Unit 6 · Topic 6.4 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 6.4: The Fundamental Theorem of Calculus and Accumulation Functions

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

Because x is already the upper limit. Inside the integral the variable runs from a to x, so it needs its own letter. Writing ax f(x)\,dx uses one letter for two jobs and is not accepted notation.

Q: Does the lower limit change the derivative?

No. Changing a adds or removes a fixed piece of area, which is a constant, and a constant has derivative 0. The lower limit does change the values of the function, though, because it sets where g equals 0.

Q: Do I need a formula for g(x) before I can find g'(x)?

No. The theorem gives g'(x) = f(x) directly. That is what makes it useful for integrands such as t3 + 1, where no area formula is available.

Q: Can an accumulation function be negative?

Yes. Area below the t-axis counts as negative, so if more of the region from a to x lies below the axis than above it, g(x) is negative.

Q: What if the upper limit is something like x2?

Then the function is a composite, and the chain rule applies: put x2 in for t, then multiply by 2x, the derivative of the upper limit.

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

Write an accumulation function as a definite integral with a variable upper limit, find its values from areas, and find its derivative with the Fundamental Theorem of Calculus.

Key vocabulary

Term	What it means	The exam trap
Definite integral	$\int_{a}^{b} f (x) d x$ , the limit of Riemann sums of $f$ on $[a, b]$ , which gives the accumulated change, the signed area between the graph of $f$ and the axis.	When part of the graph dips below the axis, that piece of area enters with its sign.
Accumulation function	A function defined by a definite integral with a variable upper limit, such as $g (x) = \int_{a}^{x} f (t) d t$ .	When a question gives you a graph, decide first whether it is the graph of the integrand or of the accumulation function, then read a number off it.
Fundamental Theorem of Calculus	If $f$ is continuous on an interval containing $a$ , then $\frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x)$ for every $x$ in that interval.	State that the integrand is continuous before you use it, and look twice at the upper limit when it is not plain $x$ .

The line y = f(t) = 3 - 0.5t for t from 0 to 7. The area under it from t = 0 to t = 3 is shaded and labeled g(3) = 6.75, and a second strip from t = 3 to t = 5 adds 2, so g(5) = 8.75.

Part 1 Think first How fast is the total growing?

Commit to a guess first. Even a wrong guess makes the idea easier to hold onto.

Predict first

Water runs into a tank through a pipe whose flow rate keeps changing. Let $W (x)$ be the number of liters that have come in during the first $x$ minutes. At the instant $x = 4$ the pipe is delivering 12 liters per minute, and by then 30 liters have come in. How fast is $W (x)$ growing at $x = 4$ ?

In short. A definite integral whose upper limit is a variable, $g (x) = \int_{a}^{x} f (t) d t$ , defines a new function, and when $f$ is continuous its derivative is $g^{'} (x) = f (x)$ .

The input of $g$ is the upper limit $x$ , and the integrand uses a different letter, such as $t$ .
The value $g (x)$ is the signed area between the graph of $f$ and the $t$ -axis from $a$ to $x$ .
If $f$ is continuous on an interval containing $a$ , then $\frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x)$ ; with an upper limit $u (x)$ , the chain rule multiplies $f (u (x))$ by $u^{'} (x)$ .

Part 2 Learn it A function made from an integral, and its derivative

4a. An integral with a moving upper limit

Until now a definite integral had two fixed limits, and its value was one number: the accumulated change from $a$ to $b$ . Now hold the lower limit fixed and let the upper limit move. Each upper limit gives a different number, so the integral becomes a function of that limit:

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

A function built this way, by accumulating $f$ from a fixed starting input, is an accumulation function.

Read it piece by piece. The input of $g$ is $x$ , the right end of the interval. Inside the integral the variable is a different letter, $t$ , because $x$ is already in use as the endpoint; $t$ just runs from $a$ to $x$ . The value $g (x)$ is the signed area between the graph of $f$ and the $t$ -axis from $t = a$ to $t = x$ , with area below the axis counted as negative.

The graph at the top shows $f (t) = 3 - 0.5 t$ with $a = 0$ . The shaded region from 0 to 3 has area 6.75, so $g (3) = 6.75$ . Sliding $x$ on to 5 adds the green strip, so $g (5) = 6.75 + 2 = 8.75$ .

When $x = a$ the interval has no width, so $g (a) = 0$ . For other inputs, when the graph of $f$ is made of lines or parts of circles, find the area with geometry.

