Area Between Curves That Cross

MATH162

Reference: Stewart 5.1 • Chapter: 5 • Section: 1

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Area Between Curves That Cross

Quick Reference: Essential Formulas

Situation	Formula	Key Point
Curves don't cross on $[a,b]$	$A = \int_a^b [f(x) - g(x)]\,dx$	One integral, top − bottom
Curves cross at $x = c$	$A = \int_a^c [\text{top}_1 - \text{bottom}_1]\,dx + \int_c^b [\text{top}_2 - \text{bottom}_2]\,dx$	Split at crossings
General formula	$A = \int_a^b \lvert f(x) - g(x) \rvert\,dx$	Absolute value ensures positive area

Warning: You cannot directly integrate $\|f(x) - g(x)\|$—you MUST split at crossing points first.

Before You Start

Test yourself on these prerequisites:

1. Can you compute an area between curves?

Find the area between $y = 4$ and $y = x^2$ from $x = -1$ to $x = 1$.

Check Your Answer

$$A = \int_{-1}^{1} (4 - x^2) \, dx = \left[4x - \frac{x^3}{3}\right]_{-1}^{1} = \frac{22}{3}$$

If this was difficult, review Area Between Curves (Vertical) first.

2. Can you find where two curves intersect?

Find where $y = \sin x$ and $y = \cos x$ intersect for $0 \leq x \leq \pi$.

Check Your Answer

$\sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}$

If this was difficult, review your trigonometry.

What Happens When Curves Switch?

So far, we've assumed one curve is always on top. But what if the curves cross? Between $x = 0$ and $x = \pi/2$, the curves $y = \cos x$ and $y = \sin x$ intersect at $x = \pi/4$. Before this point, cosine is above sine; after, sine is above cosine.

If you compute $\int_0^{\pi/2} (\cos x - \sin x) \, dx$ without checking which curve is on top, the positive area from the first half gets partially canceled by the negative contribution from the second half—giving you the net signed area, not the total area of the region.

The fix: Split the integral at crossing points, or use the absolute value formula.

Prerequisite Map

Legend: 🟡 Yellow = immediate prerequisites | 🟢 Green = this skill

Quick Reference

Property	Value
Section	Stewart §5.1
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Problem

When $f(x) \geq g(x)$ on part of $[a, b]$ but $g(x) \geq f(x)$ on another part:

$$\int_a^b [f(x) - g(x)] \, dx \neq \text{Total Area}$$

The integral gives net signed area, which can be smaller (even zero!) if positive and negative portions cancel.

The Solution: Absolute Value Formula

The total area between $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is:

$$\boxed{A = \int_a^b \vert f(x) - g(x)\vert \, dx}$$

How to Evaluate the Absolute Value Integral

The absolute value formula is conceptually correct, but you can't directly integrate $\vert f(x) - g(x)\vert $ without knowing where the expression changes sign.

Procedure:

Find crossing points — Solve $f(x) = g(x)$ for all $x$ in $[a, b]$
Split at crossings — Divide $[a, b]$ into subintervals where one curve stays on top
Integrate each piece — On each subinterval, compute $\int (\text{top} - \text{bottom}) \, dx$
Add the areas — Total area $= A_1 + A_2 + \cdots$

Visual Representation

   y
   |      f(x)
   |    ╭────╮
   |   ╱  A₁  ╲        A₁: f is above g
   |  ╱        ╲╭────  A₂: g is above f
   | ╱     ╱╲   ╲
   |╱  g(x)  ╲A₂ ╲
   └──────●───────●─── x
         c₁      c₂
   a                 b

   Total Area = A₁ + A₂ (not A₁ - A₂!)

Common Pitfalls Table

Mistake	What Goes Wrong	How to Fix
Not splitting at crossings	Get net area, not total area	Find ALL crossing points first, then split
Using same subtraction order in both integrals	One integral gives negative contribution	Switch order: if $f > g$ before, then $g > f$ after
Missing a crossing point	Incorrect partial areas	Solve $f(x) = g(x)$ carefully; check for multiple roots
Confusing net and total area	Wrong answer type	Net = $\int [f-g]$; Total = $\int \lvert f-g \rvert$
Integrating $\lvert f-g \rvert$ directly	Can't find antiderivative	MUST split at crossings first

💡 Distinguishing Net vs. Total Area

Quantity	Formula	What It Measures
Net signed area	$\int_a^b [f(x) - g(x)]\,dx$	How much $f$ exceeds $g$ "on average" (can be 0 or negative)
Total area	$\int_a^b \lvert f(x) - g(x) \rvert\,dx$	Actual geometric area between curves (always positive)

Analogy: Net is like profit/loss on a ledger (can cancel). Total is like counting all transactions (always adds up).

