of $f$ values, or a pure computation such as $\int_1^{4}\left(3x^2 - 2x\right)dx$ , or a total-quantity question with a polynomial rate and no calculator allowed.
What you’ll do: Look at the integrand first. If it is already a derivative, apply the theorem to the original function and read values from the table. If it is not, find an antiderivative, then subtract the two evaluations.
Memory hook: Top minus bottom, after antidifferentiating. Two numbers and a subtraction, and the constant never appears.
Use it on the exam: For MCQs: When a table is supplied, the integral of $f'$ is $f(b) - f(a)$ and the table is the whole solution. Do not attempt to reconstruct $f$ . For FRQs: Write the evaluation bar with both limits before substituting. The bar is the visible evidence that the theorem was used, and it earns the method point even when the arithmetic slips.

Concept Review

Two Endpoints and One Subtraction

Figure 1 — The shaded area equals the change in the antiderivative between the endpoints. The left vertical side is $F(a)$ , the right is $F(b)$ , and the area between them is their difference.

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ there, then

$\int_a^b f(x)\,dx = F(b) - F(a)$

Three consequences are worth holding at once. The choice of $F$ does not matter, because adding a constant $C$ adds it to both terms and it cancels. The result is signed, so an interval that dips below the axis produces a smaller value than the geometric area. And the theorem requires continuity on the closed interval, which is the real constraint in practice.

Situation	What to do	Why
integrand is a polynomial, sine, cosine, exponential	antidifferentiate, then subtract	the theorem applies directly
integrand is already $f'$	evaluate $f$ at the endpoints	the theorem collapses one level
integrand has a zero inside the interval	integrate piecewise if area was requested	the signed total is not the area
integrand has a vertical asymptote inside	stop, the theorem does not apply	the antiderivative is not continuous there

The shaded area equals the change in the antiderivative between the endpoints. The left vertical side is $F(a)$, the right is $F(b)$, and the area between them is their difference.

When the Integrand Is Already a Derivative

This is the highest-yield pattern in the topic, and it needs no antidifferentiation at all.

$\int_a^b f'(x)\,dx = f(b) - f(a)$

The exam pairs this with a table. A stem that supplies $f$ at selected $x$ -values and asks for $\int_0^{2} f'(x)\,dx$ is asking for one subtraction, and a student who hunts for a formula for $f$ has misread the item. The same reading applies to combinations, since the theorem is linear:

$\int_a^b \left(2f'(x) + 3\right)dx = 2\left[f(b) - f(a)\right] + 3(b-a)$

The Evaluation Bar

The conventional notation records both limits on a single vertical bar:

$\int_1^{3} 2x\,dx = \left[x^2\right]_1^{3} = 9 - 1 = 8$

Write the bar with the numbers still attached, then substitute. Collapsing the two steps into one line is where sign errors appear, especially when the lower limit is negative, because the subtraction of a negative turns into an addition and the arithmetic changes shape.

Total Accumulation and Net Change

Applied to a rate, the theorem says the integral of a rate is the net change in the quantity:

$\int_a^b Q'(t)\,dt = Q(b) - Q(a)$

Read in this form the theorem answers questions about totals without any area reasoning: the amount of water added is the change in volume, the distance travelled is the change in position when the velocity keeps one sign, and the words “net change” and “total change” are the same phrase in a different order. When the rate changes sign, the integral still gives net change, and the total distance requires a separate computation on absolute values.

Where the Theorem Refuses

The hypothesis is continuity on $[a,b]$ . If $f$ has a vertical asymptote inside the interval, no antiderivative is continuous across the whole range, and applying the subtraction anyway produces a finite number for a quantity that does not exist. The exam rarely sets this trap in an FRQ, but multiple-choice items use it, usually by offering a confident numeric answer next to “the integral diverges”.

Common Misconception Check

Myth 1: “The $+C$ has to be carried through the subtraction.” Truth: It cancels. Writing $+C$ twice and then dropping it looks untidy but costs nothing; forgetting to subtract the lower limit costs the answer.
Myth 2: “ $\int_a^b f' = f'(b) - f'(a)$ .” Truth: The theorem moves one level up, so the answer is $f(b) - f(a)$ . This is the most frequent misread of the whole unit.
Myth 3: “The integral equals the area, so it is always positive.” Truth: Signed area. A region below the axis subtracts.

