🎯⭐ INTERACTIVE LESSON

Area Between Curves

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Area Between Curves - Complete Interactive Lesson

Part 1: Area Between Two Curves

Area Between Curves

Part 1 of 7 — Foundations & Setup

Topic Overview

Part	Topic
1	Foundations & setup
2	When curves cross
3	Integrating with respect to $y$
4	Multiple regions & strategy
5	Signed vs total area
6	AP-style workshop
7	Comprehensive assessment

The Area Formula

$\boxed{A = \int_a^b [f(x) - g(x)]\,dx, \quad f(x) \geq g(x)}$

Key Fact: Always subtract top minus bottom. If you get a negative answer, you set up the subtraction in the wrong order.

Step-by-Step Strategy

Step	Action	Why
1	Find intersection points	These are limits $a$ and $b$
2	Determine which is on top	Test a point between intersections
3	Set up $\int_a^b [\text{top} - \text{bottom}]\,dx$

Worked Example

Find the area between $y = x^2$ and $y = x + 2$ .

Step 1: Intersection: $x^2 = x+2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0 \Rightarrow x=-1,\,2$

Step 2: Test $x=0$ : $f(0)=2$ , $g(0)=0$ . So $y = x +$ is on top.

Step 3–4:

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

$= \left(2+4-\frac{8}{3}\right) - \left(\frac{1}{2}-2+\frac{1}{3}\right) = \frac{10}{3}+\frac{7}{6} = \boxed{\frac{9}{2}}$

AP Tip: On the AP exam, always show intersection work. Partial credit depends on seeing correct limits.

Practice — Area Setup 🎯

Classify each setup. 🔍

Compute. ✍️

Key Takeaways — Part 1

Area = $\int_a^b(\text{top} - \text{bottom})\,dx$
Find intersections first → these are your limits
Test a point in the interval to determine which curve is on top
Area is always positive — if you get a negative value, reverse the subtraction

Part 2: When Curves Switch Position

Area Between Curves

Part 2 of 7 — When Curves Cross

Splitting the Integral

When curves switch which is on top, you must split the integral at crossing points:

$\boxed{A = \int_a^c [f(x)-g(x)]\,dx + \int_c^b [g(x)-f(x)]\,dx}$

Part 3: Integrating with Respect to y

Area Between Curves

Part 3 of 7 — Integrating with Respect to $y$

When to Use $dy$ Instead of $dx$

$\boxed{A = \int_c^d [\text{right}(y) - \text{left}(y)]\,dy}$

Part 4: Multiple Regions

Area Between Curves

Part 4 of 7 — Multiple Regions & Strategy

Multi-Region Problems

When three or more curves define a region — or when the boundary changes — break the problem into sub-regions:

$\boxed{A_{\text{total}} = A_1 + A_2 + \cdots}$

Part 5: Area with Absolute Value

Area Between Curves

Part 5 of 7 — Signed vs Total Area

Two Types of "Area"

$\boxed{\text{Signed area} = \int_a^b f(x)\,dx \qquad \text{Total area} = \int_a^b |f(x)|\,dx}$

Part 6: AP-Style Workshop

Area Between Curves

Part 6 of 7 — AP-Style Workshop

AP FRQ Pattern

Many FRQ problems give you a region $R$ and ask multiple parts about it. Here is a typical structure:

Part	What They Ask	What You Do
(a)	Find the area of $R$	$\int_a^b(\text{top}-\text{bottom})\,dx$

Part 7: Comprehensive Assessment

Area Between Curves

Part 7 of 7 — Comprehensive Assessment

Complete Formula Reference

Method	Formula
Area in $x$	$\int_a^b[f(x)-g(x)]\,dx$

Key Fact: If you integrate without splitting, positive and negative areas cancel, giving the signed area (net area), not the total area.

Clue	What to Do
Curves cross inside $[a,b]$	Find crossing point(s), split there
Function changes sign	Split where $f(x) = 0$
Graph shows intersection	Use given $x$ -value as split point

Find the area between $y = x^3$ and $y = x$ on $[-1, 1]$ .

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

Interval	Test point	Top curve
$[-1, 0]$	$x = -0.5$ : $-0.125 > -0.5$	$y = x^3$
$[0, 1]$	$x = 0.5$ : $0.5 > 0.125$	$y = x$

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

$= \left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^0 + \left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1 = \frac{1}{4} + \frac{1}{4} = \boxed{\frac{1}{2}}$

AP Tip: The AP exam loves problems where curves cross. Always check whether the "top" and "bottom" switch within the interval.