Worked Example

Problem: Find the area bounded by $y = \sin x$ and $y = \cos x$ from $x = 0$ to $x = \pi/2$.

Solution:

Step 1: Find where the curves cross.

$\sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}$ (in the given interval)

Step 2: Determine which is on top in each subinterval.

On $[0, \frac{\pi}{4}]$: Test $x = 0$: $\cos 0 = 1 > \sin 0 = 0$. So $\cos x$ is on top.
On $[\frac{\pi}{4}, \frac{\pi}{2}]$: Test $x = \frac{\pi}{2}$: $\sin \frac{\pi}{2} = 1 > \cos \frac{\pi}{2} = 0$. So $\sin x$ is on top.

Step 3: Set up split integrals.

$$A = \int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx$$

Step 4: Evaluate each integral.

$$A_1 = \left[\sin x + \cos x\right]_0^{\pi/4} = \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1) = \sqrt{2} - 1$$

$$A_2 = \left[-\cos x - \sin x\right]_{\pi/4}^{\pi/2} = (-0 - 1) - \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = -1 + \sqrt{2} = \sqrt{2} - 1$$

Step 5: Add.

$$A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2$$

Step 6: Verify.

Reasonableness check: $2\sqrt{2} - 2 \approx 2(1.41) - 2 = 0.82$. The region consists of two "lens" shapes between sine and cosine on $[0, \pi/2]$. Each lens has width $\pi/4 \approx 0.79$ and max height about $0.7$. Area ≈ $2 \times \frac{2}{3}(0.79)(0.7) \approx 0.74$. Close to $0.82$ ✓

Note: By symmetry about $x = \pi/4$, the two pieces have equal area, which we could have exploited: $A = 2A_1$.

Practice Problems

Level 1 Identifying Crossing Points

The curves $y = x$ and $y = x^3$ are graphed from $x = -2$ to $x = 2$. At which $x$-values do they intersect?

Thought Process

Set $x = x^3$, which gives $x^3 - x = 0$, factoring as $x(x^2 - 1) = x(x-1)(x+1) = 0$.

Show Answer

$x = -1, \, 0, \, 1$

These three points divide the interval $[-2, 2]$ into four subintervals where one curve is consistently above the other.

Level 2 One Crossing Point

Find the area of the region bounded by $y = 4 - x^2$ and $y = x + 2$ from $x = -2$ to $x = 2$.

Thought Process

First, find intersections: $4 - x^2 = x + 2 \Rightarrow x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1) = 0$. So $x = -2$ or $x = 1$.

The crossing is at $x = 1$ (and $x = -2$ is an endpoint).

Check which is on top in each region:

At $x = 0$: parabola gives $4$, line gives $2$. Parabola on top for $-2 < x < 1$.
At $x = 1.5$: parabola gives $4 - 2.25 = 1.75$, line gives $3.5$. Line on top for $1 < x < 2$.

Show Answer

Crossing at $x = 1$ (and $x = -2$ is the left endpoint).

$$A = \int_{-2}^{1} [(4 - x^2) - (x + 2)] \, dx + \int_{1}^{2} [(x + 2) - (4 - x^2)] \, dx$$

$$= \int_{-2}^{1} (2 - x - x^2) \, dx + \int_{1}^{2} (x^2 + x - 2) \, dx$$

$$= \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{1} + \left[\frac{x^3}{3} + \frac{x^2}{2} - 2x\right]_{1}^{2}$$

$$= \left[\left(2 - \frac{1}{2} - \frac{1}{3}\right) - \left(-4 - 2 + \frac{8}{3}\right)\right] + \left[\left(\frac{8}{3} + 2 - 4\right) - \left(\frac{1}{3} + \frac{1}{2} - 2\right)\right]$$

$$= \frac{7}{6} - \left(-\frac{10}{3}\right) + \frac{2}{3} - \left(-\frac{7}{6}\right) = \frac{7}{6} + \frac{10}{3} + \frac{2}{3} + \frac{7}{6}$$

$$= \frac{7}{3} + 4 = \frac{19}{3}$$

Level 3 Symmetric Crossing

Find the area enclosed between $y = \sin(2x)$ and $y = \cos(2x)$ from $x = 0$ to $x = \frac{\pi}{2}$.

Thought Process

$\sin(2x) = \cos(2x)$ when $\tan(2x) = 1$, so $2x = \frac{\pi}{4}$ or $2x = \frac{5\pi}{4}$, giving $x = \frac{\pi}{8}$ or $x = \frac{5\pi}{8}$.

