Area Between Curves That Cross

8 min read

Quick Reference: Essential Formulas

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 $x = 1$.

2. Can you find where two curves intersect?

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

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

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:

1. Find crossing points: solve $f(x) = g(x)$ for all $x$ in $[a, b]$
2. Split at crossings: divide $[a, b]$ into subintervals where one curve stays on top
3. Integrate each piece: on each subinterval, compute $\int (\text{top} - \text{bottom}) \, dx$
4. 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

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

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

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 $[0, 4]$?

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?

Common Misconceptions

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

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

Summary

Last updated: 2026-01-22