AP Calculus AB

## Learning Outcomes

- Apply constant multiple, sum, and difference rules for definite integrals.
- Use additivity over adjacent intervals to split or combine definite integrals.
- Reverse limits of integration correctly and account for the sign change.
- Interpret definite integrals as signed accumulation, not always ordinary area.
- Evaluate AP-style integral expressions using given integral values, graphs, or geometry.
- Avoid common sign errors when intervals are reversed, split, or below the $$x$$-axis.
- Connect definite integrals of rates to net change in a quantity.

## AP Calculus AB Syllabus

In AP Calculus AB Unit 6, definite integrals are studied as limits of Riemann sums, signed accumulation over intervals, and tools for net change. This subtopic focuses on using properties of definite integrals without always finding an antiderivative. These skills appear in both calculator and no-calculator questions, often with functions given algebraically, graphically, numerically, or in context.

- Use the constant multiple rule: $$\int_a^b kf(x)\,dx=k\int_a^b f(x)\,dx$$.
- Use sum and difference rules for integrable functions.
- Split an integral across adjacent intervals.
- Combine integrals over adjacent intervals into one integral.
- Use $$\int_a^a f(x)\,dx=0$$.
- Use $$\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx$$.
- Interpret positive and negative contributions as signed area.
- Use definite integrals of rates to find accumulated change.

## Test Your Knowledge

Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision.

1. If $$\int_1^5 f(x)\,dx=12$$ and $$\int_1^3 f(x)\,dx=7$$, what is $$\int_3^5 f(x)\,dx$$?
2. If $$\int_0^4 f(x)\,dx=-2$$ and $$\int_0^4 g(x)\,dx=5$$, what is $$\int_0^4 [3f(x)-2g(x)]\,dx$$?
3. Why is $$\int_6^2 f(x)\,dx$$ the negative of $$\int_2^6 f(x)\,dx$$?
4. A graph lies below the $$x$$-axis from $$x=0$$ to $$x=3$$ and forms a triangle with base $$3$$ and height $$4$$. What is the value of the definite integral over that interval?
5. In a motion problem, what does $$\int_a^b v(t)\,dt$$ represent, and how is it different from total distance traveled?

## Introduction

Definite integrals are not only evaluated by finding antiderivatives. On the AP Calculus AB exam, you are often expected to simplify integral expressions using properties: pulling out constants, distributing integrals across sums and differences, reversing limits, and splitting intervals. These rules are especially useful when the problem gives selected integral values, a graph, or a table rather than a formula with an easy antiderivative.

The central idea is that a definite integral measures signed accumulation over an interval. When the graph of the integrand is above the $$x$$-axis, the contribution is positive; when it is below the $$x$$-axis, the contribution is negative. The linearity and interval properties let you reorganize that signed accumulation while preserving its value.

> **Key Term: definite integral**
> A number written in the form $$\int_a^b f(x)\,dx$$ that represents the signed accumulation of $$f$$ from $$x=a$$ to $$x=b$$.

### Definite integrals as signed accumulation

A definite integral comes from the limiting process of a Riemann sum: add many products of the form “function value times small interval width,” then take the limit as the widths become very small. In symbols, one common form is

$$
\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{k=1}^n f(a+k\Delta x)\Delta x,\qquad \Delta x=\frac{b-a}{n}
$$

For this subtopic, you do not usually need to build the Riemann sum. You do need to remember what the integral means: signed accumulation.

If $$f(x)\ge 0$$ on $$[a,b]$$, then $$\int_a^b f(x)\,dx$$ equals the usual area between the graph and the $$x$$-axis. If $$f(x)\le 0$$ on $$[a,b]$$, then the integral is the negative of that area. If the graph crosses the axis, areas above and below the axis combine with signs.

> **Key Term: signed area**
> Area above the $$x$$-axis counted as positive and area below the $$x$$-axis counted as negative in a definite integral.