Only $x = \frac{\pi}{8}$ is in $[0, \frac{\pi}{2}]$.

Check: at $x = 0$, $\cos(0) = 1 > \sin(0) = 0$. So cosine is on top first. At $x = \frac{\pi}{4}$, $\sin(\frac{\pi}{2}) = 1 > \cos(\frac{\pi}{2}) = 0$. Sine is on top second.

Show Answer

Crossing at $x = \frac{\pi}{8}$.

$$A = \int_0^{\pi/8} [\cos(2x) - \sin(2x)] \, dx + \int_{\pi/8}^{\pi/2} [\sin(2x) - \cos(2x)] \, dx$$

$$= \left[\frac{\sin(2x)}{2} + \frac{\cos(2x)}{2}\right]_0^{\pi/8} + \left[-\frac{\cos(2x)}{2} - \frac{\sin(2x)}{2}\right]_{\pi/8}^{\pi/2}$$

At $x = \frac{\pi}{8}$: $\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

$$A_1 = \frac{1}{2}\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - \frac{1}{2}(0 + 1) = \frac{\sqrt{2}}{2} - \frac{1}{2}$$

$$A_2 = \left(-\frac{1}{2}(-1) - \frac{1}{2}(0)\right) - \left(-\frac{\sqrt{2}}{4} - \frac{\sqrt{2}}{4}\right) = \frac{1}{2} + \frac{\sqrt{2}}{2}$$

$$A = \left(\frac{\sqrt{2} - 1}{2}\right) + \left(\frac{1 + \sqrt{2}}{2}\right) = \sqrt{2}$$

Level 4 Multiple Crossings

Find the total area enclosed by $y = x$ and $y = x^3$ from $x = -1$ to $x = 1$.

Thought Process

Crossings: $x = x^3 \Rightarrow x(x^2 - 1) = 0 \Rightarrow x = -1, 0, 1$.

These are exactly the endpoints and midpoint! So we have two subintervals: $[-1, 0]$ and $[0, 1]$.

At $x = -0.5$: $y = x$ gives $-0.5$, $y = x^3$ gives $-0.125$. So $x^3 > x$ (less negative). At $x = 0.5$: $y = x$ gives $0.5$, $y = x^3$ gives $0.125$. So $x > x^3$.

By antisymmetry of both functions, the two regions have equal area!

Show Answer

On $[-1, 0]$: $x^3 \geq x$ (both negative, but $x^3$ is closer to zero)

On $[0, 1]$: $x \geq x^3$

$$A = \int_{-1}^{0} (x^3 - x) \, dx + \int_0^1 (x - x^3) \, dx$$

By symmetry (odd functions), both integrals have the same value:

$$A = 2\int_0^1 (x - x^3) \, dx = 2\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = 2\left(\frac{1}{2} - \frac{1}{4}\right) = 2 \cdot \frac{1}{4} = \frac{1}{2}$$

Verification: Two symmetric lens-shaped regions, each with width 1 and max height at midpoint. At $x = 0.5$: $\vert 0.5 - 0.125\vert = 0.375$. Area ≈ $2 \times \frac{2}{3}(1)(0.375) = 0.5$ ✓

🔄 Still confused about curves that cross?

Can't find crossing points? → Review Finding Intersection Points
Not sure which is on top where? → Test a point in EACH subinterval
Getting negative area? → You're computing net, not total. Split and switch subtraction order.
Confused about absolute value? → You can't integrate $\vert f-g\vert $ directly—you MUST split first

✅ Checkpoint: If you can solve Level 4 problems with multiple crossings, you understand how to handle curves that switch positions!

Level 5 Signed vs. Total Area Relationship

Let $f$ and $g$ be continuous functions on $[a, b]$ with exactly one crossing at $x = c \in (a, b)$. Suppose $\int_a^b [f(x) - g(x)] \, dx = 5$ and the total area between the curves is $13$.

Find the areas of the two regions (one where $f > g$, one where $g > f$).
If $f > g$ on $[a, c]$, what is $\int_a^c [f(x) - g(x)] \, dx$?

Thought Process

Let $A_1$ = area where $f > g$ and $A_2$ = area where $g > f$.

Total area: $A_1 + A_2 = 13$

Net signed area: $A_1 - A_2 = 5$ (since one contributes positively, one negatively to $\int [f - g] \, dx$)

Solve the system: Adding gives $2A_1 = 18$, so $A_1 = 9$. Then $A_2 = 4$.

Show Answer

Let $A_1$ = area of region where $f > g$, $A_2$ = area where $g > f$.

