Unit 6 · Topic 6.7 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 6.7: The Fundamental Theorem of Calculus and Definite Integrals

Q: Which antiderivative should I use?

Any one. A constant added to F appears in both F(b) and F(a) and cancels in the subtraction, so the simplest antiderivative, with no constant, is the usual choice.

Q: How is this different from what I did in 6.4?

In 6.4 you differentiated a function defined by an integral. Here the same fact is read the other way: that function is an antiderivative, and it leads to a way to evaluate any definite integral of a continuous function from an antiderivative.

Q: Can the answer be negative?

Yes. A definite integral is a signed accumulation, so if f is negative on most of [a, b] the result can be negative. If f is positive on the whole interval, though, a negative answer means something went wrong.

Q: What if I cannot find an antiderivative?

Some continuous functions, such as (x2), have an antiderivative that no formula you know writes down. The integral still exists. Where a question allows a calculator, write the integral with its integrand, both limits and dx, then give the calculator's value.

Q: Do I really have to check continuity every time?

On paper, a short phrase is enough: "the integrand is continuous on [1, 2]." It matters most when the integrand has a denominator or a logarithm, because those are where a break can sit inside the interval.

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

Evaluate a definite integral exactly by finding an antiderivative and applying the Fundamental Theorem of Calculus.

Key vocabulary

Term	What it means	The exam trap
Antiderivative	A function $F$ whose derivative is $f$ , so that $F^{'} (x) = f (x)$ .	Differentiate your answer before you use it; that one step catches most sign and coefficient slips.
Definite integral	$\int_{a}^{b} f (x) d x$ , the limit of Riemann sums of $f$ over $[a, b]$ , which gives the accumulated change over that interval.	Read which number is the upper limit and which is the lower before you compute anything.
Fundamental Theorem of Calculus	If $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$ , then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .	It comes with a hypothesis about the integrand. State it in your work before you use the theorem.
Accumulation function	A function defined by a definite integral with a variable limit, such as $F (x) = \int_{a}^{x} f (t) d t$ .	The $t$ inside the integral is only a placeholder; the function's input is the limit $x$ , so write the answer in terms of $x$ .

Two graphs side by side. On the left, f(x) = x squared with the region under it from x = 1 to x = 3 shaded; its area is 8.667. On the right, F(x) = x cubed over 3 climbs from 0.333 at x = 1 to 9 at x = 3, a rise of 8.667, the same number as the area.

Part 1 Think first How high does the drone climb?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A drone lifts off and climbs with upward velocity $v (t) = 2 t$ meters per second, $t$ seconds after takeoff. How many meters does it rise between $t = 1$ and $t = 4$ ?

In short. The Fundamental Theorem of Calculus says that if $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ , then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .

An antiderivative of $f$ is a function whose derivative is $f$ , and you can check one by differentiating it.
When $f$ is continuous, the accumulation function $\int_{a}^{x} f (t) d t$ is itself an antiderivative of $f$ , the one whose value at $a$ is 0.
Check that $f$ is continuous on the whole interval before you use the theorem, and subtract in the order $F (b) - F (a)$ .

Part 2 Learn it Antiderivatives and the theorem that uses them

2a. An antiderivative runs a derivative backward

An antiderivative of a function $f$ is a function $F$ whose derivative is $f$ , so $F^{'} (x) = f (x)$ . Finding one means reading a derivative rule from right to left and asking what was differentiated to give $f$ .

Every rule from Unit 2 gives you one. Since $\frac{d}{d x} (x^{5}) = 5 x^{4}$ , the function $x^{5}$ is an antiderivative of $5 x^{4}$ . Since $\frac{d}{d x} (5 \sin x) = 5 \cos x$ , the function $5 \sin x$ is an antiderivative of $5 \cos x$ .

For a power $x^{n}$ with $n \neq - 1$ , raise the exponent by 1 and divide by the new exponent. An antiderivative of $x^{2}$ is $\frac{x^{3}}{3}$ , and differentiating $\frac{x^{3}}{3}$ gives back $x^{2}$ . The one exception is $x^{- 1} = \frac{1}{x}$ , where the new exponent would be 0; for $x > 0$ , read an antiderivative off the derivative rule $\frac{d}{d x} (\ln x) = \frac{1}{x}$ instead.