For example, suppose the graph of $$f$$ forms a triangle above the $$x$$-axis from $$x=0$$ to $$x=2$$ with base $$2$$ and height $$5$$. Then

$$
\int_0^2 f(x)\,dx=\frac12(2)(5)=5
$$

But if the same triangle lies below the $$x$$-axis, then

$$
\int_0^2 f(x)\,dx=-5
$$

This sign convention is one of the most common AP traps. A definite integral is not automatically “area” in the everyday positive sense. It is signed accumulation.

> **Test Tip:**
> When using a graph, label each region as positive or negative before adding. If a region is below the $$x$$-axis, its contribution to the definite integral is negative.

The expression $$\int_a^b f(x)\,dx$$ has several parts. The function $$f(x)$$ is the integrand. The numbers $$a$$ and $$b$$ are the limits of integration, with $$a$$ the lower limit and $$b$$ the upper limit.

> **Key Term: integrand**
> The function being integrated in a definite integral, such as $$f(x)$$ in $$\int_a^b f(x)\,dx$$.
>
> **Key Term: limits of integration**
> The lower and upper values that define the interval of integration in a definite integral.

### Linearity: constants, sums, and differences

Linearity is the property that lets you handle constants, sums, and differences inside definite integrals. If $$f$$ and $$g$$ are integrable on $$[a,b]$$ and $$c$$ and $$d$$ are constants, then

$$
\int_a^b [cf(x)+dg(x)]\,dx=c\int_a^b f(x)\,dx+d\int_a^b g(x)\,dx
$$

This single rule includes three common rules:

$$
\int_a^b kf(x)\,dx=k\int_a^b f(x)\,dx
\int_a^b [f(x)+g(x)]\,dx=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx
\int_a^b [f(x)-g(x)]\,dx=\int_a^b f(x)\,dx-\int_a^b g(x)\,dx
$$

> **Key Term: linearity**
> The property that definite integrals distribute over sums and differences and allow constant multiples to be factored out.

Linearity is often tested in questions that provide integral values rather than formulas.

> **Question:**
> Suppose
>
> $$\int_0^4 f(x)\,dx=6,\qquad \int_0^4 g(x)\,dx=-3$$
>
> Find
>
> $$\int_0^4 [2f(x)-5g(x)]\,dx$$

Using linearity,

$$
\int_0^4 [2f(x)-5g(x)]\,dx=2\int_0^4 f(x)\,dx-5\int_0^4 g(x)\,dx
$$

Substitute the given values:

$$
2(6)-5(-3)=12+15=27
$$

So

$$
\int_0^4 [2f(x)-5g(x)]\,dx=27
$$

Notice that you did not need to know formulas for $$f$$ or $$g$$. This is exactly why the property is useful.

Linearity also applies when one term is a constant function. For example,

$$
\int_1^5 7\,dx=7(5-1)=28
$$

So if

$$
\int_1^5 f(x)\,dx=10
$$

then

$$
\int_1^5 [f(x)+7]\,dx=10+28=38
$$

A common mistake is to treat $$\int_a^b k\,dx$$ as just $$k$$. The interval width matters. The constant function $$y=k$$ forms a rectangle of height $$k$$ and width $$b-a$$.

> **Exam Warning:**
> Do not write $$\int_a^b k\,dx=k$$. The correct value is $$k(b-a)$$.

Linearity requires that the intervals match. You can combine

$$
\int_0^3 f(x)\,dx+\int_0^3 g(x)\,dx
$$

as

$$
\int_0^3 [f(x)+g(x)]\,dx
$$

because the limits are the same. But you cannot directly combine

$$
\int_0^3 f(x)\,dx+\int_1^4 g(x)\,dx
$$

into one integral of $$f+g$$ over a single interval without further information.