Topic-Specific Study Method

Read the Integrand Before You Antidifferentiate

Borrowed idea: The integrand tells you which route to take. If it already carries a prime, the theorem finishes in one subtraction. If it does not, antiderivative first and subtraction second. That choice takes five seconds and it decides whether the part costs one line or eight. Most students reach for the antiderivative out of habit, even when the table on the page has already answered the question. Look at the integrand, not at the limits, because the limits look identical in both routes.

Exam disguise	What you see on the page	What your brain must translate	Required execution
Table plus $\int f'$	values of $f$ at four $x$	a single subtraction	$f(\text{top}) - f(\text{bottom})$
Polynomial rate, no calculator	“how many words by the end”	antidifferentiate then subtract	bar notation with both limits
$\int \left(2f' + 4\right)$	a sum with a prime inside	linearity plus the collapse	$2\left[f(b)-f(a)\right] + 4(b-a)$
Rate that changes sign	“total distance”	abs value, split at zeros	integrate each piece positively
Asymptote inside the limits	a denominator with a zero	theorem does not apply	say so, do not compute

① Look at the integrand for a prime before anything else. A visible $f'$ means the table is the answer.

② Write the bar with both numbers on the page. When the lower limit is negative, write the parentheses explicitly, because $9 - (-1)$ and $9 - 1$ differ by more than a sign.

③ For a rate that changes sign, find the zeros first and split. The integral over the whole interval is net change, and total distance is a different request with a different setup.

Training & Error Diagnosis

Core Archetype (Simplest Form): Evaluate $\int_0^{3}\left(4x - x^2\right)dx$ .

Standard Execution:

Antiderivative. $F(x) = 2x^2 - \frac{x^3}{3}$ .

Bar. $\left[2x^2 - \frac{x^3}{3}\right]_0^{3}$ .

Top. $F(3) = 18 - 9 = 9$ .

Bottom. $F(0) = 0$ .

Answer. $9 - 0 = 9$ .

Exam Variation (Context / Parameter Shift): The twice-differentiable function $f$ has $f(0) = 10$ , $f(2) = 7$ , and $f'(2) = -8$ . Let $m(x) = 5x^3 + \int_0^{x} f'(t)\,dt$ . Find $m(2)$ , and determine whether $m$ is increasing or decreasing at $x = 2$ .

Solution: The integral of $f'$ collapses: $\int_0^{2} f'(t)\,dt = f(2) - f(0) = 7 - 10 = -3$ . So $m(2) = 5(8) + (-3) = 40 - 3 = 37$ . For the direction, differentiate: $m'(x) = 15x^2 + f'(x)$ , so $m'(2) = 60 + (-8) = 52 > 0$ and $m$ is increasing. The item uses both halves of the theorem in one stem, which is common: part one is the evaluation, part two is the derivative of the accumulation.

Common Traps & Failure Signals:

Level slip (Signal: an answer of $f'(b) - f'(a)$ ): the theorem moves one level up.
Lower limit dropped (Signal: a value equal to $F(b)$ alone): the subtraction is the theorem.
Signed and total confused (Signal: a total distance smaller than a displacement): split at the zeros.

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

Route Chosen Late: Read the integrand first.
Level Slip: You answered with $f'$ at the endpoints because the prime was the last symbol you read before the integral sign, and the theorem's own direction is easy to invert when the integrand already carries a derivative.
Limit Substitution Reversed: You put the lower limit into the antiderivative first and then subtracted the upper one, which produces the negative of the correct value, and that error is invisible on any interval where the answer happens to be small.

Past 10-Year AP Exam Appearance

Testing Frequency: Nearly Every Year; Seven of the nine released exams since 2017 contain a part whose answer is a single application of the evaluation theorem, either with an antiderivative or with a table of values for the original function
10-Year Total Count: 7 of 9 Exams · 7 FRQ Parts; Only 2021 and 2023 have no such part; in both years the integral work sat inside a part scored on some other skill
FRQ Years Appeared: 2017 (Q3a), 2018 (Q3a), 2019 (Q3b), 2022 (Q3a), 2024 (Q4c), 2025 (Q3D), 2026 (Q6C) ---

Practice Examples

MCQ 1

A Plain Evaluation

$\int_0^{2}\left(3x^2 - 2x\right)dx =$

MCQ 2

The Table Version

The function $f$ is differentiable with $f(1) = 5$ and $f(4) = 2$ . What is $\int_1^{4} f'(x)\,dx$ ?