Always check by differentiating. A guess is an antiderivative exactly when its derivative is the function you started with. The check catches the usual slip with sine and cosine, where one of the two derivatives carries a minus sign: $\frac{d}{d x} (\cos x) = - \sin x$ , while $\frac{d}{d x} (\sin x) = \cos x$ .

An antiderivative is never the only one. $x^{5} + 7$ and $x^{5} - 2$ also have derivative $5 x^{4}$ , because the derivative of a constant is 0. On an interval, any two antiderivatives of the same function differ by a constant.

Quick check · Match

No calculator. Tap a function, then an antiderivative of it.

2b. The accumulation function is an antiderivative

In topic 6.4 you met functions built from an integral, such as $F (x) = \int_{a}^{x} f (t) d t$ , and learned that $F^{'} (x) = f (x)$ when $f$ is continuous. Read that sentence again with this lesson's word in it: if $f$ is continuous on an interval containing $a$ , then $F (x) = \int_{a}^{x} f (t) d t$ is an antiderivative of $f$ on that interval.

So every continuous function has an antiderivative, even one no rule gives you. And this antiderivative has one feature the others lack:

F (a) = \int_{a}^{a} f (t) d t = 0.

That fixes which antiderivative it is. Suppose you want $F (x) = \int_{1}^{x} 2 t d t$ . Since $F^{'} (x) = 2 x$ , $F$ is $x^{2}$ plus some constant $C$ . Since $F (1) = 0$ , $1 + C = 0$ and $C = - 1$ . So

\int_{1}^{x} 2 t d t = x^{2} - 1.

Check it: the derivative is $2 x$ , and the value at $x = 1$ is 0.

Look at what that says about any other antiderivative $G$ of $f$ . $G$ and $F$ differ by a constant, $G (x) = F (x) + C$ , and at $x = a$ the constant is $G (a)$ . So $F (x) = G (x) - G (a)$ : the accumulation function is any antiderivative minus its own value at the lower limit.

Quick check · Multiple choice

No calculator. Let $F (x) = \int_{2}^{x} 3 t^{2} d t$ . Which expression is $F (x)$ ?

2c. The Fundamental Theorem: F(b) minus F(a)

Put $x = b$ into the last line of the chunk before. The accumulation function at $b$ is the definite integral, so for any antiderivative $F$ :

Fundamental Theorem of Calculus. If $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$ , then $\int_{a}^{b} f (x) d x = F (b) - F (a) .$

That turns an area or an accumulation into two function values and a subtraction. For $\int_{0}^{2} x^{3} d x$ : an antiderivative of $x^{3}$ is $F (x) = \frac{x^{4}}{4}$ , so the integral is $F (2) - F (0) = 4 - 0 = 4$ .

Three things to hold on to.

Any antiderivative works. With $\frac{x^{4}}{4} + 10$ you get $14 - 10 = 4$ . The constant appears in both values and cancels, so use the simplest one.
Upper limit first. It is $F (b) - F (a)$ , the value at the top of the integral sign minus the value at the bottom. Reversed, you get the opposite sign.
Check the hypothesis. The theorem needs $f$ continuous on all of $[a, b]$ . A function such as $\frac{1}{x}$ breaks at $x = 0$ , so the theorem says nothing about an integral of it over an interval that contains 0.

Quick check · Calculate

No calculator. Evaluate $\int_{1}^{3} (3 x^{2} + 2 x) d x$ .

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Subtracting values of the integrand, $f (b) - f (a)$ , instead of values of an antiderivative.	$f (b) - f (a)$ is the change in $f$ . The integral is the change in a function whose rate is $f$ , which is $F$ .	Write $F$ first and check that $F^{'} = f$ . Then evaluate $F$ , not $f$ , at both limits.
Writing $F (a) - F (b)$ , or dropping $F (a)$ because it looks small.	The theorem is $F (b) - F (a)$ in that order, and $F (a)$ is 0 only for some antiderivatives.	Write both values in a line, upper limit first, before you do any arithmetic: $F (b) - F (a)$ .
Using the theorem on an integrand that is not continuous on the interval, such as $\frac{1}{x^{2}}$ on $[- 1, 1]$ .	Continuity on $[a, b]$ is the theorem's hypothesis. Without it, $F (b) - F (a)$ can be a number that is not the integral at all.	Before you antidifferentiate, find where the integrand breaks and make sure no break lies in $[a, b]$ .

