Unit 6 · Topic 6.6 · about 30 minutes · 7 parts · 13 scored checks, 3 of them multiple choice

AP Calculus AB 6.6: Applying Properties of Definite Integrals

Q: Is a definite integral the same as the area under the curve?

Only where the graph is on or above the x-axis. Below the axis the definite integral counts area as negative. For total area, add the areas of the pieces as positive numbers.

Q: Why does reversing the limits change the sign?

In a Riemann sum each width x is the change in x from the start of a piece to its end. Going from b down to a makes every width negative, so every product in the sum, and the sum itself, changes sign.

Q: Can I use these properties when I do not know a formula for the function?

Yes, and that is often the point. Given the values of a few integrals, the constant multiple, sum, reversal and adjacent-interval properties give the values of others without any formula for the function.

Q: Does the value of a function at a jump change the integral?

No. A single input has no width, so its value adds no area. Integrate each side of the jump as its own function and add the two pieces.

Q: Do I need to memorize area formulas for this?

You need the ones for rectangles, triangles, trapezoids and circles, including half and quarter circles. A graph such as y = r2 - x2 is the top half of a circle of radius r centered at the origin.

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

Calculate a definite integral from the signed areas of familiar shapes and from the properties of definite integrals.

Key vocabulary

Term	What it means	The exam trap
Definite integral	The number $\int_{a}^{b} f (x) d x$ , the limit of Riemann sums for $f$ on $[a, b]$ ; where $f \geq 0$ it is the area between the graph and the $x$ -axis.	When part of the graph is below the axis, the integral and the total area are different numbers. Read which one the question asks for.
Limits of integration	The numbers $a$ and $b$ in $\int_{a}^{b} f (x) d x$ : the lower limit $a$ , where the integral starts, and the upper limit $b$ , where it ends.	The lower limit does not have to be the smaller number. Check which limit is on the bottom before you use a value you were given.
Adjacent intervals	Two intervals that share an endpoint, such as $[a, b]$ and $[b, c]$ , so that together they make one interval.	Before you add two integrals, check that their intervals really meet at a shared endpoint.
Removable discontinuity	A point $x = c$ where $\lim_{x \to c} f (x)$ exists but $f (c)$ is not defined or not equal to the limit: a hole in the graph.	Being defined at $c$ does not settle whether $f$ is continuous at $c$ . Check the definition of continuity there.
Jump discontinuity	A point where both one-sided limits of $f$ exist but are not equal, so the graph jumps from one height to another.	Work out which formula applies on each side of the jump before you compute anything on either piece.

Graph of f, equal to 2 for x from 0 up to 2 and to 5 minus x from 2 to 6, so it jumps from 2 up to 3 at x = 2. The shaded pieces are a rectangle of area 4, a triangle above the axis of area 4.5, and a triangle below the axis of signed area -0.5, which add to 8.

Part 1 Think first How far did the scooter go?

Make a guess before the lesson starts. Even a wrong guess helps the answer stick.

Predict first

An electric scooter starts from rest and speeds up steadily, reaching 6 meters per second after 4 seconds. It then holds 6 meters per second for 3 more seconds. You have no formula for its velocity, only that description. How far does it travel in those 7 seconds?

In short. A definite integral $\int_{a}^{b} f (x) d x$ can often be evaluated without an antiderivative, from the signed areas of shapes and from integrals whose values you already know.

Area above the $x$ -axis counts as positive and area below counts as negative, so rectangles, triangles, trapezoids and circles give many integrals.
Constants factor out, integrals of sums and differences split, reversing the limits changes the sign, and integrals over adjacent intervals add.
A function with a removable or a jump discontinuity still has a definite integral: a single point adds no area, and a jump is handled by splitting the interval there.