MCQ 3

A Combination

If $\int_0^{3} f'(x)\,dx = 8$ , then $\int_0^{3}\left(2f'(x) + 5\right)dx =$

FRQ 4

Table, Collapse, and a Direction

The functions $f$ and $g$ are twice differentiable. Selected values are $f(0) = 12$ , $f(3) = 4$ , $f'(0) = -5$ , $f'(3) = -2$ , $g(0) = 1$ , and $g(3) = 6$ .

(a) Let $h(x) = \int_0^{x} f'(t)\,dt$ . Find $h(3)$ .

(b) Let $k(x) = 2x^2 + \int_0^{x}\left(3f'(t) - g'(t)\right)dt$ . Find $k(3)$ .

(c) Find the average rate of change of $f$ on $[0,3]$ , and explain why the Mean Value Theorem guarantees a point $c$ in $(0,3)$ with $f'(c)$ equal to that rate.

Cheatsheet

Core Reference

	Statement	Use
evaluation	$\int_a^b f = F(b) - F(a)$	the base case
collapse	$\int_a^b f' = f(b) - f(a)$	table problems
linearity	$\int_a^b (2f' + k) = 2\left[f(b)-f(a)\right] + k(b-a)$	combined integrands
net change	$\int_a^b Q' = Q(b) - Q(a)$	rate to total
refusal	asymptote inside $[a,b]$	the theorem does not apply

Sentences That Score

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Unit 6Fundamental Theorem of Calculus, Part 2

Concept 4.150Built for May 2027 AP Exam

Fundamental Theorem of Calculus, Part 2

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: “Evaluate $\int_0^{2}\left(2f'(x) + 4\right)dx$ ” with a table of $f$ values, or a pure computation such as $\int_1^{4}\left(3x^2 - 2x\right)dx$ , or a total-quantity question with a polynomial rate and no calculator allowed.
What you’ll do: Look at the integrand first. If it is already a derivative, apply the theorem to the original function and read values from the table. If it is not, find an antiderivative, then subtract the two evaluations.
Memory hook: Top minus bottom, after antidifferentiating. Two numbers and a subtraction, and the constant never appears.
Use it on the exam: For MCQs: When a table is supplied, the integral of $f'$ is $f(b) - f(a)$ and the table is the whole solution. Do not attempt to reconstruct $f$ . For FRQs: Write the evaluation bar with both limits before substituting. The bar is the visible evidence that the theorem was used, and it earns the method point even when the arithmetic slips.

Concept Review

Two Endpoints and One Subtraction

Figure 1 — The shaded area equals the change in the antiderivative between the endpoints. The left vertical side is $F(a)$ , the right is $F(b)$ , and the area between them is their difference.

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ there, then

$\int_a^b f(x)\,dx = F(b) - F(a)$

Situation	What to do	Why
integrand is a polynomial, sine, cosine, exponential	antidifferentiate, then subtract	the theorem applies directly
integrand is already $f'$	evaluate $f$ at the endpoints	the theorem collapses one level
integrand has a zero inside the interval	integrate piecewise if area was requested	the signed total is not the area
integrand has a vertical asymptote inside	stop, the theorem does not apply	the antiderivative is not continuous there

The shaded area equals the change in the antiderivative between the endpoints. The left vertical side is $F(a)$, the right is $F(b)$, and the area between them is their difference.

When the Integrand Is Already a Derivative

This is the highest-yield pattern in the topic, and it needs no antidifferentiation at all.

$\int_a^b f'(x)\,dx = f(b) - f(a)$

$\int_a^b \left(2f'(x) + 3\right)dx = 2\left[f(b) - f(a)\right] + 3(b-a)$

The Evaluation Bar

The conventional notation records both limits on a single vertical bar:

$\int_1^{3} 2x\,dx = \left[x^2\right]_1^{3} = 9 - 1 = 8$

Total Accumulation and Net Change

Applied to a rate, the theorem says the integral of a rate is the net change in the quantity:

$\int_a^b Q'(t)\,dt = Q(b) - Q(a)$

Where the Theorem Refuses

Common Misconception Check

Myth 1: “The $+C$ has to be carried through the subtraction.” Truth: It cancels. Writing $+C$ twice and then dropping it looks untidy but costs nothing; forgetting to subtract the lower limit costs the answer.
Myth 2: “ $\int_a^b f' = f'(b) - f'(a)$ .” Truth: The theorem moves one level up, so the answer is $f(b) - f(a)$ . This is the most frequent misread of the whole unit.
Myth 3: “The integral equals the area, so it is always positive.” Truth: Signed area. A region below the axis subtracts.