### Additivity over intervals and changing limits

Additivity over intervals says that an integral over a larger interval can be split into integrals over smaller adjacent intervals. If $$a<c<b$$, then

$$
\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx
$$

Equivalently,

$$
\int_a^c f(x)\,dx+\int_c^b f(x)\,dx=\int_a^b f(x)\,dx
$$

> **Key Term: additivity over intervals**
> The property that allows a definite integral over an interval to be split into, or built from, definite integrals over adjacent intervals.

This is heavily tested because it allows you to find missing interval values.

> **Question:**
> Suppose
>
> $$\int_1^8 f(x)\,dx=20,\qquad \int_1^5 f(x)\,dx=12$$
>
> Find
>
> $$\int_5^8 f(x)\,dx$$

By additivity,

$$
\int_1^8 f(x)\,dx=\int_1^5 f(x)\,dx+\int_5^8 f(x)\,dx
$$

So

$$
20=12+\int_5^8 f(x)\,dx
$$

Therefore,

$$
\int_5^8 f(x)\,dx=8
$$

Two other properties work closely with additivity:

$$
\int_a^a f(x)\,dx=0
$$

and

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

The first says that an interval with zero width has zero accumulation. The second says that reversing the direction of integration changes the sign.

> **Key Term: reversed limits**
> Limits of integration written in the opposite direction, causing the definite integral to change sign.

For example, if

$$
\int_2^7 f(x)\,dx=9
$$

then

$$
\int_7^2 f(x)\,dx=-9
$$

This sign change is not optional. It follows from the orientation of the interval.

Additivity can also be used when the limits are not written in increasing order, but then you must handle signs carefully.

> **Question:**
> Given
>
> $$\int_0^2 f(x)\,dx=4,\qquad \int_2^6 f(x)\,dx=-7$$
>
> Find
>
> $$\int_6^0 f(x)\,dx$$

First combine in the increasing direction:

$$
\int_0^6 f(x)\,dx=\int_0^2 f(x)\,dx+\int_2^6 f(x)\,dx=4+(-7)=-3
$$

Now reverse the limits:

$$
\int_6^0 f(x)\,dx=-\int_0^6 f(x)\,dx=-(-3)=3
$$

So

$$
\int_6^0 f(x)\,dx=3
$$

> **Test Tip:**
> If the requested integral has reversed limits, it is often safer to first compute the integral in left-to-right order, then change the sign at the end.

Additivity is also useful with piecewise functions. Suppose

$$
f(x)=
\begin{cases}
2x, & 0\le x\le 1\\
3, & 1<x\le 4
\end{cases}
$$

Then

$$
\int_0^4 f(x)\,dx=\int_0^1 2x\,dx+\int_1^4 3\,dx
$$

Evaluate each part:

$$
\int_0^1 2x\,dx=\left.x^2\right|_0^1=1
$$

and

$$
\int_1^4 3\,dx=3(4-1)=9
$$

Therefore,

$$
\int_0^4 f(x)\,dx=10
$$

This method is common when a function changes formula, sign, or geometric shape at a point.

### Combining properties in AP-style problems

AP Calculus AB questions often combine linearity, additivity, reversed limits, and signed area in one problem. You need a clear sequence of steps rather than a memorized trick.

Start by checking:

1. Are the limits the same? If so, linearity may apply directly.
2. Are the intervals adjacent? If so, additivity may apply.
3. Are any limits reversed? If so, change the sign.
4. Is the graph above or below the axis? If using geometry, assign signs.
5. Is the integrand a rate? If so, interpret the integral as net change.

Here is a typical symbolic example.

