Properties of definite integrals — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Zero interval rule, reversal of limits rule, constant multiple and sum/difference rules, interval additivity, odd/even function symmetry, comparison properties, and the Mean Value Theorem for Definite Integrals for exam application.

You should already know: Definition of the definite integral as a limit of Riemann sums. The Fundamental Theorem of Calculus. Basic identification of even and odd functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Properties of definite integrals?

Properties of definite integrals are a set of algebraic and geometric rules for manipulating and simplifying definite integrals, part of Unit 6: Integration and Accumulation of Change, which accounts for 10-15% of the AP Calculus BC exam weight per the official College Board CED. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. The definite integral ${\int }_a^{b}f(x)dx$ is defined as the net signed area between $f(x)$ and the $x$-axis from $x=a$ to $x=b$, and all these properties follow directly from the Riemann sum definition of the integral. Instead of computing full Riemann sums or antiderivatives for every problem, these rules let you quickly evaluate, bound, or rearrange integral expressions to solve for unknown values. These properties are foundational, as every subsequent integration topic (from substitution to volume applications) relies on correct use of these rules.

2. Basic Algebraic Properties of Definite Integrals

All basic algebraic properties of definite integrals follow directly from the limit definition of the Riemann sum, as they describe how the limit behaves when we adjust bounds or combine integrands. The core rules are:

1. Zero interval rule: If the upper and lower bounds are equal, the interval has width zero, so ${\int }_a^{a}f(x)dx=0$ for any integrable $f(x)$. Geometrically, there is no area over a single point.
2. Reversal of limits rule: Swapping the upper and lower bounds flips the sign of every subinterval width in the Riemann sum, so ${\int }_b^{a}f(x)dx=-{\int }_a^{b}f(x)dx$.
3. Constant multiple rule: Constants factor out of integrals because limits pull out constants: ${\int }_a^{b}kf(x)dx=k{\int }_a^{b}f(x)dx$ for any constant $k$.
4. Sum/difference rule: The integral of a sum is the sum of the integrals: ${\int }_a^b\left[f(x)\pm g(x)\right]dx={\int }_a^{b}f(x)dx\pm {\int }_a^{b}g(x)dx$.
5. Additivity over intervals: For any three real numbers $a,b,c$, ${\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx={\int }_a^{b}f(x)dx$, regardless of where $c$ falls relative to $a$ and $b$. This is the most frequently used rule for finding unknown integrals from given values.

Worked Example

Given that ${\int }_2^{6}f(x)dx=8$ and ${\int }_4^{2}f(x)dx=3$, find the value of ${\int }_4^{6}5f(x)dx$.

1. First apply the reversal of limits rule to the second given integral: ${\int }_2^{4}f(x)dx=-{\int }_4^{2}f(x)dx=-3$.
2. Use additivity to split the known integral from 2 to 6: ${\int }_2^{6}f(x)dx={\int }_2^{4}f(x)dx+{\int }_4^{6}f(x)dx=8$.
3. Substitute the known value of ${\int }_2^{4}f(x)dx$: $-3+{\int }_4^{6}f(x)dx=8$, so ${\int }_4^{6}f(x)dx=11$.
4. Apply the constant multiple rule: ${\int }_4^{6}5f(x)dx=5\times 11=55$.

Exam tip: If you are given multiple integral values and asked for an unknown, always write down the additivity rule explicitly first, mapping which bounds you need to combine, to avoid sign errors from limit reversal.

3. Symmetry Properties for Even and Odd Functions

Symmetry properties allow you to evaluate or simplify integrals over symmetric intervals (centered at $x=0$, of the form $[-a,a]$) without computing full antiderivatives, which saves significant time on AP multiple-choice questions. Recall that an even function satisfies $f(-x)=f(x)$ (symmetric across the y-axis) and an odd function satisfies $f(-x)=-f(x)$ (symmetric about the origin).

To derive the rule for odd functions, split the integral: ${\int }_{-a}^{a}f(x)dx={\int }_{-a}^{0}f(x)dx+{\int }_0^{a}f(x)dx$. Substitute $u=-x$ into the first integral: ${\int }_{-a}^{0}f(x)dx={\int }_a^{0}f(-u)(-du)={\int }_0^a-f(u)du=-{\int }_0^{a}f(u)du$. Adding to the second integral gives $-{\int }_0^{a}f(u)du+{\int }_0^{a}f(x)dx=0$. For even functions, the same substitution gives ${\int }_{-a}^{0}f(x)dx={\int }_0^{a}f(u)du$, so the total integral is $2{\int }_0^{a}f(x)dx$. The final rules are: $$ \text{Odd } f: \int_{-a}^a f(x) dx = 0 \quad \quad \text{Even } f: \int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx $$

Worked Example

Evaluate ${\int }_{-3}^3\left(5x^7-4x^3+2x-6\sin x\right)dx$ without integrating term-by-term.