Topic-Specific Study Method

Read the Integrand Before You Antidifferentiate

Exam disguise	What you see on the page	What your brain must translate	Required execution
Table plus $\int f'$	values of $f$ at four $x$	a single subtraction	$f(\text{top}) - f(\text{bottom})$
Polynomial rate, no calculator	“how many words by the end”	antidifferentiate then subtract	bar notation with both limits
$\int \left(2f' + 4\right)$	a sum with a prime inside	linearity plus the collapse	$2\left[f(b)-f(a)\right] + 4(b-a)$
Rate that changes sign	“total distance”	abs value, split at zeros	integrate each piece positively
Asymptote inside the limits	a denominator with a zero	theorem does not apply	say so, do not compute

① Look at the integrand for a prime before anything else. A visible $f'$ means the table is the answer.

② Write the bar with both numbers on the page. When the lower limit is negative, write the parentheses explicitly, because $9 - (-1)$ and $9 - 1$ differ by more than a sign.

③ For a rate that changes sign, find the zeros first and split. The integral over the whole interval is net change, and total distance is a different request with a different setup.

Training & Error Diagnosis

Core Archetype (Simplest Form): Evaluate $\int_0^{3}\left(4x - x^2\right)dx$ .

Standard Execution:

Antiderivative. $F(x) = 2x^2 - \frac{x^3}{3}$ .

Bar. $\left[2x^2 - \frac{x^3}{3}\right]_0^{3}$ .

Top. $F(3) = 18 - 9 = 9$ .

Bottom. $F(0) = 0$ .

Answer. $9 - 0 = 9$ .

Solution: The integral of $f'$ collapses: $\int_0^{2} f'(t)\,dt = f(2) - f(0) = 7 - 10 = -3$ . So $m(2) = 5(8) + (-3) = 40 - 3 = 37$ . For the direction, differentiate: $m'(x) = 15x^2 + f'(x)$ , so $m'(2) = 60 + (-8) = 52 > 0$ and $m$ is increasing. The item uses both halves of the theorem in one stem, which is common: part one is the evaluation, part two is the derivative of the accumulation.

Common Traps & Failure Signals:

Level slip (Signal: an answer of $f'(b) - f'(a)$ ): the theorem moves one level up.
Lower limit dropped (Signal: a value equal to $F(b)$ alone): the subtraction is the theorem.
Signed and total confused (Signal: a total distance smaller than a displacement): split at the zeros.

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

Route Chosen Late: Read the integrand first.
Level Slip: You answered with $f'$ at the endpoints because the prime was the last symbol you read before the integral sign, and the theorem's own direction is easy to invert when the integrand already carries a derivative.
Limit Substitution Reversed: You put the lower limit into the antiderivative first and then subtracted the upper one, which produces the negative of the correct value, and that error is invisible on any interval where the answer happens to be small.

Past 10-Year AP Exam Appearance

Testing Frequency: Nearly Every Year; Seven of the nine released exams since 2017 contain a part whose answer is a single application of the evaluation theorem, either with an antiderivative or with a table of values for the original function
10-Year Total Count: 7 of 9 Exams · 7 FRQ Parts; Only 2021 and 2023 have no such part; in both years the integral work sat inside a part scored on some other skill
FRQ Years Appeared: 2017 (Q3a), 2018 (Q3a), 2019 (Q3b), 2022 (Q3a), 2024 (Q4c), 2025 (Q3D), 2026 (Q6C) ---

Practice Examples

MCQ 1

A Plain Evaluation

$\int_0^{2}\left(3x^2 - 2x\right)dx =$

MCQ 2

The Table Version

The function $f$ is differentiable with $f(1) = 5$ and $f(4) = 2$ . What is $\int_1^{4} f'(x)\,dx$ ?

MCQ 3

A Combination

If $\int_0^{3} f'(x)\,dx = 8$ , then $\int_0^{3}\left(2f'(x) + 5\right)dx =$

FRQ 4

Table, Collapse, and a Direction

The functions $f$ and $g$ are twice differentiable. Selected values are $f(0) = 12$ , $f(3) = 4$ , $f'(0) = -5$ , $f'(3) = -2$ , $g(0) = 1$ , and $g(3) = 6$ .

(a) Let $h(x) = \int_0^{x} f'(t)\,dt$ . Find $h(3)$ .

(b) Let $k(x) = 2x^2 + \int_0^{x}\left(3f'(t) - g'(t)\right)dt$ . Find $k(3)$ .