> **Question:**
> Suppose
>
> $$\int_0^3 f(x)\,dx=5,\qquad \int_3^7 f(x)\,dx=-2,\qquad \int_0^7 g(x)\,dx=4$$
>
> Find
>
> $$\int_7^0 [2f(x)+g(x)]\,dx$$

First handle the reversed limits:

$$
\int_7^0 [2f(x)+g(x)]\,dx=-\int_0^7 [2f(x)+g(x)]\,dx
$$

Use linearity:

$$
-\int_0^7 [2f(x)+g(x)]\,dx=-\left(2\int_0^7 f(x)\,dx+\int_0^7 g(x)\,dx\right)
$$

Find $$\int_0^7 f(x)\,dx$$ using additivity:

$$
\int_0^7 f(x)\,dx=\int_0^3 f(x)\,dx+\int_3^7 f(x)\,dx=5+(-2)=3
$$

Now substitute:

$$
-\left(2(3)+4\right)=-10
$$

So

$$
\int_7^0 [2f(x)+g(x)]\,dx=-10
$$

This problem tests several basic properties at once. The arithmetic is simple, but the order matters.

Definite integrals also appear in context. If $$v(t)$$ is velocity, then

$$
\int_a^b v(t)\,dt
$$

is displacement, or net change in position. If velocity is sometimes positive and sometimes negative, the positive and negative parts can cancel. Total distance traveled is different:

$$
\int_a^b |v(t)|\,dt
$$

That absolute value removes the sign of velocity and counts all motion as positive distance.

> **Key Term: net change**
> The total signed change in a quantity over an interval, often found by integrating a rate of change.

For example, if a particle has velocity $$v(t)$$ and

$$
\int_0^5 v(t)\,dt=-8
$$

then its position changed by $$-8$$ units over the interval. That does not necessarily mean it traveled $$8$$ units total; it may have moved back and forth. To find total distance, you would need to integrate speed, $$|v(t)|$$, often splitting the interval where velocity changes sign.

> **Exam Warning:**
> In motion problems, do not confuse displacement with total distance. $$\int_a^b v(t)\,dt$$ gives net change in position; $$\int_a^b |v(t)|\,dt$$ gives total distance traveled.

Graphs are another common setting. Suppose the graph of $$f$$ consists of:

- a semicircle above the $$x$$-axis from $$x=-2$$ to $$x=2$$ with radius $$2$$;
- a triangle below the $$x$$-axis from $$x=2$$ to $$x=5$$ with base $$3$$ and height $$4$$.

Then

$$
\int_{-2}^2 f(x)\,dx=\frac12\pi(2)^2=2\pi
$$

and

$$
\int_2^5 f(x)\,dx=-\frac12(3)(4)=-6
$$

By additivity,

$$
\int_{-2}^5 f(x)\,dx=2\pi-6
$$

If the question instead asks for

$$
\int_5^{-2} f(x)\,dx
$$

then reverse the sign:

$$
\int_5^{-2} f(x)\,dx=-(2\pi-6)=6-2\pi
$$

A final useful idea is that these properties apply to integrable functions. On the AP Calculus AB exam, most functions you meet in this setting are continuous, piecewise continuous, or clearly defined by graphs or tables. Continuity is enough to guarantee integrability, and bounded piecewise continuous functions can also be integrated.

> **Key Term: integrable**
> A function for which the definite integral over the interval exists.

## Key Point Checklist

This article has covered the following key knowledge points:

- A definite integral represents signed accumulation over an interval.
- Area above the $$x$$-axis contributes positively to a definite integral.
- Area below the $$x$$-axis contributes negatively to a definite integral.
- Linearity allows constants to be factored out of definite integrals.
- Linearity allows integrals of sums and differences to be separated.
- The constant function rule is $$\int_a^b k\,dx=k(b-a)$$.
- Additivity over intervals allows $$\int_a^b f(x)\,dx$$ to be split at an interior point.
- Reversing limits changes the sign of a definite integral.
- An integral over a zero-width interval equals zero.
- Graph-based integral questions often require geometry plus signed area.
- In context, integrating a rate gives net change.
- In motion, displacement and total distance are different integral quantities.

## Key Terms and Concepts

- definite integral
- signed area
- integrand
- limits of integration
- linearity
- additivity over intervals
- reversed limits
- net change
- integrable


Applying properties of definite integrals - Linearity, additivity over intervals