Practice — Splitting Integrals 🎯

Identify the strategy. 🔍

Calculate. ✍️

Key Takeaways — Part 2

When curves cross, split the integral at each crossing point
Signed area allows cancellation; total area does not
Always use |top − bottom| on each subinterval
Test a point in each subinterval to determine which curve is on top

A=∫cd[right(y)−left(y)]dy

Use $dx$ when...	Use $dy$ when...
Curves are functions of $x$	Curves are functions of $y$ (e.g., $x = y^2$ )
"Top minus bottom" is clear	"Right minus left" is simpler
Region splits in $x$	One integral in $y$ avoids splitting

Key Fact: Integrating wrt $y$ means the limits are $y$ -values and you subtract right minus left.

Find the area between $x = y^2$ and $x = 4$ .

Intersections: $y^2 = 4 \Rightarrow y = \pm 2$

Right: $x = 4$ . Left: $x = y^2$ .

$A = \int_{-2}^{2}(4-y^2)\,dy = 2\int_0^2(4-y^2)\,dy = 2\left[4y-\frac{y^3}{3}\right]_0^2 = 2\left(8-\frac{8}{3}\right) = \boxed{\frac{32}{3}}$

$y = \sqrt{x}$ and $y = \frac{x}{4}$ — compare setups.

In $x$	In $y$
Need to find crossings in $x$ : $\sqrt{x} = x/4 \Rightarrow x=0,16$	Rewrite: $x=y^2$ and $x=4y$
$\int_0^{16}(\sqrt{x}-\frac{x}{4})\,dx$
Both give $\frac{32}{3}$	Often simpler in $y$

Practice — Integrating in $y$ 🎯

Choose the best approach. 🔍

Calculate. ✍️

Key Takeaways — Part 3

Use $dy$ when curves are naturally functions of $y$
Subtract right minus left (not top minus bottom)
Limits are $y$ -values of intersection points
Integrating in $y$ can turn a two-integral problem into a single integral

Situation	Strategy
Three curves form a triangle	Find all 3 vertices, integrate each edge
Boundary changes at a point	Split into sub-integrals
Mix of horizontal and vertical bounds	Choose $dx$ or $dy$ for each piece
Given a graph with shaded region	Identify each boundary segment

Area bounded by $y = x$ , $y = 2-x$ , and $y = 0$ .

Vertices: $(0,0)$ , $(2,0)$ , $(1,1)$ (where $x = 2-x \Rightarrow x=1$ ).

Split at $x=1$ :

$[0,1]$ : top is $y=x$ , bottom is $y=0$
$[1,2]$ : top is $y=2-x$ , bottom is $y=0$

$A = \int_0^1 x\,dx + \int_1^2(2-x)\,dx = \frac{1}{2} + \frac{1}{2} = \boxed{1}$

Area enclosed by $y = x^2$ , $y = 2x$ , and $y = 4$ .

Intersections: $x^2 = 2x$ at $(0,0)$ and $(2,4)$ . $2x = 4$ at $x=2$ . $x^2=4$ at $x=2$ .

All three curves meet at $(2,4)$ . Region: between $y=x^2$ and $y=2x$ from $x=0$ to $x=2$ .

$A = \int_0^2(2x-x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \boxed{\frac{4}{3}}$

AP Tip: On the AP exam, sketch the region before integrating. Even a rough sketch prevents choosing the wrong boundaries.

Practice — Multi-Region 🎯

Choose the setup. 🔍

Key Takeaways — Part 4

Complex regions: break into simpler sub-regions
Sketch first to identify which curves bound each piece
Choosing $dx$ vs $dy$ can reduce the number of integrals needed
Verify with geometry when possible (triangles, rectangles)

	Signed Area	Total Area
Can be negative?	Yes	No
Cancellation?	Positive and negative cancel	No cancellation
Physical meaning	Net displacement	Total distance
Formula	$\int_a^b f(x)\,dx$	$\int_a^b

Key Fact: The AP exam frequently asks you to distinguish between these. "Total area" and "area of the region" always mean the positive (absolute value) version.

$f(x) = x^2 - 4$ on $[-3, 3]$ . Find signed and total area.

Signed: $\int_{-3}^3(x^2-4)\,dx = 2\int_0^3(x^2-4)\,dx = 2[\frac{x^3}{3}-4x]_0^3 = 2(9-12) = -6$

Total: Split at $x = \pm 2$ (where $x^2 - 4 = 0$ ):

$[-3,-2]$ : $f > 0$ , area $= \int_{-3}^{-2}(x^2-4)\,dx = \frac{7}{3}$
$[-2,2]$ : $f < 0$ , area $= \int_{-2}^2(4-x^2)\,dx = \frac{32}{3}$
$[2,3]$ : $f > 0$ , area $= \int_2^3(x^2-4)\,dx = \frac{7}{3}$

$\text{Total} = \frac{7}{3} + \frac{32}{3} + \frac{7}{3} = \boxed{\frac{46}{3}}$

AP Tip: When the problem says "area enclosed by the curve and the $x$ -axis," it means total area (always positive).

Practice — Signed vs Total 🎯

Classify each statement. 🔍

Calculate. ✍️

Key Takeaways — Part 5

Signed area allows cancellation (can be negative)
Total area uses absolute value (always positive)
AP exam: "area of the region" = total area
Odd functions on symmetric intervals have signed area $= 0$

Region $R$ is bounded by $y = \sqrt{x}$ , $y = 0$ , and $x = 4$ .

(a) Area of $R$ :

$A = \int_0^4 \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^4 = \frac{2}{3}(8) = \boxed{\frac{16}{3}}$

(b) $R$ has cross-sections perpendicular to $x$ -axis that are squares. Volume:

Side $= \sqrt{x}$ . $A(x) = (\sqrt{x})^2 = x$ .

$V = \int_0^4 x\,dx = \left[\frac{x^2}{2}\right]_0^4 = 8$

(c) Rotate $R$ about $x$ -axis. Volume:

$V = \pi\int_0^4(\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \boxed{8\pi}$

AP Tip: For "write but do not evaluate," you earn full credit for a correct integral with correct limits. Do NOT simplify the integrand.

AP-Style Area Problems 🎯

AP setup decisions. 🔍

AP Challenge. ✍️

Key Takeaways — Part 6

AP FRQs often define a region $R$ and ask area, cross-section, and revolution questions
"Write but do not evaluate" = show integral with limits, do not compute
Always show intersection work for partial credit
Practice the full sequence: intersections → setup → evaluate

Mistake	Correction
Subtracting bottom minus top	Always check which is on top at a test point
Forgetting to split at crossings	Total area never cancels — split where curves cross
Using $x$ -limits with $dy$ integral	Match limits to variable of integration
Not showing intersection work	AP graders need to see $f(x)=g(x)$ and solution
Confusing signed and total area	"Area of the region" = total; net change = signed

Quiz — Foundations 🎯

Quiz — Advanced 🎯

Final classification. 🔍

Final Challenge. ✍️

Area Between Curves — Complete!

You’ve mastered:

Part	Topic
1	Foundations & setup
2	When curves cross
3	Integrating with respect to $y$
4	Multiple regions & strategy
5	Signed vs total area
6	AP-style workshop
7	Comprehensive assessment

You’re ready for AP-level area between curves problems!

Area Between Curves

How to Spot a Split

Worked Example

Key Takeaways — Part 2

Worked Example

Conversion Example

Key Takeaways — Part 3

Decision Guide

Worked Example 1

Worked Example 2

Key Takeaways — Part 4

Worked Example

Key Takeaways — Part 5

Worked AP Problem

Key Takeaways — Part 6

Top AP Mistakes

Area Between Curves — Complete!

Area Between Curves

Area Between Curves - Complete Interactive Lesson

Part 1: Area Between Two Curves

Area Between Curves

Topic Overview

The Area Formula

Step-by-Step Strategy

Worked Example

Key Takeaways — Part 1

Part 2: When Curves Switch Position

Area Between Curves

Splitting the Integral

Part 3: Integrating with Respect to y

Area Between Curves

When to Use dydydy Instead of dxdxdx

Part 4: Multiple Regions

Area Between Curves

Multi-Region Problems

Part 5: Area with Absolute Value

Area Between Curves

Two Types of "Area"

Part 6: AP-Style Workshop

Area Between Curves

AP FRQ Pattern

Part 7: Comprehensive Assessment

Area Between Curves

Complete Formula Reference

How to Spot a Split

Worked Example

Key Takeaways — Part 2

Worked Example

Conversion Example

Key Takeaways — Part 3

Decision Guide

Worked Example 1

Worked Example 2

Key Takeaways — Part 4

Worked Example

Key Takeaways — Part 5

Worked AP Problem

Key Takeaways — Part 6

Top AP Mistakes

Area Between Curves — Complete!

When to Use $dy$ Instead of $dx$