(c) Find the average rate of change of $f$ on $[0,3]$ , and explain why the Mean Value Theorem guarantees a point $c$ in $(0,3)$ with $f'(c)$ equal to that rate.

Cheatsheet

Core Reference

	Statement	Use
evaluation	$\int_a^b f = F(b) - F(a)$	the base case
collapse	$\int_a^b f' = f(b) - f(a)$	table problems
linearity	$\int_a^b (2f' + k) = 2\left[f(b)-f(a)\right] + k(b-a)$	combined integrands
net change	$\int_a^b Q' = Q(b) - Q(a)$	rate to total
refusal	asymptote inside $[a,b]$	the theorem does not apply

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

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Start Topic Practice Next Topic: 6.6 Antiderivatives for basic functions

Unit 6Fundamental Theorem of Calculus, Part 2

Concept 4.150Built for May 2027 AP Exam

Fundamental Theorem of Calculus, Part 2

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: “Evaluate $\int_0^{2}\left(2f'(x) + 4\right)dx$ ” with a table of $f$ values, or a pure computation such as $\int_1^{4}\left(3x^2 - 2x\right)dx$ , or a total-quantity question with a polynomial rate and no calculator allowed.
What you’ll do: Look at the integrand first. If it is already a derivative, apply the theorem to the original function and read values from the table. If it is not, find an antiderivative, then subtract the two evaluations.
Memory hook: Top minus bottom, after antidifferentiating. Two numbers and a subtraction, and the constant never appears.
Use it on the exam: For MCQs: When a table is supplied, the integral of $f'$ is $f(b) - f(a)$ and the table is the whole solution. Do not attempt to reconstruct $f$ . For FRQs: Write the evaluation bar with both limits before substituting. The bar is the visible evidence that the theorem was used, and it earns the method point even when the arithmetic slips.

Concept Review

Two Endpoints and One Subtraction

Figure 1 — The shaded area equals the change in the antiderivative between the endpoints. The left vertical side is $F(a)$ , the right is $F(b)$ , and the area between them is their difference.

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ there, then

$\int_a^b f(x)\,dx = F(b) - F(a)$

Situation	What to do	Why
integrand is a polynomial, sine, cosine, exponential	antidifferentiate, then subtract	the theorem applies directly
integrand is already $f'$	evaluate $f$ at the endpoints	the theorem collapses one level
integrand has a zero inside the interval	integrate piecewise if area was requested	the signed total is not the area
integrand has a vertical asymptote inside	stop, the theorem does not apply	the antiderivative is not continuous there

The shaded area equals the change in the antiderivative between the endpoints. The left vertical side is $F(a)$, the right is $F(b)$, and the area between them is their difference.

When the Integrand Is Already a Derivative

This is the highest-yield pattern in the topic, and it needs no antidifferentiation at all.

$\int_a^b f'(x)\,dx = f(b) - f(a)$

$\int_a^b \left(2f'(x) + 3\right)dx = 2\left[f(b) - f(a)\right] + 3(b-a)$

The Evaluation Bar

The conventional notation records both limits on a single vertical bar:

$\int_1^{3} 2x\,dx = \left[x^2\right]_1^{3} = 9 - 1 = 8$

Total Accumulation and Net Change

Applied to a rate, the theorem says the integral of a rate is the net change in the quantity:

$\int_a^b Q'(t)\,dt = Q(b) - Q(a)$

Where the Theorem Refuses

Common Misconception Check

Myth 1: “The $+C$ has to be carried through the subtraction.” Truth: It cancels. Writing $+C$ twice and then dropping it looks untidy but costs nothing; forgetting to subtract the lower limit costs the answer.
Myth 2: “ $\int_a^b f' = f'(b) - f'(a)$ .” Truth: The theorem moves one level up, so the answer is $f(b) - f(a)$ . This is the most frequent misread of the whole unit.
Myth 3: “The integral equals the area, so it is always positive.” Truth: Signed area. A region below the axis subtracts.

Topic-Specific Study Method

Read the Integrand Before You Antidifferentiate

Exam disguise	What you see on the page	What your brain must translate	Required execution
Table plus $\int f'$	values of $f$ at four $x$	a single subtraction	$f(\text{top}) - f(\text{bottom})$
Polynomial rate, no calculator	“how many words by the end”	antidifferentiate then subtract	bar notation with both limits
$\int \left(2f' + 4\right)$	a sum with a prime inside	linearity plus the collapse	$2\left[f(b)-f(a)\right] + 4(b-a)$
Rate that changes sign	“total distance”	abs value, split at zeros	integrate each piece positively
Asymptote inside the limits	a denominator with a zero	theorem does not apply	say so, do not compute

① Look at the integrand for a prime before anything else. A visible $f'$ means the table is the answer.