Part 2 Learn it Shapes first, then the rules that combine them

6a. A definite integral from shapes you know

When the graph of $f$ is made of line segments and pieces of circles, you can evaluate $\int_{a}^{b} f (x) d x$ with geometry. The definite integral measures the region between the graph and the $x$ -axis from $x = a$ to $x = b$ , with one adjustment: area above the axis counts as positive and area below the axis counts as negative.

Find each piece's area with a formula you know, give it its sign, and add.

Rectangle: base times height.
Triangle: $\frac{1}{2}$ times base times height.
Trapezoid: the average of the two parallel sides times the width.
Circle of radius $r$ : $π r^{2}$ , so a semicircle is $\frac{1}{2} π r^{2}$ and a quarter circle is $\frac{1}{4} π r^{2}$ .

Example: $\int_{- 2}^{2} \sqrt{4 - x^{2}} d x$ . The graph of $y = \sqrt{4 - x^{2}}$ is the top half of the circle of radius 2 centered at the origin, so the region is a semicircle above the axis and the integral is $\frac{1}{2} π (2)^{2} = 2 π$ .

With a sign change: $\int_{0}^{5} (2 - x) d x$ . The line crosses the axis at $x = 2$ . From 0 to 2 it bounds a triangle above the axis with area $\frac{1}{2} (2) (2) = 2$ . From 2 to 5 it bounds a triangle below the axis with area $\frac{1}{2} (3) (3) = 4.5$ , which counts as $- 4.5$ . So $\int_{0}^{5} (2 - x) d x = 2 - 4.5 = - 2.5$ . The total area of the triangles, 6.5, is a different number.

Quick check · Calculate

No calculator. Use geometry to evaluate $\int_{1}^{7} (x - 3) d x$ .

Your answer

6b. Constant multiples and sums

Two properties let you take an integral apart and use values you already know.

Constant multiple: $\int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x$ . Multiplying every height by $k$ multiplies every signed area by $k$ .

Sum: $\int_{a}^{b} (f (x) + g (x)) d x = \int_{a}^{b} f (x) d x + \int_{a}^{b} g (x) d x$ . Stacking one function's heights on top of another's adds their signed areas. Put the two together, with $k = - 1$ , and a difference splits the same way.

A constant on its own is a rectangle: $\int_{a}^{b} c d x = c (b - a)$ , its height times the width of the interval, not just $c$ .

Example. Suppose $\int_{1}^{4} f (x) d x = 6$ and $\int_{1}^{4} g (x) d x = - 2$ . Then

\int_{1}^{4} (5 f (x) - 3 g (x) + 2) d x = 5 (6) - 3 (- 2) + 2 (4 - 1) = 42.

There is no such rule for a product. In general $\int_{a}^{b} f (x) g (x) d x$ is not $\int_{a}^{b} f (x) d x \cdot \int_{a}^{b} g (x) d x$ : with $f (x) = g (x) = 1$ on $[0, 2]$ , the integral of the product is 2, while the product of the integrals is $2 \cdot 2 = 4$ .

Quick check · Multiple choice

No calculator. Suppose $\int_{2}^{6} f (x) d x = 7$ and $\int_{2}^{6} g (x) d x = - 3$ . What is $\int_{2}^{6} (4 f (x) - g (x) + 5) d x$ ?

6c. Reversing the limits and joining intervals

Reversal of limits: $\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x$ . In a Riemann sum each width $Δ x$ is the change in $x$ across a piece. Running from $b$ back to $a$ makes every width negative, so the whole sum changes sign. An interval of no width gives nothing: $\int_{a}^{a} f (x) d x = 0$ .

Adjacent intervals: $\int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x = \int_{a}^{c} f (x) d x$ . The region from $a$ to $b$ and the region from $b$ to $c$ together make the region from $a$ to $c$ .