Part 3 Watch it done Evaluating a definite integral with the theorem

Watch each step, then do the last one yourself.

Evaluate $\int_{1}^{2} (6 x^{2} - \frac{4}{x^{2}}) d x$ .

Step 1 · Check the hypothesis
The integrand $6 x^{2} - \frac{4}{x^{2}}$ is defined for every $x$ except 0, and 0 is not in $[1, 2]$ . So it is continuous on $[1, 2]$ , and the Fundamental Theorem of Calculus applies.
Why this step?
Why check continuity before you look for an antiderivative?
Continuity on $[a, b]$ is what the theorem needs. On $[- 1, 2]$ the same integrand breaks at $x = 0$ , and subtracting antiderivative values there would give a number the theorem does not back up.
Step 2 · Find an antiderivative
Write $- \frac{4}{x^{2}}$ as $- 4 x^{- 2}$ . Raising the exponent to $- 1$ and dividing by $- 1$ gives $4 x^{- 1}$ . So $F (x) = 2 x^{3} + \frac{4}{x}$ , and the check works: $F^{'} (x) = 6 x^{2} - \frac{4}{x^{2}}$ .
Step 3 · Evaluate at both limits
$F (2) = 2 (8) + \frac{4}{2} = 18$ and $F (1) = 2 (1) + \frac{4}{1} = 6$ .
Your step · Calculate

No calculator. Finish it. With $F (x) = 2 x^{3} + \frac{4}{x}$ , $F (2) = 18$ and $F (1) = 6$ , what is $\int_{1}^{2} (6 x^{2} - \frac{4}{x^{2}}) d x$ ?

Your answer

Part 4 Try it A leak and a claim

Try it · Calculate

No calculator. Water leaks from a barrel at a rate of $r (t) = 6 \sqrt{t}$ liters per hour, $t$ hours after noon. How many liters leak out from $t = 1$ to $t = 4$ ?

Your answer

Try it · Decide and justify

No calculator. A classmate writes $\int_{- 1}^{1} \frac{1}{x^{2}} d x = F (1) - F (- 1) = - 1 - 1 = - 2$ , using the antiderivative $F (x) = - \frac{1}{x}$ . Is the classmate's work right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

On the exam free response is written by hand, so do this one on paper first. Nothing you type here leaves this page. Then compare with the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. A tank holds 50 liters of water at time $t = 0$ . For $0 \leq t \leq 4$ , water flows into the tank at the rate $W (t) = 6 \sqrt{t}$ liters per minute and drains out at the rate $D (t) = 2 t + 5$ liters per minute, where $t$ is in minutes.

(a) Use the Fundamental Theorem of Calculus to find $\int_{0}^{4} W (t) d t$ . Show the antiderivative you use. Using correct units, interpret the meaning of the value in the context of the problem.

(b) Find the amount of water in the tank at time $t = 4$ . Show the work that leads to your answer.

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

Model answer

(a) $W (t) = 6 t^{1 / 2}$ , and an antiderivative is $4 t^{3 / 2}$ , since $\frac{d}{d t} (4 t^{3 / 2}) = 6 t^{1 / 2}$ . $W$ is continuous on $[0, 4]$ , so $\int_{0}^{4} W (t) d t = 4 (4)^{3 / 2} - 4 (0)^{3 / 2} = 32$ . During the first 4 minutes, 32 liters of water flow into the tank.

(b) Let $A (t)$ be the amount of water in the tank, in liters, at time $t$ . Then $A (4) = 50 + \int_{0}^{4} (W (t) - D (t)) d t$ . An antiderivative of $D (t) = 2 t + 5$ is $t^{2} + 5 t$ , and $D$ is continuous on $[0, 4]$ , so $\int_{0}^{4} D (t) d t = (4^{2} + 5 (4)) - (0^{2} + 5 (0)) = 36$ . So $A (4) = 50 + 32 - 36 = 46$ liters.

(c) The amount in the tank is $A (t) = 50 + \int_{0}^{t} (W (s) - D (s)) d s$ , so $A^{'} (t) = W (t) - D (t)$ . $A^{'} (1) = W (1) - D (1) = 6 - 7 = - 1 < 0$ , so the amount of water is decreasing at $t = 1$ .