② Write the bar with both numbers on the page. When the lower limit is negative, write the parentheses explicitly, because $9 - (-1)$ and $9 - 1$ differ by more than a sign.

③ For a rate that changes sign, find the zeros first and split. The integral over the whole interval is net change, and total distance is a different request with a different setup.

Training & Error Diagnosis

Core Archetype (Simplest Form): Evaluate $\int_0^{3}\left(4x - x^2\right)dx$ .

Standard Execution:

Antiderivative. $F(x) = 2x^2 - \frac{x^3}{3}$ .

Bar. $\left[2x^2 - \frac{x^3}{3}\right]_0^{3}$ .

Top. $F(3) = 18 - 9 = 9$ .

Bottom. $F(0) = 0$ .

Answer. $9 - 0 = 9$ .

Solution: The integral of $f'$ collapses: $\int_0^{2} f'(t)\,dt = f(2) - f(0) = 7 - 10 = -3$ . So $m(2) = 5(8) + (-3) = 40 - 3 = 37$ . For the direction, differentiate: $m'(x) = 15x^2 + f'(x)$ , so $m'(2) = 60 + (-8) = 52 > 0$ and $m$ is increasing. The item uses both halves of the theorem in one stem, which is common: part one is the evaluation, part two is the derivative of the accumulation.

Common Traps & Failure Signals:

Level slip (Signal: an answer of $f'(b) - f'(a)$ ): the theorem moves one level up.
Lower limit dropped (Signal: a value equal to $F(b)$ alone): the subtraction is the theorem.
Signed and total confused (Signal: a total distance smaller than a displacement): split at the zeros.

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

Route Chosen Late: Read the integrand first.
Level Slip: You answered with $f'$ at the endpoints because the prime was the last symbol you read before the integral sign, and the theorem's own direction is easy to invert when the integrand already carries a derivative.
Limit Substitution Reversed: You put the lower limit into the antiderivative first and then subtracted the upper one, which produces the negative of the correct value, and that error is invisible on any interval where the answer happens to be small.

Past 10-Year AP Exam Appearance

Testing Frequency: Nearly Every Year; Seven of the nine released exams since 2017 contain a part whose answer is a single application of the evaluation theorem, either with an antiderivative or with a table of values for the original function
10-Year Total Count: 7 of 9 Exams · 7 FRQ Parts; Only 2021 and 2023 have no such part; in both years the integral work sat inside a part scored on some other skill
FRQ Years Appeared: 2017 (Q3a), 2018 (Q3a), 2019 (Q3b), 2022 (Q3a), 2024 (Q4c), 2025 (Q3D), 2026 (Q6C) ---

Practice Examples

MCQ 1

A Plain Evaluation

$\int_0^{2}\left(3x^2 - 2x\right)dx =$

MCQ 2

The Table Version

The function $f$ is differentiable with $f(1) = 5$ and $f(4) = 2$ . What is $\int_1^{4} f'(x)\,dx$ ?

MCQ 3

A Combination

If $\int_0^{3} f'(x)\,dx = 8$ , then $\int_0^{3}\left(2f'(x) + 5\right)dx =$

FRQ 4

Table, Collapse, and a Direction

The functions $f$ and $g$ are twice differentiable. Selected values are $f(0) = 12$ , $f(3) = 4$ , $f'(0) = -5$ , $f'(3) = -2$ , $g(0) = 1$ , and $g(3) = 6$ .

(a) Let $h(x) = \int_0^{x} f'(t)\,dt$ . Find $h(3)$ .

(b) Let $k(x) = 2x^2 + \int_0^{x}\left(3f'(t) - g'(t)\right)dt$ . Find $k(3)$ .

(c) Find the average rate of change of $f$ on $[0,3]$ , and explain why the Mean Value Theorem guarantees a point $c$ in $(0,3)$ with $f'(c)$ equal to that rate.

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Core Reference

	Statement	Use
evaluation	$\int_a^b f = F(b) - F(a)$	the base case
collapse	$\int_a^b f' = f(b) - f(a)$	table problems
linearity	$\int_a^b (2f' + k) = 2\left[f(b)-f(a)\right] + k(b-a)$	combined integrands
net change	$\int_a^b Q' = Q(b) - Q(a)$	rate to total
refusal	asymptote inside $[a,b]$	the theorem does not apply

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