Properties of Definite Integrals

MATH161

Reference: Stewart 4.2 • Chapter: 4 • Section: 2

Navigation: Wiki Home > Skills > Properties of Definite Integrals

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 Area

Interpretation: 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 rectangles

Application: 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

Level 1 Applying the Constant Rule

Evaluate $\int_2^7 5\,dx$ using Property 1.

Thought Process

Property 1 states $\int_a^b c\,dx = c(b-a)$. Here $c = 5$, $a = 2$, $b = 7$.

Show Answer

$$\int_2^7 5\,dx = 5(7-2) = 5 \times 5 = 25$$

Level 2 Combining Integrals

Given that $\int_0^3 f(x)\,dx = 7$ and $\int_0^3 g(x)\,dx = -2$, evaluate:

$\int_0^3 [f(x) + g(x)]\,dx$
$\int_0^3 [2f(x) - 3g(x)]\,dx$

Thought Process

Use the sum rule (Property 2) and constant multiple rule (Property 3). For part (b), apply both rules: first separate the integral, then factor out the constants.

Show Answer

(a) By Property 2: $$\int_0^3 [f(x) + g(x)]\,dx = \int_0^3 f(x)\,dx + \int_0^3 g(x)\,dx = 7 + (-2) = 5$$

(b) By Properties 2, 3, and 4: $$\int_0^3 [2f(x) - 3g(x)]\,dx = 2\int_0^3 f(x)\,dx - 3\int_0^3 g(x)\,dx$$ $$= 2(7) - 3(-2) = 14 + 6 = 20$$

Level 3 Splitting and Combining Intervals

Given:

$\int_0^5 f(x)\,dx = 12$
$\int_0^8 f(x)\,dx = 19$

Find $\int_5^8 f(x)\,dx$.

Thought Process

Property 5 says $\int_0^5 f + \int_5^8 f = \int_0^8 f$. Solve for the unknown piece.

Show Answer

By Property 5 (additivity): $$\int_0^5 f(x)\,dx + \int_5^8 f(x)\,dx = \int_0^8 f(x)\,dx$$

Substituting: $$12 + \int_5^8 f(x)\,dx = 19$$

Therefore: $$\int_5^8 f(x)\,dx = 19 - 12 = 7$$

Level 4 Estimating with Bounds

Without evaluating the integral, show that:

$$2 \leq \int_1^3 \sqrt{x}\,dx \leq 2\sqrt{3}$$

Thought Process

Property 8 says if $m \leq f(x) \leq M$ on $[a,b]$, then $m(b-a) \leq \int_a^b f(x)\,dx \leq M(b-a)$.

Find the minimum and maximum values of $f(x) = \sqrt{x}$ on $[1, 3]$. Since $\sqrt{x}$ is increasing, the min is at $x = 1$ and the max is at $x = 3$.

Show Answer

On the interval $[1, 3]$, since $f(x) = \sqrt{x}$ is increasing:

Minimum value: $m = f(1) = \sqrt{1} = 1$
Maximum value: $M = f(3) = \sqrt{3}$

By Property 8: $$m(b-a) \leq \int_1^3 \sqrt{x}\,dx \leq M(b-a)$$ $$1 \cdot (3-1) \leq \int_1^3 \sqrt{x}\,dx \leq \sqrt{3} \cdot (3-1)$$ $$2 \leq \int_1^3 \sqrt{x}\,dx \leq 2\sqrt{3}$$

This bounds the integral between approximately $2$ and $3.46$.

(The actual value is $\frac{2}{3}(3\sqrt{3} - 1) \approx 2.80$, which lies between our bounds.)

Level 5 Working with Reversed Limits and Multiple Properties

Given:

$\int_1^4 f(x)\,dx = 8$
$\int_4^6 f(x)\,dx = -3$
$\int_6^1 g(x)\,dx = 5$

Evaluate $\int_1^6 [2f(x) - g(x)]\,dx$.

Thought Process

First, handle the reversed limits on the third integral: $\int_6^1 g = -\int_1^6 g$, so $\int_1^6 g = -5$.

Next, combine the $f$ integrals using additivity: $\int_1^6 f = \int_1^4 f + \int_4^6 f$.

Finally, apply linearity: $\int_1^6 [2f - g] = 2\int_1^6 f - \int_1^6 g$.

Show Answer

Step 1: Find $\int_1^6 g(x)\,dx$

Using the reversed limits convention: $$\int_6^1 g(x)\,dx = -\int_1^6 g(x)\,dx$$ $$5 = -\int_1^6 g(x)\,dx$$ $$\int_1^6 g(x)\,dx = -5$$

Step 2: Find $\int_1^6 f(x)\,dx$

Using additivity (Property 5): $$\int_1^6 f(x)\,dx = \int_1^4 f(x)\,dx + \int_4^6 f(x)\,dx = 8 + (-3) = 5$$

Step 3: Apply linearity $$\int_1^6 [2f(x) - g(x)]\,dx = 2\int_1^6 f(x)\,dx - \int_1^6 g(x)\,dx$$ $$= 2(5) - (-5) = 10 + 5 = 15$$

Mastery Checklist

[ ] I can state and apply the constant rule $\int_a^b c\,dx = c(b-a)$
[ ] I can split integrals of sums: $\int(f+g) = \int f + \int g$
[ ] I can factor constants out of integrals: $\int cf = c\int f$
[ ] I can combine integrals over adjacent intervals using additivity
[ ] I can handle reversed limits: $\int_a^b f = -\int_b^a f$
[ ] I can bound an integral using Property 8 without evaluating it
[ ] I can combine multiple properties to solve complex problems

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—$(a+b) + (c+d) = a + b + c + d$, 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.

Previous	Up	Next
Definite Integral Definition	Skills Index	FTC Part 1

Last updated: 2026-01-22