With reversal, the adjacent-interval property holds whatever order $a$ , $b$ and $c$ come in, as long as $f$ is integrable on an interval that contains all three. That makes it a tool for finding a missing piece. Suppose $\int_{0}^{10} f (x) d x = 12$ and $\int_{0}^{4} f (x) d x = 5$ . The pieces from 0 to 4 and from 4 to 10 make up 0 to 10, so

\int_{4}^{10} f (x) d x = \int_{0}^{10} f (x) d x - \int_{0}^{4} f (x) d x = 12 - 5 = 7,

and reversing the limits, $\int_{10}^{4} f (x) d x = - 7$ .

Before you combine values, write out the limits of every integral you know and of the one you want. The missing piece is usually easy to see once they are side by side.

Quick check · Match

No calculator. Suppose $\int_{0}^{5} f (x) d x = 8$ and $\int_{0}^{3} f (x) d x = 6$ . Tap an integral, then its value.

6d. Holes and jumps

The definite integral was defined for continuous functions, but the definition extends to a function with a removable discontinuity or a jump discontinuity. Two facts make that work.

A single point has no width. Changing a function's value at one input, or leaving it with no value there, changes no area. Take $p (x) = \frac{x^{2} - 1}{x - 1}$ , which is not defined at $x = 1$ . Everywhere else $p (x) = x + 1$ , so $\int_{0}^{3} p (x) d x = \int_{0}^{3} (x + 1) d x$ . That region is a trapezoid with parallel sides 1 and 4 and width 3, so the integral is $\frac{1 + 4}{2} \cdot 3 = 7.5$ .

A jump splits the interval. Integrate each side of the jump as its own function, then add the pieces with the adjacent-interval property. The graph at the top of this lesson jumps at $x = 2$ . The rectangle on the left, the triangle above the axis and the triangle below it give $4 + 4.5 + (- 0.5) = 8$ from 0 to 6.

Which side's formula holds at the jump itself makes no difference, for the same reason as before: it is one point, and one point adds no area.

Quick check · Decide and justify

No calculator. Let $g (x) = 1$ for $0 \leq x < 3$ and $g (x) = 4$ for $3 \leq x \leq 5$ . A classmate says $\int_{0}^{5} g (x) d x = 11$ , even though $g$ is not continuous at $x = 3$ . Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Adding the area of a region below the $x$ -axis as a positive number in a definite integral.	The definite integral counts area below the axis as negative, so that region subtracts.	Give each piece its sign before you add. Save the all-positive total for a question that asks for area.
Writing $\int_{a}^{b} c d x = c$ for a constant $c$ .	A constant is a height. Over $[a, b]$ it makes a rectangle whose width is $b - a$ .	Write $\int_{a}^{b} c d x = c (b - a)$ , height times width.
Using a given value of $\int_{a}^{b} f (x) d x$ for $\int_{b}^{a} f (x) d x$ .	Running the limits backward changes the sign of the integral.	Read which limit is on the bottom, and write $\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x$ before you substitute a number.

Part 3 Watch it done Three properties and two shapes in one integral

Watch each step, then do the last one yourself.

Let $f (x) = x + 1$ for $0 \leq x < 2$ and $f (x) = 6 - 2 x$ for $2 \leq x \leq 5$ . Find $\int_{5}^{0} 3 f (x) d x$ .

Step 1 · Reverse the limits and factor out the 3
$\int_{5}^{0} 3 f (x) d x = - \int_{0}^{5} 3 f (x) d x = - 3 \int_{0}^{5} f (x) d x$ .
Step 2 · Split at the jump
As $x$ approaches 2 from the left, $f (x)$ approaches 3, but $f (2) = 6 - 4 = 2$ . That is a jump, so integrate each side on its own: $\int_{0}^{5} f (x) d x = \int_{0}^{2} (x + 1) d x + \int_{2}^{5} (6 - 2 x) d x$ .
Why this step?
Why may the first piece use $x + 1$ all the way to $x = 2$ , when $f (2)$ is 2 and not 3?
The value at a single input sits over no width, so it adds no area. That is why the definite integral extends to a function with a jump: each side is integrated as its own function, and which side owns the point $x = 2$ makes no difference.
Step 3 · Left piece: a trapezoid
From 0 to 2, $y = x + 1$ rises from 1 to 3 above the axis, so $\int_{0}^{2} (x + 1) d x = \frac{1 + 3}{2} \cdot 2 = 4$ .
Step 4 · Right piece: two triangles
$6 - 2 x = 0$ at $x = 3$ . From 2 to 3 the region is a triangle above the axis with base 1 and height 2, area 1. From 3 to 5 it is a triangle below the axis with base 2 and height 4, area 4, which counts as $- 4$ . So $\int_{2}^{5} (6 - 2 x) d x = 1 - 4 = - 3$ .
Your step · Calculate

