Area Between Curves

Area Between Curves

Part 1 of 7 — Foundations & Setup

Topic Overview

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

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 $y=x+2$ 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

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:

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

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

How to Spot a Split

Worked Example

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

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

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

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

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

Area Between Curves

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

When to Use $dy$ Instead of $dx$

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

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

Worked Example

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

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

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

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

Conversion Example

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

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

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:

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

Decision Guide

Worked Example 1

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 $y=0$

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

Worked Example 2

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. 🔍

Compute. ✍️

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)

Area Between Curves

Part 5 of 7 — Signed vs Total Area

Two Types of "Area"

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

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.

Worked Example

$f(x) = x^2 - 4$ 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 $= 0$

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:

Worked AP Problem

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

Area Between Curves

Part 7 of 7 — Comprehensive Assessment

Complete Formula Reference

Top AP Mistakes

Quiz — Foundations 🎯

Quiz — Advanced 🎯

Final classification. 🔍

Final Challenge. ✍️

Area Between Curves — Complete!

You’ve mastered:

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