Signed integral vs total area

Overview

Big Picture

A definite integral and a total area answer different questions.

A definite integral keeps signs. Total geometric area counts the size of every region.

You already know how to evaluate a definite integral. Here, the extra decision is whether to split the interval before calculating area.

Key idea

The Difference

For limits written from left to right, $a<b$ :

Position of a region	Contribution to $\int_a^b f(x)\,dx$	Geometric area
Above the $x$ -axis	Positive	Positive
Below the $x$ -axis	Negative	Positive

A zero-size region has zero area. Geometric area is never negative.

Above-axis and below-axis contributions can cancel in a definite integral. They cannot cancel when finding total geometric area: both regions take up space.

Method

Step by Step

To find the total area between a curve and the $x$ -axis over an interval:

Find the zeros: solve $f(x)=0$ , keeping those inside the interval. Check which are crossings.
Split at crossings: each resulting region must stay on one side of the axis. Splitting at a zero where the curve only touches the axis is harmless, but unnecessary.
Integrate separately: evaluate $F(\text{right})-F(\text{left})$ for each region.
Interpret each sign: a negative result represents a below-axis region. Its area is the magnitude of that result.
Add all the non-negative region areas.

For example, if the separate signed contributions are $I_1$ and $I_2$ :

\text{signed integral}=I_1+I_2, \qquad \text{total area}=|I_1|+|I_2|.

The area formula works here because each piece stays on one side of the axis.

Key idea

Why

Integration adds signed contributions: above the axis adds; below the axis subtracts. Geometric area measures size, so being below the axis does not make a region's area negative.

Convert each region to its area before adding. Converting only after adding loses area whenever opposite signs have cancelled.

Worked example

Worked Example 1

Find the signed integral and total area for $y=2x-4$ , from $x=0$ to $x=4$ .

Locate the crossing: $2x-4=0$ , so $x=2$ .

Redrawn for this page from the resource's text schematic.
Interval	0 to 2	2 to 4
Curve position	below	above
Signed contribution	−4	+4
Region area	4	4

An antiderivative is $F(x)=x^2-4x$ . Split at $2$ :

\int_0^2(2x-4)\,dx=[x^2-4x]_0^2=-4,

\int_2^4(2x-4)\,dx=[x^2-4x]_2^4=0-(-4)=4.

The signed integral is $-4+4=\boxed{0}$ .

The total area is $4+4=\boxed{8\text{ square units}}$ .

Zero signed integral does not mean zero area. Counting only one region would give $4$ , which misses half the required area.

Worked example

Worked Example 2

Find the signed integral and total area for $y=x-1$ , from $x=0$ to $x=4$ .

The crossing is $x=1$ . The curve is below the axis on $[0,1]$ and above it on $[1,4]$ .

Using $F(x)=\frac{x^2}{2}-x$ :

\int_0^1(x-1)\,dx= \left[\frac{x^2}{2}-x\right]_0^1=-\frac12,

\int_1^4(x-1)\,dx= \left[\frac{x^2}{2}-x\right]_1^4 =4-\left(-\frac12\right)=\frac92.

The signed integral is $-\frac12+\frac92=\boxed{4}$ .

The total area is $\frac12+\frac92=\boxed{5\text{ square units}}$ . The unequal regions do not cancel completely, but cancellation still reduces the signed answer.

Common mistake

The Trap

“One definite integral = total area” can fail.

In Example 2, one integral gives $4$ , but the total area is $5$ . Even taking the absolute value of that whole integral gives $|4|=4$ , still wrong.

Entirely above the axis: the integral gives the area.
Entirely below the axis: the magnitude of the integral gives the area.
Crossing the axis: split first, find each region's area, then add.

Before finishing, check: Have I counted every region? Is my final geometric area non-negative?

Practice

Your Turn

For Questions 1–4, find both the signed definite integral and the total geometric area. Show any split points. Give areas in square units.

$y=3x$ , from $x=-2$ to $x=2$ .
$y=2x-6$ , from $x=0$ to $x=5$ . Find the crossing before integrating.
$y=x-5$ , from $x=1$ to $x=3$ . Decide whether any split is needed.
$y=x^2-1$ , from $x=-2$ to $x=2$ . Find all crossings inside the interval.
A continuous curve is below the axis for $0<x<2$ and above it for $2<x<5$ . You are given

\int_0^2 f(x)\,dx=-7,\qquad \int_2^5 f(x)\,dx=3.

Find the signed integral from $0$ to $5$ and the total area. Explain the error in: “The total area is $|-7+3|=4$ .”

Answers

Fully Worked Answers

Show the answers

1. Equal regions

$3x=0$ gives a crossing at $x=0$ . The graph is below the axis to the left and above it to the right.

\int_{-2}^0 3x\,dx=\left[\frac32x^2\right]_{-2}^0=0-6=-6,

\int_0^2 3x\,dx=\left[\frac32x^2\right]_0^2=6-0=6.

Signed integral: $-6+6=\boxed{0}$ .

Total area: $|-6|+|6|=\boxed{12\text{ square units}}$ .

The below-axis region has area $6$ , and the above-axis region contributes another $6$ .

2. Unequal regions

$2x-6=0$ gives $x=3$ . Split there: below the axis on $[0,3]$ , above on $[3,5]$ .

\int_0^3(2x-6)\,dx=[x^2-6x]_0^3=-9,

\int_3^5(2x-6)\,dx=[x^2-6x]_3^5=-5-(-9)=4.

Signed integral: $-9+4=\boxed{-5}$ .

Total area: $9+4=\boxed{13\text{ square units}}$ .

The negative contribution becomes a positive region area before adding.

3. Entirely below the axis

$x-5=0$ gives $x=5$ , outside $[1,3]$ . Throughout the required interval, $x-5<0$ , so no split is needed.

\int_1^3(x-5)\,dx =\left[\frac{x^2}{2}-5x\right]_1^3 =-\frac{21}{2}-\left(-\frac92\right) =\boxed{-6}.

Total area: $|-6|=\boxed{6\text{ square units}}$ .

Taking the magnitude of the whole integral works here because the entire region is below the axis.

4. Three regions

$x^2-1=0$ gives $x=-1$ and $x=1$ . Both lie inside the interval. The curve is above, then below, then above the axis.

Use $F(x)=\frac{x^3}{3}-x$ :

\int_{-2}^{-1}(x^2-1)\,dx =\frac23-\left(-\frac23\right)=\frac43,

\int_{-1}^{1}(x^2-1)\,dx =-\frac23-\frac23=-\frac43,

\int_1^2(x^2-1)\,dx =\frac23-\left(-\frac23\right)=\frac43.

Signed integral: $\frac43-\frac43+\frac43=\boxed{\frac43}$ .

Total area: $\frac43+\frac43+\frac43=\boxed{4\text{ square units}}$ .

The middle region contributes negatively to the integral, but its area is positive. All three regions must be included.

5. Given signed contributions

The regions are already separated at $x=2$ .

Signed integral: $\int_0^5 f(x)\,dx=-7+3=\boxed{-4}$ .

Total area: $|-7|+|3|=7+3=\boxed{10\text{ square units}}$ .

The proposed answer takes the magnitude after cancellation has happened. Instead, count the below-axis area of $7$ and the above-axis area of $3$ separately, then add.