No calculator. Finish it. Let $f (x) = x + 1$ for $0 \leq x < 2$ and $f (x) = 6 - 2 x$ for $2 \leq x \leq 5$ , so that $\int_{0}^{2} f (x) d x = 4$ and $\int_{2}^{5} f (x) d x = - 3$ . What is $\int_{5}^{0} 3 f (x) d x$ ?

Your answer

Part 4 Try it A hole, and a set of claims

Try it · Spot the error

No calculator. Let $h (x) = \frac{x^{2} - 4}{x - 2}$ for $x \neq 2$ , and let $h (2) = 7$ . A student evaluates $\int_{0}^{4} h (x) d x$ . Tap the first line with a mistake.

Try it · Sort it

No calculator. Here $f$ and $g$ are continuous on an interval that contains $a$ , $b$ and $c$ . Sort each statement: is it true for every such $f$ and $g$ , or not true in general? Tap a card, then a bin.

True for every $f$ and $g$

Not true in general

Part 5 Explain it Answer it the way the exam asks

On the exam free response is handwritten, 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 3, from free-response Part B (calculator not permitted).

Free-response practice · Question 3 style

No calculator. Water flows into and out of a tank. For $0 \leq t \leq 10$ , the net rate at which water flows into the tank is $R (t)$ liters per minute, where $t$ is measured in minutes, $R (t) = 2 t$ for $0 \leq t < 4$ , and $R (t) = t - 8$ for $4 \leq t \leq 10$ .

(a) The function $R$ is not continuous at $t = 4$ . Explain why $\int_{0}^{10} R (t) d t$ is still defined.

(b) Use geometry and the properties of definite integrals to find $\int_{0}^{10} R (t) d t$ . Show the work that leads to your answer. Using correct units, interpret the meaning of this value in the context of the problem.

(c) A hose adds water at a constant 1.5 liters per minute for the whole 10 minutes, so the net rate becomes $R (t) + 1.5$ . Find $\int_{0}^{10} (R (t) + 1.5) d t$ , and identify the property of definite integrals that you use.

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

Model answer

(a) As $t$ approaches 4 from the left, $R (t)$ approaches 8, but $R (4) = - 4$ , so $R$ has a jump discontinuity at $t = 4$ . On each side of the jump $R$ is a line. The definite integral extends to a function with a jump discontinuity: integrate each side and add, $\int_{0}^{10} R (t) d t = \int_{0}^{4} R (t) d t + \int_{4}^{10} R (t) d t$ .

(b) $\int_{0}^{4} 2 t d t$ is a triangle with base 4 and height 8, so it is 16. On $[4, 10]$ , $t - 8 = 0$ at $t = 8$ : from 4 to 8 a triangle below the axis with base 4 and height 4 gives $- 8$ , and from 8 to 10 a triangle above with base 2 and height 2 gives 2. So $\int_{0}^{10} R (t) d t = 16 + (- 8) + 2 = 10$ liters. Over the 10 minutes, the amount of water in the tank increased by 10 liters.

(c) By the sum property, $\int_{0}^{10} (R (t) + 1.5) d t = \int_{0}^{10} R (t) d t + \int_{0}^{10} 1.5 d t = 10 + 1.5 (10) = 25$ liters.

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

(a) Identifies the jump at $t = 4$ and says the definite integral extends to a function with a jump discontinuity, integrated on each side and added.
(b) Splits the integral at $t = 4$ and gets 16 for $\int_{0}^{4} R (t) d t$ .
(b) Counts the triangle from 4 to 8 as $- 8$ and the one from 8 to 10 as 2, and gets 10.
(b) Says the amount of water in the tank increased by 10 liters over the 10 minutes.
(c) Identifies the sum property, uses it with $\int_{0}^{10} 1.5 d t = 15$ , and gets 25 liters.

Part 6 Check yourself Integrals from a table of known values

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

The functions $f$ and $g$ are continuous on $[0, 7]$ . Their graphs are not given, but the values of three definite integrals are.

Known definite integrals
Integral	Value
$\int_{0}^{3} f (x) d x$	4
$\int_{0}^{7} f (x) d x$	$- 2$
$\int_{3}^{7} g (x) d x$	5

Case question 1 · Calculate

No calculator. What is $\int_{3}^{7} f (x) d x$ ?

Your answer

Case question 2 · Calculate

No calculator. What is $\int_{7}^{3} (2 f (x) - g (x)) d x$ ?

Your answer

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

Check yourself · Fill the statement

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

For $0 \leq x \leq 4$ , the region between the graph of $y = - \sqrt{16 - x^{2}}$ and the $x$ -axis is a of radius 4 that lies below the axis, so $\int_{0}^{4} - \sqrt{16 - x^{2}} d x$ equals .

Check yourself · Multiple choice

No calculator. The function $f$ is continuous on $[0, 10]$ . Which expression is equal to $\int_{2}^{5} f (x) d x$ for every such $f$ ?

Part 7 Keep it Bring back earlier ideas

One question from earlier in this unit and one from Unit 1, whose kinds of discontinuity this lesson's last idea leans on.

Bring it back · Calculate

No calculator. Let $g (x) = \int_{1}^{x} (t^{2} + 5) d t$ . What is $g^{'} (2)$ ?

Your answer

Bring it back · Multiple choice

No calculator. Let $k (x) = \frac{x^{2} + x - 6}{x - 2}$ for $x \neq 2$ , and let $k (2) = 1$ . What kind of discontinuity does $k$ have at $x = 2$ ?

In three lines

A definite integral is a signed area: area above the axis counts as positive and area below counts as negative.
Constants factor out, sums and differences split, reversed limits change the sign, and integrals over adjacent intervals add.
A removable or jump discontinuity does not stop a definite integral: one point adds no area, and a jump splits the interval.

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 you build an integral from pieces, write each piece as an integral with its limits and its signed value, such as $\int_{3}^{5} f (x) d x = - 4$ , before you add or multiply. The property you used is then visible in your work, and a sign slip is easy to find.

Questions students ask

Is a definite integral the same as the area under the curve?

Only where the graph is on or above the $x$ -axis. Below the axis the definite integral counts area as negative. For total area, add the areas of the pieces as positive numbers.

Why does reversing the limits change the sign?

In a Riemann sum each width $Δ x$ is the change in $x$ from the start of a piece to its end. Going from $b$ down to $a$ makes every width negative, so every product in the sum, and the sum itself, changes sign.

Can I use these properties when I do not know a formula for the function?

Yes, and that is often the point. Given the values of a few integrals, the constant multiple, sum, reversal and adjacent-interval properties give the values of others without any formula for the function.

Does the value of a function at a jump change the integral?

No. A single input has no width, so its value adds no area. Integrate each side of the jump as its own function and add the two pieces.

Do I need to memorize area formulas for this?

You need the ones for rectangles, triangles, trapezoids and circles, including half and quarter circles. A graph such as $y = \sqrt{r^{2} - x^{2}}$ is the top half of a circle of radius $r$ centered at the origin.

Course alignment, for teachers

AP Calculus AB topic 6.6, 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.