Properties of Definite Integrals — AP Calculus AB
1. Basic Algebraic Properties of Definite Integrals ★★☆☆☆ ⏱ 4 min
These four core rules follow directly from the Riemann sum definition of the definite integral, and let you combine known integral values to find unknowns, a common AP exam problem type.
- Reversal of bounds: swapping lower and upper bounds flips the sign of the integral
- Zero-length integral: when bounds are equal, the integral equals 0
- Constant multiple rule: scaling a function by a constant scales the integral by the same constant
- Sum/difference rule: the integral of a sum/difference equals the sum/difference of individual integrals
\int_b^a f(x)dx = -\int_a^b f(x)dx
\int_a^a f(x)dx = 0
\int_a^b k f(x)dx = k \int_a^b f(x)dx
\int_a^b \left[f(x) \pm g(x)\right]dx = \int_a^b f(x)dx \pm \int_a^b g(x)dx
Exam tip: Always reverse bounds and apply the negative sign before pulling out constants or splitting integrals to avoid sign errors.
2. Additivity of Integration Over Intervals ★★☆☆☆ ⏱ 3 min
Additivity lets you split an integral over a large interval into the sum of smaller integrals over adjacent intervals. This is especially useful for piecewise functions and finding unknown integrals from given related values.
The rule holds for any three real numbers $a, b, c$, regardless of whether $c$ is between $a$ and $b$:
\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx
Intuition: the total net area from $a$ to $b$ is the sum of the net area from $a$ to $c$ and from $c$ to $b$. If $c$ is outside $[a,b]$, the reversal of bounds rule automatically adjusts the sign to keep the equality true.
Exam tip: If you need an integral over a subinterval and know the full integral value, always rearrange additivity to solve for the unknown instead of guessing bound order.
3. Symmetry Properties for Even and Odd Functions ★★★☆☆ ⏱ 3 min
Symmetry is a powerful time-saving tool on the AP exam, letting you evaluate integrals over symmetric intervals without computing any antiderivatives. Recall that an even function satisfies $f(-x)=f(x)$ (symmetric over the y-axis) and an odd function satisfies $f(-x)=-f(x)$ (symmetric about the origin).
For integrals over the symmetric interval $[-a, a]$, the rules are:
- Odd function: Net area above the axis on the positive side cancels net area on the negative side, so $\int_{-a}^a f(x) dx = 0$
- Even function: Area left of the y-axis equals area right of the y-axis, so $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$
\int_{-a}^a f(x) dx = 0 \quad (\text{for odd } f)
\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx \quad (\text{for even } f)
Exam tip: Always check for symmetry before starting to integrate—if the interval is symmetric, you can often eliminate all computation immediately.
4. Comparison and Bounding Properties ★★★☆☆ ⏱ 3 min
Comparison properties let you bound the value of an integral when you cannot compute its exact value, a common multiple-choice question type on the AP exam.
The core comparison rule states that if $f(x) \leq g(x)$ for all $x$ in $[a,b]$, then:
\int_a^b f(x) dx \leq \int_a^b g(x) dx
A useful special case is the bound property, which uses the minimum and maximum of $f(x)$ on $[a,b]$ to bound the integral. If $m = \min f(x)$ and $M = \max f(x)$ on $[a,b]$, then:
m(b-a) \leq \int_a^b f(x) dx \leq M(b-a)
Intuition: The net area under $f(x)$ is between the area of a rectangle of height $m$ and a rectangle of height $M$, both with width equal to the interval length $b-a$.
Exam tip: If the function is monotonic on the interval, the minimum and maximum are always at the endpoints, so you do not need to find critical points.
5. AP-Style Concept Check ★★★☆☆ ⏱ 1 min
Common Pitfalls
Why: Students mix up the order of bounds for the second integral when $c$ is between $a$ and $b$.
Why: Students confuse even and odd symmetry rules, or do not confirm the function type before applying the rule.
Why: Students pull out the constant first and forget that reversing bounds flips the sign of the entire integral.
Why: Students remember the symmetry rule but forget it only applies to intervals symmetric around 0.
Why: Students confuse net area (what definite integrals measure) with total area.