1. Check the parity (odd/even nature) of each term: any power $x^n$ is odd if $n$ is odd, and $\sin (kx)$ is always odd for constant $k$.
2. Each term in the integrand is odd: $5x^7$ (odd), $-4x^3$ (odd), $2x$ (odd), $-6\sin x$ (odd).
3. The sum of any number of odd functions is also odd, so the entire integrand is an odd function.
4. The interval $[-3,3]$ is symmetric around 0, so by the odd function symmetry rule, ${\int }_{-3}^3\left(5x^7-4x^3+2x-6\sin x\right)dx=0$.

Exam tip: Always check for symmetry before expanding or integrating a polynomial or trigonometric function over a symmetric interval — the answer is often 0 for an odd integrand, saving you 2+ minutes of unnecessary computation.

4. Comparison Properties and the Mean Value Theorem for Integrals

Comparison properties let you bound the value of a definite integral without computing it exactly, a common question type on AP MCQ. The core comparison rules are:

1. If $f(x)\ge g(x)$ for all $x\in [a,b]$ and $a<b$, then ${\int }_a^{b}f(x)dx\ge {\int }_a^{b}g(x)dx$.
2. If $m\le f(x)\le M$ for all $x\in [a,b]$ and $a<b$, then $m(b-a)\le {\int }_a^{b}f(x)dx\le M(b-a)$.

From the second rule, we get the Mean Value Theorem (MVT) for Integrals: if $f(x)$ is continuous on $[a,b]$, then there exists at least one point $c\in [a,b]$ such that: $$ f(c) = \frac{1}{b-a} \int_a^b f(x) dx $$ The value $f(c)$ is the average value of $f(x)$ on $[a,b]$. Geometrically, this means the net area under $f(x)$ equals the area of a rectangle with height $f(c)$ and width $(b-a)$.

Worked Example

For $f(x)=e^{-x^2}$ on $[0,2]$, find the value of $c$ guaranteed by the MVT for Integrals, rounded to one decimal place. Use the approximation ${\int }_0^2{e}^{-x^2}dx\approx 0.882$.

1. By the MVT for Integrals, $f(c)=\frac{1}{2-0}{\int }_0^2{e}^{-x^2}dx$.
2. Substitute the given approximation for the integral: $f(c)=\frac{0.882}{2}=0.441$.
3. Set $f(c)=e^{-c^2}=0.441$, take the natural logarithm of both sides: $-c^2=\ln (0.441)\approx -0.818$.
4. Solve for $c$ (which is positive in $[0,2]$): $c=\sqrt{0.818}\approx 0.9$, which falls in the interval $[0,2]$ as required.

Exam tip: When asked for the average value of a function on an interval, never forget to divide the integral by $(b-a)$ — this is one of the most commonly missed points on AP FRQ questions.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming additivity only works when $c$ is between $a$ and $b$, so ${\int }_2^{6}f(x)dx={\int }_2^{8}f(x)dx+{\int }_8^{6}f(x)dx$ is written as invalid. Why: Students memorize additivity from geometric examples where $c$ is between the bounds, but the rule holds for any order of $a,b,c$. Correct move: If the upper bound of the first integral matches the lower bound of the second integral, additivity holds automatically, and the sign will be correct after applying the reversal rule.
• Wrong move: Assuming a polynomial with leading odd degree is entirely odd, ignoring constant or even-degree terms, so ${\int }_{-2}^2(x^3+3x^2)dx=0$ is claimed incorrectly. Why: Students check only the leading term for parity instead of all terms. Correct move: Check every term in the integrand separately for parity before applying the symmetry rule.
• Wrong move: Forgetting to change the sign when reversing integral limits, e.g., writing ${\int }_5^{2}f(x)dx={\int }_2^{5}f(x)dx$. Why: Students associate integrals with positive area and forget definite integrals measure net signed area. Correct move: Every time you write an integral with the upper bound smaller than the lower bound, immediately add a negative sign when reversing the limits.
• Wrong move: Separating the integral of a product into the product of integrals, e.g., ${\int }_0^{2}x{e}^{x}dx=\left({\int }_0^{2}xdx\right)\left({\int }_0^2{e}^{x}dx\right)$. Why: Students incorrectly generalize the constant multiple and sum rules to products. Correct move: Memorize that the integral of a product is not the product of integrals; use integration by parts or substitution to evaluate products instead of this shortcut.
• Wrong move: Claiming that if $f(x)>0$ on $(a,b)$, then ${\int }_a^{b}f(x)dx>0$ even when $b<a$. Why: Students remember the comparison rule for $a<b$ and forget reversing bounds flips the sign. Correct move: Always confirm $a<b$ before applying comparison properties; if $b<a$, reverse the bounds and flip the inequality sign.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Given that ${\int }_2^{5}f(x)dx=7$ and ${\int }_1^{5}f(x)dx=11$, what is the value of ${\int }_1^{2}3f(x)dx$? A) $4$ B) $12$ C) $18$ D) $54$

Worked Solution: Use the additivity rule to rearrange the given integrals: ${\int }_1^{5}f(x)dx={\int }_1^{2}f(x)dx+{\int }_2^{5}f(x)dx$. Substitute the known values: $11={\int }_1^{2}f(x)dx+7$, so ${\int }_1^{2}f(x)dx=4$. Next apply the constant multiple rule: ${\int }_1^{2}3f(x)dx=3\times 4=12$. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=3x^5+2x^3-4x\cos x$. (a) Show that $f(x)$ is an odd function. (b) Use the symmetry property of definite integrals to simplify ${\int }_{-2\pi }^{2\pi }f(x)dx$. Do not integrate term-by-term. (c) Given that ${\int }_0^{2\pi }f(x)dx=k$, write ${\int }_{-2\pi }^{0}f(x)dx$ in terms of $k$.

Worked Solution: (a) To confirm $f(x)$ is odd, verify $f(-x)=-f(x)$: $$ \begin{align*} f(-x) &= 3(-x)^5 + 2(-x)^3 - 4(-x)\cos(-x) \ &= -3x^5 - 2x^3 + 4x \cos x = -\left(3x^5 + 2x^3 - 4x \cos x\right) = -f(x) \end{align*} $$ This satisfies the definition of an odd function.

(b) The interval $[-2\pi ,2\pi ]$ is symmetric about the origin, and $f(x)$ is odd. By the symmetry rule for odd functions, ${\int }_{-2\pi }^{2\pi }f(x)dx=0$.

(c) Split the full integral: ${\int }_{-2\pi }^{2\pi }f(x)dx={\int }_{-2\pi }^{0}f(x)dx+{\int }_0^{2\pi }f(x)dx=0$. Rearranging gives ${\int }_{-2\pi }^{0}f(x)dx=-{\int }_0^{2\pi }f(x)dx=-k$.

Question 3 (Application / Real-World Style)

The power $P(t)$, measured in kilowatts (kW), drawn by a factory over a 24-hour period starting at midnight ($t=0$) is given by $P(t)=2t+10\sin \left(\frac{\pi t}{12}\right)$, where $t$ is hours after midnight. Total energy consumed $E$ is given by $E={\int }_0^{24}P(t)dt$. Use properties of definite integrals to find $E$, in kilowatt-hours (kWh).

Worked Solution: Split the integral using the sum rule: $E={\int }_0^{24}2tdt+{\int }_0^{24}10\sin \left(\frac{\pi t}{12}\right)dt$. For the second integral, substitute $u=t-12$, shifting the interval to $[-12,12]$, giving: $$ 10 \int_{-12}^{12} \sin\left(\frac{\pi(u+12)}{12}\right) du = 10 \int_{-12}^{12} \sin\left(\frac{\pi u}{12} + \pi\right) du = -10 \int_{-12}^{12} \sin\left(\frac{\pi u}{12}\right) du $$ $\sin \left(\frac{\pi u}{12}\right)$ is odd, so the integral over the symmetric interval $[-12,12]$ is 0, so the entire second term equals 0. Evaluate the first term: ${\int }_0^{24}2tdt=[t^2{]}_0^{24}=576$. The factory consumes 576 kWh of energy over the 24-hour period.

7. Quick Reference Cheatsheet

8. What's Next

Mastering properties of definite integrals is an essential prerequisite for every integration technique that comes next in Unit 6. Immediately after this topic, you will learn u-substitution for definite integrals, where you will need to correctly adjust bounds and apply reversal and additivity rules to simplify substituted integrals. This topic also forms the foundation for later topics including integration by parts, improper integrals, and applications of integration such as finding net area, volume, and average value of functions over an interval. Without solid command of the sign rules and additivity, you will make frequent avoidable errors when manipulating integrals in these future topics, especially in FRQ questions where multiple steps of integral rearrangement are required.

U-Substitution for Definite Integrals Integration by Parts Average Value of a Function Improper Integrals