Unit 6 · Topic 6.6 Beta

Properties of Definite Integrals

Definite integrals add over adjacent intervals, change sign when the limits are reversed, are 0 over a single point, and split over sums and constant multiples (but not products).

Practice 1: Implementing Mathematical ProcessesPractice 2: Connecting Representations

Question set for this topic

Part 1 · Hook

Why this matters

A free-response question gives you three integrals in a table and asks for a fourth you were never told. No formula for

f

, no graph, just numbers. The properties of definite integrals let you build the answer the way you would piece together a trip from its legs: adding, subtracting and reversing.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The region between $y = -3$ and the x-axis on $[0, 2]$ gives $\int_0^2 (-3)\,dx$ $=$

$-6$
6
$-3$
0

Show the answer

Signed area: the rectangle of area 6 lies below the axis.

Correct: $-6$ :
6:
$-3$ :
0:

2. Which limit of sums is $\int_0^2 x\,dx$ ?

$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{2i}{n} \cdot \frac{2}{n}$
$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{i}{n} \cdot \frac{2}{n}$
$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{2i}{n}$
$\sum_{i=1}^{2} i$

Show the answer

$\Delta x = \frac{2}{n}$ and $x_i = \frac{2i}{n}$ .

Correct: $\lim_{n \to \infty} \sum_{i=1}^{n} \frac{2i}{n} \cdot \frac{2}{n}$ :
$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{i}{n} \cdot \frac{2}{n}$ :
$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{2i}{n}$ :
$\sum_{i=1}^{2} i$ :

Part 4 · See it

See it first

A positive curve y = f(x) over points a, b and c on the x-axis. The region under the curve from a to b is shaded crimson and from b to c blue. The equation reads: the integral of f from a to c equals the integral from a to b plus the integral from b to c. — Additivity: the integral from a to c is the integral from a to b plus the integral from b to c. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Riemann sums are built by adding width times heightanything true of sums (splitting a sum, pulling out a constant) is true of integrals
Two intervals meet end to endtheir integrals add: $\int_a^b f$ $+ \int_b^c f$ $= \int_a^c f$
An interval is run backwardevery width is negative, so $\int_b^a f$ $= -\int_a^b f$ , and $\int_a^a f = 0$
A constant k is integratedthe result is k times the width: $\int_a^b k\,dx$ $= k(b - a)$

Part 6 · Key ideas

Key ideas

Additivity: $\int_a^b f(x)\,dx$ $+ \int_b^c f(x)\,dx$ $= \int_a^c f(x)\,dx$ , for any order of $a$ , $b$ , $c$ .
Reversed limits: $\int_b^a f(x)\,dx$ $= -\int_a^b f(x)\,dx$ , and $\int_a^a f(x)\,dx$ $= 0$ .
Linearity: $\int_a^b (kf(x) + g(x))\,dx$ $= k\int_a^b f(x)\,dx$ $+ \int_a^b g(x)\,dx$ , and $\int_a^b k\,dx$ $= k(b - a)$ .
A bounded function with a jump or a hole can still be integrated: split at the break and use geometry on each piece.

Part 7 · Misconception

A common mistake

The wrong idea: $\int_3^7 (f(x) + 4)\,dx$ $= \int_3^7 f(x)\,dx$ $+ 4$ .

What actually happens: The constant 4 is a height, so it is integrated over the interval too: $\int_3^7 4\,dx$ $= 4 \cdot (7 - 3)$ $= 16$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Known values of definite integrals

The functions

f

and

g

are continuous. The table gives four definite integrals.

Definite integrals of f and g
Integral	Value
$\int_0^3 f(x)\,dx$	5
$\int_0^7 f(x)\,dx$	12
$\int_0^3 g(x)\,dx$	4
$\int_3^7 g(x)\,dx$	$-2$

1. Find $\int_3^7 f(x)\,dx$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Additivity: $\int_0^7 f$ $= \int_0^3 f$ $+ \int_3^7 f$ , so $\int_3^7 f(x)\,dx$ $= 12 - 5$ $= 7$ .

Answer: 7

2. Find $\int_7^3 \left(2f(x) - g(x)\right)dx$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\int_3^7 (2f - g)\,dx$ $= 2(7) - (-2)$ $= 16$ . Reversing the limits changes the sign: $\int_7^3 (2f(x) - g(x))\,dx$ $= -16$ .

Answer: -16

3. Find $\int_3^7 \left(f(x) + 4\right)dx$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\int_3^7 f(x)\,dx$ $+ \int_3^7 4\,dx$ $= 7$ $+ 4(7 - 3)$ $= 7$ $+ 16$ $= 23$ .

Answer: 23

4. Find $\int_0^7 g(x)\,dx$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\int_0^7 g$ $= \int_0^3 g$ $+ \int_3^7 g$ $= 4$ $+ (-2)$ $= 2$ .

Answer: 2

5. For continuous functions $f$ and $g$ , which statement is not true in general?

$\int_b^a f(x)\,dx$ $= -\int_a^b f(x)\,dx$
$\int_a^b fg\,dx$ $= \int_a^b f\,dx \cdot \int_a^b g\,dx$
$\int_a^b \left(f(x) + g(x)\right)dx$ $= \int_a^b f(x)\,dx$ $+ \int_a^b g(x)\,dx$
$\int_a^b 5f(x)\,dx$ $= 5\int_a^b f(x)\,dx$

Show the answer

Sums and constant multiples pass through an integral; products do not. On $[0, 2]$ with $f = g = 1$ : $\int_0^2 1\,dx$ $= 2$ , but $2 \cdot 2 = 4$ .

$\int_b^a f(x)\,dx$ $= -\int_a^b f(x)\,dx$ : This is true: running the interval backward makes every width negative.
Correct: $\int_a^b fg\,dx$ $= \int_a^b f\,dx \cdot \int_a^b g\,dx$ : Correct, this is false: with $f = g = 1$ on $[0, 2]$ the left side is 2 and the right side is 4. Integrals split over sums, not over products.
$\int_a^b \left(f(x) + g(x)\right)dx$ $= \int_a^b f(x)\,dx$ $+ \int_a^b g(x)\,dx$ : This is true: every Riemann sum of f + g splits into a sum for f plus a sum for g.
$\int_a^b 5f(x)\,dx$ $= 5\int_a^b f(x)\,dx$ : This is true: a constant factor comes out of every Riemann sum, so it comes out of the integral.

6. $f$ is positive on $[0, 2]$ , and the region between its graph and the x-axis there has area 3. Let $g(x)$ $= \int_2^x f(t)\,dt$ . What is $g(0)$ ?

3
0
$-6$
$-3$

Show the answer

$g(0)$ $= \int_2^0 f(t)\,dt$ $= -\int_0^2 f(t)\,dt$ $= -3$ . With the upper limit below the lower limit, a positive integrand gives a negative value.

3: The area is 3, but $g(0)$ runs from 2 back to 0: reversed limits give $-3$ .
0: $g$ is 0 at $x = 2$ , its lower limit, not at $x = 0$ .
$-6$ : Reversing the limits changes only the sign: -3.
Correct: $-3$ : Correct: $g(0)$ $= \int_2^0 f(t)\,dt$ $= -\int_0^2 f(t)\,dt$ $= -3$ .

Part 9 · Summary

Summary

Definite integrals add over adjacent intervals, change sign when the limits are reversed, are 0 over a single point, and split over sums and constant multiples (but not products). Use these with geometry, splitting at jumps, to find integrals from tables and graphs.

Part 10 · Up next

What comes next

Part 11 · Connections