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

(a) Writes the antiderivative $4 t^{3 / 2}$ and evaluates it at both limits to get 32.
(a) Says 32 liters of water flow into the tank from $t = 0$ to $t = 4$ minutes, with units.
(b) Writes $50 + \int_{0}^{4} (W (t) - D (t)) d t$ , with the initial 50, both limits and $d t$ .
(b) Uses the antiderivative $t^{2} + 5 t$ to get 36 liters drained, and gets 46 liters.
(c) Compares $W (1) = 6$ with $D (1) = 7$ and concludes decreasing because $W (1) - D (1) < 0$ .

Part 6 Check yourself An antiderivative from a table

The two case questions use the same table. Answer them from it.

The function $f$ is continuous for all real numbers, and $F$ is an antiderivative of $f$ . Some values of $F$ are in the table.

Values of an antiderivative F of f
$x$	$F (x)$
0	4
1	1
2	$- 3$
3	2
4	6
5	5

Case question 1 · Calculate

No calculator. What is the value of $\int_{1}^{4} f (x) d x$ ?

Your answer

Case question 2 · Multiple choice

No calculator. Let $G (x) = \int_{2}^{x} f (t) d t$ . What is $G (5)$ ?

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

Check yourself · Spot the error

No calculator. A student evaluates $\int_{0}^{π / 2} (\cos x + 2 x) d x$ . Tap the first line with a mistake.

Check yourself · Sort it

No calculator. Sort each function: is it an antiderivative of $2 x$ , or not?

An antiderivative of $2 x$

Not an antiderivative of $2 x$

Part 7 Keep it Bring back earlier ideas

The theorem leans on a property of integrals from 6.6 and on the derivative rules from Unit 2. Pulling them back up is what makes them last.

Bring it back · Calculate

No calculator. If $\int_{1}^{5} g (x) d x = 9$ and $\int_{1}^{3} g (x) d x = 4$ , what is $\int_{3}^{5} 2 g (x) d x$ ?

Your answer

Bring it back · Multiple choice

No calculator. If $f (x) = 3 \sin x - 2 \cos x$ , what is $f^{'} (x)$ ?

In three lines

An antiderivative of f is a function whose derivative is f, and differentiating it is the check.
For continuous f, the integral of f from a to x is the antiderivative that is 0 at a.
If f is continuous on [a, b] and F is any antiderivative, the integral from a to b is F(b) minus F(a).

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.

On a no-calculator integral, write the antiderivative you are using and then $F (b) - F (a)$ with both values before you simplify, so a reader can follow every step. On a calculator integral, write the integral itself, with its integrand, both limits and $d x$ , before you give the number to three decimal places.

Questions students ask

Which antiderivative should I use?

Any one. A constant added to $F$ appears in both $F (b)$ and $F (a)$ and cancels in the subtraction, so the simplest antiderivative, with no constant, is the usual choice.

How is this different from what I did in 6.4?

In 6.4 you differentiated a function defined by an integral. Here the same fact is read the other way: that function is an antiderivative, and it leads to a way to evaluate any definite integral of a continuous function from an antiderivative.

Can the answer be negative?

Yes. A definite integral is a signed accumulation, so if $f$ is negative on most of $[a, b]$ the result can be negative. If $f$ is positive on the whole interval, though, a negative answer means something went wrong.

What if I cannot find an antiderivative?

Some continuous functions, such as $\sin (x^{2})$ , have an antiderivative that no formula you know writes down. The integral still exists. Where a question allows a calculator, write the integral with its integrand, both limits and $d x$ , then give the calculator's value.

Do I really have to check continuity every time?

On paper, a short phrase is enough: "the integrand is continuous on $[1, 2]$ ." It matters most when the integrand has a denominator or a logarithm, because those are where a break can sit inside the interval.

Course alignment, for teachers

AP Calculus AB topic 6.7, Unit 6: Integration and Accumulation of Change.

Skill 3.D: Apply an appropriate mathematical definition, theorem, or test.

This lesson maps all 3 essential knowledge statements of the topic to the part that teaches each one: 3 are 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 How high does the drone climb?

Part 2 Learn it Antiderivatives and the theorem that uses them

2a. An antiderivative runs a derivative backward

2b. The accumulation function is an antiderivative

2c. The Fundamental Theorem: F(b) minus F(a)

Part 3 Watch it done Evaluating a definite integral with the theorem

Part 4 Try it A leak and a claim

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself An antiderivative from a table

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic