Properties of definite integrals

AP Calculus BC· AP Calculus BC CED — Integration and Accumulation of Change· 14 min read

1. Basic Algebraic Properties of Definite Integrals★★☆☆☆⏱ 4 min

All basic algebraic properties of definite integrals follow directly from the limit definition of the Riemann sum. These rules let you rearrange, simplify, and solve for unknown integral values without computing full Riemann sums or antiderivatives.

Zero interval rule: If upper and lower bounds are equal, $\int_{a}^{a} f (x) d x = 0$ for any integrable $f (x)$ . Geometrically, there is no area over a single point.
Reversal of limits rule: Swapping upper and lower bounds flips the sign of the integral: $\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x$ .
Constant multiple rule: Constants factor out of integrals: $\int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x$ for any constant $k$ .
Sum/difference rule: The integral of a sum/difference equals the sum/difference of integrals: $\int_{a}^{b} [f (x) \pm g (x)] d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x$ .
Additivity over intervals: For any three real numbers $a, b, c$ , $\int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x = \int_{a}^{b} f (x) d x$ , regardless of the order of $a, b, c$ .

📐 Worked Example

Given that $\int_{2}^{6} f (x) d x = 8$ and $\int_{4}^{2} f (x) d x = 3$ , find the value of $\int_{4}^{6} 5 f (x) d x$ .

1
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 and solve for $\int_{4}^{6} f (x) d x$ :
$-3 + \int_4^6 f(x) dx = 8 \implies \int_4^6 f(x) dx = 11$
4
Apply the constant multiple rule to get the final result:
$\int_4^6 5 f(x) dx = 5 \times 11 = 55$

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.

2. Symmetry Properties for Even and Odd Functions★★☆☆☆⏱ 4 min

Symmetry properties let you simplify and evaluate integrals over symmetric intervals centered at $x = 0$ , i.e., $[- a, a]$ , without computing full antiderivatives, saving significant time on 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).

🔬 Derivation

Goal:

Derive symmetry rules for integrals of odd/even functions over $[- a, a]$

Starting from:

Split the integral into two halves: $\int_{- a}^{a} f (x) d x = \int_{- a}^{0} f (x) d x + \int_{0}^{a} f (x) d x$

1
For an odd function, substitute $u = - x$ into the first integral: $d u = - d x$ , when $x = - a$ , $u = a$ ; when $x = 0$ , $u = 0$
$\int_{-a}^0 f(x) dx = \int_a^0 f(-u) (-du) = \int_0^a -f(u) du = - \int_0^a f(u) du$
2
Add the two integrals to get the total for odd $f$ :
$- \int_0^a f(u) du + \int_0^a f(x) dx = 0$
3
For an even function, the same substitution gives $f (- u) = f (u)$ , so the first integral becomes $\int_{0}^{a} f (u) d u$ , so total integral is:
$\int_0^a f(u) du + \int_0^a f(x) dx = 2 \int_0^a f(x) dx$

Result:

The final symmetry rules are: for odd $f$ , $\int_{- a}^{a} f (x) d x = 0$ ; for even $f$ , $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ .

📐 Worked Example

Evaluate $\int_{- 3}^{3} (5 x^{7} - 4 x^{3} + 2 x - 6 sin x) d x$ without integrating term-by-term.

1
Check parity of each term: any power $x^{n}$ is odd if $n$ is odd, and $sin (k x)$ is always odd for constant $k$ .
2
All terms in the integrand are odd: $5 x^{7}$ (odd), $- 4 x^{3}$ (odd), $2 x$ (odd), $- 6 sin x$ (odd). The sum of odd functions is odd, so the entire integrand is odd.
3
The interval $[- 3, 3]$ is symmetric around 0, so by the odd function symmetry rule, the integral equals 0.
$\int_{-3}^3 \left( 5x^7 - 4x^3 + 2x - 6 \sin x \right) dx = 0$

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.

3. Comparison Properties and the Mean Value Theorem for Integrals★★★☆☆⏱ 3 min

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

If $f (x) \geq g (x)$ for all $x \in [a, b]$ and $a < b$ , then $\int_{a}^{b} f (x) d x \geq \int_{a}^{b} g (x) d x$ .
If $m \leq f (x) \leq M$ for all $x \in [a, b]$ and $a < b$ , then $m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)$ .

📘 Definition

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 $f (c) = \frac{1}{b - a} \int_{a}^{b} f (x) d x$ . The value $f (c)$ is the average value of $f (x)$ on $[a, b]$ , and 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 $\int_{0}^{2} e^{- x^{2}} d x \approx 0.882$ .

1
By the MVT for Integrals, $f (c)$ equals the average value, which is:
$f(c) = \frac{1}{b-a} \int_a^b f(x) dx = \frac{1}{2-0} \int_0^2 e^{-x^2} dx$
2
Substitute the given approximation for the integral:
$f(c) = \frac{0.882}{2} = 0.441$
3
Set $f (c) = e^{- c^{2}} = 0.441$ , then take the natural logarithm of both sides:
$-c^2 = \ln(0.441) \approx -0.818$
4
Solve for positive $c$ (since $c$ must be in $[0, 2]$ ):
$c = \sqrt{0.818} \approx 0.9$

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.

4. AP-Style Worked Practice Problems★★★☆☆⏱ 3 min

📐 Worked Example

Given that $\int_{2}^{5} f (x) d x = 7$ and $\int_{1}^{5} f (x) d x = 11$ , what is the value of $\int_{1}^{2} 3 f (x) d x$ ?
Options: A) 4, B) 12, C) 18, D) 54

1
Use additivity to split the integral over $[1, 5]$ :
$\int_1^5 f(x) dx = \int_1^2 f(x) dx + \int_2^5 f(x) dx$
2
Substitute the known values and solve for $\int_{1}^{2} f (x) d x$ :
$11 = \int_1^2 f(x) dx + 7 \implies \int_1^2 f(x) dx = 4$
3
Apply the constant multiple rule to get the final result:
$\int_1^2 3 f(x) dx = 3 \times 4 = 12$
4
The correct answer is option B.

📐 Worked Example

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 $P (t) = 2 t + 10 sin (\frac{π t}{12})$ , where $t$ is hours after midnight. Total energy consumed $E$ is given by $E = \int_{0}^{24} P (t) d t$ . Use properties of definite integrals to find $E$ , in kilowatt-hours (kWh).

1
Split the integral using the sum rule:
$E = \int_0^{24} 2t dt + \int_0^{24} 10 \sin\left(\frac{\pi t}{12}\right) dt$
2
For the second integral, substitute $u = t - 12$ , shifting the interval to $[- 12, 12]$ :
$10 \int_{-12}^{12} \sin\left(\frac{\pi(u+12)}{12}\right) du = 10 \int_{-12}^{12} \sin\left(\frac{\pi u}{12} + \pi\right) du = -10 \int_{-12}^{12} \sin\left(\frac{\pi u}{12}\right) du$
3
$sin (\frac{π u}{12})$ is an odd function, so the integral over the symmetric interval $[- 12, 12]$ equals 0, making the entire second term 0.
4
Evaluate the first term using the Fundamental Theorem of Calculus:
$\int_0^{24} 2t dt = \left[t^2\right]_0^{24} = 24^2 - 0 = 576$
5
The total energy consumed is 576 kWh.

5. Common Pitfalls

Wrong move:

Claiming additivity only works when $c$ is between $a$ and $b$ , marking $\int_{2}^{6} f (x) d x = \int_{2}^{8} f (x) d x + \int_{8}^{6} f (x) d x$ as invalid.

Why:

Students memorize additivity from geometric examples where $c$ is between 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, incorrectly claiming $\int_{- 2}^{2} (x^{3} + 3 x^{2}) d x = 0$ .

Why:

Students check only the leading term for parity instead of all terms in the integrand.

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 $\int_{5}^{2} f (x) d x = \int_{2}^{5} f (x) d x$ .

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., $\int_{0}^{2} x e^{x} d x = (\int_{0}^{2} x d x) (\int_{0}^{2} e^{x} d x)$ .

Why:

Students incorrectly generalize the constant multiple and sum rules to products of functions.

Correct move:

Memorize that the integral of a product is not the product of integrals; use integration by parts or substitution to evaluate products.

Wrong move:

Claiming that if $f (x) > 0$ on $(a, b)$ , then $\int_{a}^{b} f (x) d x > 0$ even when $b < a$ .

Why:

Students remember the comparison rule for $a < b$ and forget reversing bounds flips the sign of the integral.

Correct move:

Always confirm $a < b$ before applying comparison properties; if $b < a$ , reverse the bounds and flip the inequality sign.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Zero Interval	$\int_{a}^{a} f (x) d x = 0$	Always true for any integrable $f$
Reversal of Limits	$\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x$	Always true, flips the sign of the integral
Constant Multiple	$\int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x$	$k$ must be constant with respect to $x$
Sum/Difference Rule	$\int_{a}^{b} [f (x) \pm g (x)] d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x$	Applies to any finite sum of integrable functions
Additivity over Intervals	$\int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x = \int_{a}^{b} f (x) d x$	Holds for any order of $a, b, c$
Odd Function Symmetry	$\int_{- a}^{a} f (x) d x = 0$	Only applies if $f (- x) = - f (x)$ and $a > 0$
Even Function Symmetry	$\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$	Only applies if $f (- x) = f (x)$ and $a > 0$
MVT for Integrals (Average Value)	$f (c) = \frac{1}{b - a} \int_{a}^{b} f (x) d x$	$c \in [a, b]$ exists if $f$ is continuous on $[a, b]$
Comparison Rule	If $f (x) \geq g (x)$ on $[a, b], a < b$ , then $\int_{a}^{b} f (x) d x \geq \int_{a}^{b} g (x) d x$	Reverse the inequality if $a > b$ after reversing limits

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Find unknown integral value from given values
2022 · FRQ
Average value of a continuous function
2021 · MCQ
Simplify integral over symmetric interval

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.