Finding the Area Between Curves That Intersect at More Than Two Points: AP Calculus AB/BC Study Guide
Introduction
Welcome back to AP Calculus! Today, we're going on a mathemagical journey to find areas between curves that play hardtoget by intersecting at more than two points. 🎢 If you think calculus is tough, just wait until you see these curves! But don't worry, we'll guide you through it as smoothly as solving an integral.
The Curvy World of Finding Areas
In calculus, determining the area between curves that intersect at multiple points is a thrilling ride. It involves calculating the definite integral of the absolute value of the difference between two curves over a given interval. Think of it like finding the space between two roller coasters on a graph—without the stomachchurning drops! 🎢📈
The formula for the area between two curves, ( y = f(x) ) and ( y = g(x) ), from ( x = a ) to ( x = b ) is given by:
[ \text{Area} = \int_{a}^{b} \left f(x)  g(x) \right , dx ]
Using the absolute value ensures that any region below the xaxis gets its fair share of the area count—because math is all about equality, right? Remember, which curve is on top can change over different intervals, so pay close attention to those intersections and curve behaviors.
StepbyStep Strategy for Curve Intersection Areas
Here's a roadmap to help you find the area between intersecting curves like a pro:
 Identify the Points of Intersection: Start by setting the two equations equal to each other. This is like solving a whoworeitbest contest.
 Graph the Functions: Optional but super helpful. Seeing the curves' dance helps you decide which moves to make.
 Choose Your Approach: Decide whether to use vertical slices (topbottom) or horizontal slices (rightleft). Think of slicing a pie, vertically or horizontally—just don’t eat it...yet.
 Set Up the Integral: Use different intervals based on those intersection points. Integrals are like calculus sandwiches; build them carefully!
 Evaluate the Integrals: Finally, chow down on those integrals to find the area. Remember, calories don't count in math. 😉
Let's Dive Into an Example!
Consider these two dance partners of math: ( y = x^3  2x^2 ) and ( y = x^2  2x ), and the interval ( 0 \le x \le 2 ).

Find the Points of Intersection: We set the equations equal to each other: [ x^3  2x^2 = x^2  2x ] Simplify and solve: [ x^3  3x^2 + 2x = 0 ] Factor out the common term: [ x(x  1)(x  2) = 0 ] So, the intersection points are ( x = 0, 1, 2 ).

Graph the Functions: Plot the curves to see the beautiful chaos. It looks like the blue curve is catching and overtaking the red curve.

Identify the Approach: From the graph, between ( 0 \le x \le 1 ), the blue curve ( ( y = x^3  2x^2 ) ) is on top. Between ( 1 \le x \le 2 ), the red curve ( ( y = x^2  2x ) ) takes over. Vertical slicing it is! 🎂

Set Up the Integral: Here’s how we build our integrals: [ \text{Area} = \int_{0}^{1} [(x^3  2x^2)  (x^2  2x)] , dx + \int_{1}^{2} [(x^2  2x)  (x^3  2x^2)] , dx ]

Evaluate the Integrals: Crunch the numbers (without getting crumbs everywhere!): [ \int_{0}^{1} (x^3  3x^2 + 2x) , dx + \int_{1}^{2} (x^3 + 3x^2  2x) , dx ] Do the math: [ \left[ \frac{1}{4}x^4  x^3 + x^2 \right]{0}^{1} + \left[ \frac{1}{4}x^4 + x^3  x^2 \right]{1}^{2} ] Simplify: [ \left( \frac{1}{4}(1)^4  (1)^3 + (1)^2 \right)  \left( \frac{1}{4}(0)^4  (0)^3 + (0)^2 \right) + \left( \frac{1}{4}(2)^4 + (2)^3  (2)^2 \right)  \left( \frac{1}{4}(1)^4 + (1)^3  (1)^2 \right) ] [ = \frac{1}{4}  1 + 1 + \left[ 4 + 8  4 \right]  \left[ \frac{1}{4} + 1  1 \right] ] [ = \frac{1}{4}  1 + 1 + 0 + \frac{1}{4} = \frac{1}{2} \text{ unit}^2 ]
And there it is! The area between the curves is ( \frac{1}{2} \text{ unit}^2 ).
Time for Some Practice!
Give it a whirl and find the area between ( 3x^3  x^2  10x ) and ( x^2 + 2x ) over the interval ( 2 \le x \le 2 ).
Solution to Practice Problem

Set the equations equal to find intersections: [ 3x^3  x^2  10x = x^2 + 2x ] Simplify: [ 3x^3  12x = 0 ] Factor: [ 3x(x^2  4) = 0 ] [ 3x(x  2)(x + 2) = 0 ] Intersections are at ( x = 2, 0, 2 ).

Visualize the curves.

Set up the integrals (vertical slices): [ \text{Area} = \int_{2}^{0} [(3x^3  x^2  10x)  (x^2 + 2x)] , dx + \int_{0}^{2} [(x^2 + 2x)  (3x^3  x^2  10x)] , dx ]

Evaluate: [ \text{Area} = \int_{2}^{0} (3x^3  12x) , dx + \int_{0}^{2} (3x^3 + 12x) , dx ] Simplify and evaluate: [ = \left[ \frac{3}{4}x^4  6x^2 \right]{2}^{0} + \left[ \frac{3}{4}x^4 + 6x^2 \right]{0}^{2} = 24 \text{ unit}^2 ]
Boom! You've nailed it!
Key Terms to Review
 Area Between Two Curves: The region enclosed by two curves on a coordinate plane, calculated by integrating the difference of their functions over an interval.
 Horizontal Slices: Dividing a shape into parallel strips perpendicular to the xaxis for area or volume calculations.
 Integrate: Finding the area under a curve by summing infinitesimal rectangles; it’s like collecting microareas to form a macroarea.
 Intersection Points: The points where two or more curves meet on a graph.
 Vertical Slices: Dividing up an area into thin vertical strips for easier calculation.
Conclusion
We hope this guide turned the mysterious process of finding areas between intersecting curves into a breeze! Remember to pay close attention to intersection points and the behavior of the curves. With some practice, you'll be sketching integrals and calculating areas like a pro. 📚✏️ Happy integrating!