Properties of Definite Integrals
Rules for Combining and Comparing Integrals
Once you know what a definite integral means, you need tools to manipulate integrals algebraically. The properties in this section let you break complex integrals into simpler pieces, factor out constants, combine integrals over adjacent intervals, and bound the value of an integral without computing it exactly.
These properties are not arbitrary rules: they follow directly from the definition of the integral as a limit of sums. If you understand why $\sum(a_i + b_i) = \sum a_i + \sum b_i$, you already understand why $\int(f + g) = \int f + \int g$.
The key insight: Integrals inherit the linearity of sums. Since the integral is a limit of sums, any property that works for finite sums extends to integrals.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Integration |
| Course | MATH161 |
| Section | Stewart 4.2 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
Key Concepts
Basic Properties
Property 1: Integral of a Constant
$$\int_a^b c\,dx = c(b - a)$$
Interpretation: The area of a rectangle with height $c$ and width $b - a$.
Property 2: Sum Rule
$$\int_a^b [f(x) + g(x)]\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx$$
Why it works: $\sum[f(x_i) + g(x_i)]\Delta x = \sum f(x_i)\Delta x + \sum g(x_i)\Delta x$
Property 3: Constant Multiple Rule
$$\int_a^b cf(x)\,dx = c\int_a^b f(x)\,dx$$
Why it works: Constants can be factored out of sums.
Property 4: Difference Rule
$$\int_a^b [f(x) - g(x)]\,dx = \int_a^b f(x)\,dx - \int_a^b g(x)\,dx$$
This follows from Properties 2 and 3 with $c = -1$.
Interval Properties
Property 5: Additivity over Intervals
$$\int_a^c f(x)\,dx + \int_c^b f(x)\,dx = \int_a^b f(x)\,dx$$
y
| _____
| / \
| / | \
| / A₁ | A₂ \
+----|--|----|-→ x
a c b
A₁ + A₂ = Total AreaInterpretation: The area from $a$ to $b$ equals the area from $a$ to $c$ plus the area from $c$ to $b$.
Convention: Reversed Limits
$$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$$
$$\int_a^a f(x)\,dx = 0$$
These conventions ensure Property 5 works for any arrangement of $a$, $b$, and $c$.
Comparison Properties
Property 6: Non-negativity
If $f(x) \geq 0$ for all $x$ in $[a, b]$, then: $$\int_a^b f(x)\,dx \geq 0$$
Property 7: Ordering
If $f(x) \geq g(x)$ for all $x$ in $[a, b]$, then: $$\int_a^b f(x)\,dx \geq \int_a^b g(x)\,dx$$
Property 8: Bounding
If $m \leq f(x) \leq M$ for all $x$ in $[a, b]$, then:
$$\boxed{m(b-a) \leq \int_a^b f(x)\,dx \leq M(b-a)}$$
y
| _____
M +----/-----\---- ← upper bound
| / \
| / curve \
m +--------------- ← lower bound
|
+----|------|-→ x
a b
Area between bounding rectanglesApplication: You can estimate an integral without computing it exactly.
Summary Table
| Property | Statement |
|---|---|
| Constant | $\int_a^b c\,dx = c(b-a)$ |
| Sum | $\int_a^b (f+g)\,dx = \int_a^b f\,dx + \int_a^b g\,dx$ |
| Constant Multiple | $\int_a^b cf\,dx = c\int_a^b f\,dx$ |
| Additivity | $\int_a^c f\,dx + \int_c^b f\,dx = \int_a^b f\,dx$ |
| Reversed Limits | $\int_a^b f\,dx = -\int_b^a f\,dx$ |
| Bounds | $m(b-a) \leq \int_a^b f\,dx \leq M(b-a)$ |
Practice Problems
Evaluate $\int_2^7 5\,dx$ using Property 1.
Given that $\int_0^3 f(x)\,dx = 7$ and $\int_0^3 g(x)\,dx = -2$, evaluate:
- $\int_0^3 [f(x) + g(x)]\,dx$
- $\int_0^3 [2f(x) - 3g(x)]\,dx$
Given:
- $\int_0^5 f(x)\,dx = 12$
- $\int_0^8 f(x)\,dx = 19$
Find $\int_5^8 f(x)\,dx$.
Without evaluating the integral, show that:
$$2 \leq \int_1^3 \sqrt{x}\,dx \leq 2\sqrt{3}$$
Given:
- $\int_1^4 f(x)\,dx = 8$
- $\int_4^6 f(x)\,dx = -3$
- $\int_6^1 g(x)\,dx = 5$
Evaluate $\int_1^6 [2f(x) - g(x)]\,dx$.
Common Misconceptions
the integral of a product factors into the product of two integrals.
This is the multiplicative-not-additive error. The linearity properties (sum rule and constant multiple rule) do not extend to products. The sum rule $\int(f+g) = \int f + \int g$ holds because addition distributes through limits of sums; multiplication does not. For instance, $\int_0^1 x \cdot e^x\,dx \neq \left(\int_0^1 x\,dx\right)\left(\int_0^1 e^x\,dx\right)$: the left side requires integration by parts while the right side gives $\tfrac{1}{2}(e-1) \approx 0.859$, which is incorrect.
the additivity property $\int_a^c f + \int_c^b f = \int_a^b f$ allows combining any two integrals with shared limits.
This is the concept-image-conflicts-definition error. The additivity property requires the intervals to be adjacent and non-overlapping; specifically, the upper limit of the first integral must equal the lower limit of the second. The equality $\int_0^3 f\,dx + \int_2^5 f\,dx \neq \int_0^5 f\,dx$ because the intervals overlap on $[2,3]$, counting that piece twice on the left side.
Mastery Checklist
Mental Model
Integrals as “Super-Sums”: Think of the integral as an enhanced version of addition. Just as you can rearrange and factor ordinary sums (e.g., $(a+b) + (c+d) = a + b + c + d$ and $3(a+b) = 3a + 3b$) you can do the same with integrals. The properties are exactly the rules you already know for sums, extended to continuous “sums” over intervals. If it works for $\sum$, it works for $\int$.
Connections
Looking back:
- The Definite Integral Definition as a limit of sums explains why these properties hold
- Limit Laws ensure the properties transfer through the limit
Looking ahead:
- The Fundamental Theorem of Calculus provides a much faster way to evaluate integrals
- Property 5 (additivity) is essential for Integration by Parts and other techniques
CCI-style insight: A common error is applying properties when limits don’t match. For instance, $\int_0^3 f\,dx + \int_2^5 f\,dx \neq \int_0^5 f\,dx$ because the intervals overlap at $[2,3]$. Always check that intervals are adjacent before combining.
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|---|---|---|
| Definite Integral Definition | Skills Index | FTC Part 1 |
Last updated: 2026-01-22