System of equations:

$A_1 + A_2 = 13$ (total area)
$A_1 - A_2 = 5$ (net signed integral)

Adding: $2A_1 = 18 \Rightarrow A_1 = 9$

Subtracting: $2A_2 = 8 \Rightarrow A_2 = 4$

(a) The two regions have areas $9$ and $4$.

(b) Since $f > g$ on $[a, c]$, the integral $\int_a^c [f(x) - g(x)] \, dx = A_1 = 9$.

CCI-Style Conceptual Questions

Question 1: A student calculates $\int_0^4 [f(x) - g(x)] \, dx = 0$. Does this mean the curves $f$ and $g$ enclose zero area on $[0, 4]$?

Answer

No! It means the net signed area is zero—the positive and negative contributions exactly cancel. The curves could enclose substantial area, with equal amounts above and below each other. To find total area, you'd need to compute $\int_0^4 \vert f(x) - g(x)\vert \, dx$.

Question 2: When setting up the split integral for curves that cross at $x = c$, why do we need to change the order of subtraction (from $f - g$ to $g - f$) in the second integral?

Answer

Area must be positive. If $f > g$ on $[a, c]$, then $f - g > 0$ there, giving positive area. But on $[c, b]$ where $g > f$, the expression $f - g$ is negative. To get positive area, we compute $g - f$ instead (or equivalently, take the absolute value).

Mastery Checklist

[ ] I can find crossing points by solving $f(x) = g(x)$
[ ] I can determine which curve is on top in each subinterval
[ ] I understand the difference between net signed area and total area
[ ] I can set up and evaluate split integrals with correct subtraction order
[ ] I can recognize when curves might have multiple crossings
[ ] I can verify each piece gives positive area before summing

✅ All boxes checked? You've mastered area between curves! Ready for applications in physics.

Exam Strategy Tips

🎯 How to handle crossing curves on exams

Step 0 — Find ALL crossings:

Solve $f(x) = g(x)$ completely. Don't stop at the first solution.
For trigonometric functions, remember there may be multiple solutions in the interval.

Step 1 — Make a sign chart:

List all crossings and endpoints: $a, c_1, c_2, ..., b$
Test ONE point in each subinterval to determine which function is larger

Step 2 — Write split integrals:

Each subinterval gets its own integral
Subtraction order changes based on which function is on top

Step 3 — Sanity check each piece:

Each individual integral should give a POSITIVE number
If you get a negative, you subtracted wrong in that piece

Step 4 — Add up all pieces:

Total area = sum of all positive pieces
Double-check: total should be larger than any single piece

Time-saver: If problem has symmetry (e.g., odd functions), compute one piece and multiply.

Mental Model

The Accountant's View:

Think of the area between curves as a business's daily profit/loss. On days when revenue exceeds costs ($f > g$), you gain area. On days when costs exceed revenue ($g > f$), you lose area. The ordinary integral gives your net profit—gains minus losses. But if you want to know the total volume of transactions (total area), you need to add up the absolute values: $\vert $gains$\vert $ + $\vert $losses$\vert $.

Connections

Looking back:

Area Between Curves (Vertical) — The basic formula when one curve stays on top

Looking ahead:

Net vs. Total Distance — When velocity is sometimes negative, $\int v \, dt$ gives displacement, but $\int \vert v\vert \, dt$ gives total distance traveled. Same principle!
Work Done by Variable Forces — Work can be positive or negative depending on force direction

📚 Historical Note

The distinction between "net" and "total" area reflects a deeper mathematical concept: signed measure. When Lebesgue developed modern integration theory in the early 1900s, he formalized the idea that integration naturally produces signed quantities.

The physical analogy is compelling: if you walk 10 steps forward and 10 steps back, your net displacement is 0, but your total distance traveled is 20 steps. The integral $\int v\,dt$ gives displacement (signed); the integral $\int \vert v\vert \,dt$ gives distance (unsigned).

This same principle appears throughout physics and economics: work done by friction (can be negative), profit/loss accounting, and electric charge balance all involve signed vs. unsigned quantities.

Summary

Concept	Key Point
Problem	Curves cross, so top/bottom switch
Solution	Split at crossing points; integrate each piece separately
Net area formula	$\int_a^b [f(x) - g(x)]\,dx$ — can be zero or negative
Total area formula	$\int_a^b \lvert f(x) - g(x) \rvert\,dx$ — always positive
Can't integrate directly	Absolute value requires splitting first
Each piece	Should give positive area; if negative, switch subtraction order

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Area (Horizontal)	Section 5.1	Volumes (Disk/Washer)

Last updated: 2026-01-22