Quick check · Calculate

No calculator. Let $f (t) = 6 - 2 t$ , and let $g (x) = \int_{1}^{x} f (t) d t$ . What is $g (3)$ ?

Your answer

4b. The derivative of an accumulation function

How fast does $g (x) = \int_{a}^{x} f (t) d t$ change as $x$ moves? Push $x$ a little further, to $x + h$ . The new area is the old area plus a thin strip from $x$ to $x + h$ . If $f$ is continuous, its height hardly changes across that strip, so the strip's area is close to $f (x) \cdot h$ . Then

\frac{g (x + h) - g (x)}{h} \approx \frac{f (x) \cdot h}{h} = f (x),

and the approximation gets better as $h \to 0$ . The limit of that difference quotient is the derivative, so

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

This is the Fundamental Theorem of Calculus for accumulation functions. It has a hypothesis: $f$ is continuous on an interval containing $a$ , and $x$ is in that interval. Say so before you use it.

The derivative is the integrand with $x$ put in for $t$ . The lower limit $a$ does not appear in it, because moving $a$ only adds or removes a fixed piece of area, a constant. You also never need a formula for $g$ . For $g (x) = \int_{1}^{x} \sqrt{t^{3} + 1} d t$ no area formula is in sight, yet $\sqrt{t^{3} + 1}$ is continuous for $t \geq - 1$ , so $g^{'} (x) = \sqrt{x^{3} + 1}$ there, and $g^{'} (2) = \sqrt{9} = 3$ .

Quick check · Calculate

No calculator. Let $g (x) = \int_{0}^{x} \sqrt{t^{2} + 9} d t$ . What is $g^{'} (4)$ ?

Your answer

4c. When the upper limit is a function of x

Sometimes the upper limit is not $x$ but an expression in $x$ , such as $x^{2}$ or $5 x$ . Then the function is a composite. Let $G (u) = \int_{a}^{u} f (t) d t$ be the outer function; then

\int_{a}^{x^{2}} f (t) d t = G (x^{2}) .

The chain rule from Unit 3 gives $G^{'} (x^{2}) \cdot 2 x$ , and the theorem says $G^{'} (u) = f (u)$ . So, when $f$ is continuous on an interval containing $a$ and $x^{2}$ ,

\frac{d}{d x} (\int_{a}^{x^{2}} f (t) d t) = f (x^{2}) \cdot 2 x .

In words: put the upper limit in for $t$ , then multiply by the derivative of the upper limit. Both moves are needed. Two slips are common. One leaves out the factor $2 x$ . The other puts plain $x$ into the integrand when the upper limit is $x^{2}$ .

For example, let $P (x) = \int_{0}^{5 x} \cos (t^{2}) d t$ . The integrand is continuous everywhere, the upper limit $5 x$ goes in for $t$ , and its derivative, 5, multiplies the result:

P^{'} (x) = \cos ((5 x)^{2}) \cdot 5 = 5 \cos (25 x^{2}) .

When the upper limit is plain $x$ , its derivative is 1, which is why no extra factor appeared in the last section.

Quick check · Multiple choice

No calculator. Let $H (x) = \int_{1}^{x^{3}} \frac{1}{1 + t^{2}} d t$ . Which of the following is $H^{'} (x)$ ?

Common mistakes

The mistake	Why it is wrong	Do this instead
Reading the value $g (x)$ of $g (x) = \int_{a}^{x} f (t) d t$ as the height $f (x)$ .	$g (x)$ is the accumulated area from $a$ to $x$ . The height $f (x)$ is how fast that area is growing, which is $g^{'} (x)$ .	For a value of $g$ , find an area from $a$ to $x$ . For a value of $g^{'}$ , read the height of $f$ .
Writing $g^{'} (x) = f (x) - f (a)$ , or differentiating the integrand to get $f^{'} (x)$ .	The theorem says the derivative of $\int_{a}^{x} f (t) d t$ is exactly $f (x)$ . The lower limit only adds a constant to $g$ , and nothing gets differentiated.	Check that $f$ is continuous, then write the integrand with $x$ in place of $t$ .
Dropping the chain-rule factor when the upper limit is an expression such as $x^{2}$ .	$\int_{a}^{x^{2}} f (t) d t$ is a composite function, so its derivative is $f (x^{2}) \cdot 2 x$ , not $f (x^{2})$ .	Put the upper limit in for $t$ , then multiply by the derivative of the upper limit.

Part 3 Watch it done Values and a derivative from a semicircle

Watch each step, then do the last one yourself.

Let $g (x) = \int_{- 3}^{x} \sqrt{9 - t^{2}} d t$ for $- 3 \leq x \leq 3$ . Find $g (0)$ and $g (3)$ , and find $g^{'} (x)$ .

Step 1 · Read the integrand as a graph
$y = \sqrt{9 - t^{2}}$ is the upper half of the circle $t^{2} + y^{2} = 9$ , which has radius 3 and center at the origin. It runs from $t = - 3$ to $t = 3$ .
Step 2 · Find g(0) from the area
$g (0) = \int_{- 3}^{0} \sqrt{9 - t^{2}} d t$ is the area of the quarter of that circle from $t = - 3$ to $t = 0$ : $g (0) = \frac{1}{4} π (3)^{2} = \frac{9 π}{4}$ .
Step 3 · Find g(3) from the area
$g (3) = \int_{- 3}^{3} \sqrt{9 - t^{2}} d t$ is the area of the whole upper half: $g (3) = \frac{1}{2} π (3)^{2} = \frac{9 π}{2}$ .
Step 4 · Check the hypothesis, then differentiate
$\sqrt{9 - t^{2}}$ is continuous on $[- 3, 3]$ , an interval that contains the lower limit $- 3$ . By the Fundamental Theorem of Calculus, $g^{'} (x) = \sqrt{9 - x^{2}}$ for $x$ in that interval.
Why this step?
Why state that $\sqrt{9 - t^{2}}$ is continuous on $[- 3, 3]$ before writing $g^{'} (x)$ ?
The theorem promises $g^{'} (x) = f (x)$ only when $f$ is continuous on an interval containing the lower limit and $x$ . Here the integrand is defined only for $- 3 \leq t \leq 3$ , so naming that interval also says where the result holds.
Your step · Calculate

No calculator. Finish it. For $g (x) = \int_{- 3}^{x} \sqrt{9 - t^{2}} d t$ , where $g^{'} (x) = \sqrt{9 - x^{2}}$ on $[- 3, 3]$ , what is $g^{'} (\sqrt{5})$ ?

Your answer

Part 4 Try it A claim to judge and a derivation to check

Try it · Decide and justify

No calculator. Let $f (t) = 6 - 3 t$ and $g (x) = \int_{0}^{x} f (t) d t$ . A classmate notices that $f (2) = 0$ and says that $g (2) = 0$ too. Is the classmate right?

Which reason supports your answer?

Try it · Spot the error

No calculator. A student finds $k^{'} (2)$ for $k (x) = \int_{0}^{x^{2}} \sqrt{t + 5} d t$ . Tap the first line with a mistake.

Part 5 Explain it Write it the way the exam asks

Free response is handwritten on the exam, so work this on paper first. Nothing you type here leaves this page. Then open the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 5, from free-response Part B (calculator not permitted).

Free-response practice · Question 5 style

No calculator. A tank holds 40 liters of water at time $t = 0$ . For $0 \leq t \leq 6$ , water flows into the tank at a rate of $r (t)$ liters per hour, where $t$ is measured in hours, $r (t) = 2 t$ for $0 \leq t \leq 3$ , and $r (t) = 12 - 2 t$ for $3 < t \leq 6$ . No water leaves the tank.

(a) Write an expression involving an integral for $W (x)$ , the number of liters of water in the tank at time $t = x$ hours.

(b) Find $W (4)$ . Show the work that leads to your answer.

(c) Find $W^{'} (4)$ , and justify your answer. Using correct units, interpret the meaning of $W^{'} (4)$ in the context of the problem.

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

Model answer

(a) $W (x) = 40 + \int_{0}^{x} r (t) d t$ .

(b) $W (4) = 40 + \int_{0}^{4} r (t) d t$ . From $t = 0$ to $t = 3$ the graph of $r$ and the $t$ -axis bound a triangle of area $\frac{1}{2} (3) (6) = 9$ . From $t = 3$ to $t = 4$ they bound a trapezoid of area $\frac{6 + 4}{2} (1) = 5$ . So $W (4) = 40 + 9 + 5 = 54$ liters.

(c) Both pieces of $r$ equal 6 at $t = 3$ , so $r$ is continuous on $0 \leq t \leq 6$ . By the Fundamental Theorem of Calculus, $W^{'} (x) = r (x)$ , so $W^{'} (4) = r (4) = 12 - 8 = 4$ . At time $t = 4$ hours, the amount of water in the tank is increasing at a rate of 4 liters per hour.

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

Writes $W (x) = 40 + \int_{0}^{x} r (t) d t$ , with the starting 40 liters and the upper limit $x$ .
Finds $\int_{0}^{4} r (t) d t = 9 + 5 = 14$ from the areas, and reports $W (4) = 54$ liters.
States that $r$ is continuous and uses the Fundamental Theorem of Calculus to get $W^{'} (4) = r (4) = 4$ .
Interprets $W^{'} (4)$ : at $t = 4$ the water in the tank is increasing at 4 liters per hour.

Part 6 Check yourself A function given by line segments

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

The function $f$ is continuous on $0 \leq t \leq 8$ , and its graph consists of line segments joining the points in the table. Let $g (x) = \int_{0}^{x} f (t) d t$ .

Points the graph of f passes through, joined by line segments
$t$	$f (t)$
0	4
2	4
4	0
6	$- 2$
8	$- 2$

Case question 1 · Calculate

No calculator. What is $g (6)$ ?

Your answer

Case question 2 · Multiple choice

No calculator. What is $g^{'} (7)$ ?

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

Check yourself · Fill the statement

No calculator. The function $f$ is continuous everywhere. Tap a blank, then tap an answer.

For $g (x) = \int_{3}^{x} f (t) d t$ , the value $g (3)$ is , and the derivative $g^{'} (5)$ is .

Check yourself · Match

No calculator. Each function of $x$ is defined by an integral. Tap a function, then its derivative.

Part 7 Keep it Bring back earlier ideas

This lesson leans on the units of accumulated change from 6.1 and on the chain rule from Unit 3. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. Water leaks from a barrel at a rate of $L (t)$ gallons per minute, where $t$ is measured in minutes. In what units is the area of the region between the graph of $L$ and the $t$ -axis from $t = 0$ to $t = 10$ measured?

Bring it back · Calculate

No calculator. Let $h (x) = (x^{2} + 1)^{3}$ . What is $h^{'} (1)$ ?

Your answer

In three lines

A definite integral with a variable upper limit, $\int_{a}^{x} f (t) d t$ , defines a new function of $x$ .
Its value at $x$ is the signed area from $a$ to $x$ , so it is 0 at $x = a$ .
For continuous $f$ its derivative is $f (x)$ ; with an upper limit $u (x)$ , multiply $f (u (x))$ by $u^{'} (x)$ .

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.

To differentiate a function defined by an integral, first say the integrand is continuous, then write the integrand at the upper limit times the derivative of the upper limit. To find a value of an accumulation function, write the integral with both limits and its $d t$ , then find the signed area, and give units when there is a context.

Questions students ask

Why is the variable inside the integral

t

and not

x

Because $x$ is already the upper limit. Inside the integral the variable runs from $a$ to $x$ , so it needs its own letter. Writing $\int_{a}^{x} f (x) d x$ uses one letter for two jobs and is not accepted notation.

Does the lower limit change the derivative?

No. Changing $a$ adds or removes a fixed piece of area, which is a constant, and a constant has derivative 0. The lower limit does change the values of the function, though, because it sets where $g$ equals 0.

Do I need a formula for

g (x)

before I can find

g^{'} (x)

No. The theorem gives $g^{'} (x) = f (x)$ directly. That is what makes it useful for integrands such as $\sqrt{t^{3} + 1}$ , where no area formula is available.

Can an accumulation function be negative?

Yes. Area below the $t$ -axis counts as negative, so if more of the region from $a$ to $x$ lies below the axis than above it, $g (x)$ is negative.

What if the upper limit is something like

x^{2}

Then the function is a composite, and the chain rule applies: put $x^{2}$ in for $t$ , then multiply by $2 x$ , the derivative of the upper limit.

Course alignment, for teachers

AP Calculus AB topic 6.4, Unit 6: Integration and Accumulation of Change.

Skill 1.D: Identify an appropriate mathematical rule or procedure based on the relationship between concepts (e.g., rate of change and accumulation) or processes (e.g., differentiation and its inverse process, anti-differentiation) to solve problems.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 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 fast is the total growing?

Part 2 Learn it A function made from an integral, and its derivative

4a. An integral with a moving upper limit

4b. The derivative of an accumulation function

4c. When the upper limit is a function of x

Part 3 Watch it done Values and a derivative from a semicircle

Part 4 Try it A claim to judge and a derivation to check

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A function given by line segments

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic