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8.6 Finding the Area Between Curves That Intersect at More Than Two Points

2 min readfebruary 15, 2024

Welcome back to AP Calculus with Fiveable! In the last two study guides, we discussed how to find the area between curves expressed as functions of xx and as functions of yy. Today, we’re going to apply what we know from those key topics to curves that intersect at more than two points! 🧠

📏 Understanding Finding Area Between Curves

In calculus, finding the area between curves intersecting at more than two points is crucial. This technique involves calculating the definite integral of the absolute value of the difference between the two curves over a given interval. The formula for the area between two curves y=f(x)y=f(x) and y=g(x)y=g(x) from x=ax=a to x=bx=b is given by:

Area =abf(x)g(x)dx Area \ = \int_{a}^{b} |f(x)-g(x)| \,dx

Any negative regions are guaranteed to be considered by the absolute value. We can calculate the area between the curves inside the given interval using this approach.

🪜 Finding Area Between Intersecting Curves Steps

Here are some steps you can take when approaching questions asking you to solve for the area between two curves that intersect at more than two points:

  1. 👀 Identify the points of intersection by setting the two equations equal to each other.
  2. 📈 Optional, but highly recommend: graph the functions.
  3. 🤔 Identify what approach to take (top-bottom with vertical slices, or right-left with horizontal slices).
  4. 🔢 Set up the integral with different intervals based on intersection points.
  5. 💯 Evaluate the integrals to find the area.

Let’s go step by step and walk through an example!

🤓 Area Between Curves Walkthrough

Find the area between the curves y=x32x2y = x^3-2x^2 and y=x22xy=x^2-2x between the interval 0x20\le x \le 2.

👀 Step 1) Identify the points of intersection.

To find the points of intersection, we set our two equations equal to each other, like so:

x32x2=x22xx^3−2x^2=x^2−2x

Now simplify to find the points! 🕵️

x33x2+2x=0x^3−3x^2+2x=0
x(x1)(x2)=0x(x−1)(x−2)=0

We now know that these two functions intersect at x=0,1,2x=0, 1, 2.

📈 Step 2) Graph the functions.

This helps you visualize what is going on and identify which approach to take in the next step.

Untitled

Image Created with Desmos

🤔 Step 3) Identify what approach to take

Taking a look at the graph, the curves are on top of one another. Therefore, we can use vertical slices and subtract the bottom function from the top function.

In the interval 0x10 \le x \le1, we have x32x2x^3−2x^2 as our f(x)f(x) in the blue and x22xx^2−2x as our g(x)g(x) in the red.

Vice versa is true in the interval 1x21\le x \le2, since the red curve in the graph above is on top of the blue curve.

🔢 Step 4) Set up the integral with different intervals

So we set up our integral accordingly!

Area=01((x32x2)(x22x))dx+    12((x22x)(x32x2))dx\text{Area}=\int_{0}^{1}((x^3−2x^2)−(x^2−2x))dx+\ \ \ \ \\\int_{1}^{2}((x^2-2x)−(x^3−2x^2))dx

💯 Step 5) Evaluate the integrals to find the area.

Area=01((x33x2+2x)dx+    12(x3+3x22x)dx\text{Area}=\int_{0}^{1}((x^3−3x^2+2x)dx+\ \ \ \ \\\int_{1}^{2}(-x^3+3x^2-2x)dx
Area=(14x4x3+x2)01+(14x4+x3x2)12\text{Area}=(\tfrac{1}{4}x^4−x^3+x^2)\Big|_0^1+(−\tfrac{1}{4}x^4+x^3-x^2)\Big|_1^2
Area=12 unit2\text{Area}= \tfrac{1}{2}\ unit^2

Great work! 🙏


👩‍🏫 Area Between Curves Practice

Your turn to try a question!

Find the area between 3x2x210x3x^2-x^2-10x and x2+2x-x^2+2x between the interval 2x2-2 \le x \le 2.

✏️ Solution to Practice Problem

First, set the two equations equal to find the points of intersection. ⬇️

3x3x210x=x2+2x3x^3-x^2-10x = -x^2+2x
3x312x=03x^3-12x=0
3x(x24)=03x(x^2-4) =0
3x(x2)(x+2)=03x(x-2)(x+2)=0

\therefore our points of intersection are -2, 0, and 2.

Anddd…here’s the graph of these two functions, so you can determine which approach to take.

Untitled

Image Created with Desmos

Since the functions are on top of each other, we can set up the integrals with vertical slices in mind.

Area=20((3x3x210x)(x2+2x))dx+    02((x2+2x)(3x3x210x))dx\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
Area=20(3x312x)dx+    02(3x3+12x)dx\text{Area}=\int_{-2}^{0}(3x^3−12x)dx+\ \ \ \ \\\\\int_{0}^{2}(-3x^3+12x)dx
=34x4122x220+34x4+122x202=24=\tfrac{3}{4}x^4−\tfrac{12}{2}x^2\Big|_{-2}^{0}+ \tfrac{-3}{4}x^4+\tfrac{12}{2}x^2\Big|_0^2 = 24
Area=24 unit2\text{Area} = 24 \ \text{unit}^2

Great work!


🕺Closing

This walkthrough and practice problems aimed to demystify the process, providing step-by-step guidance on how to approach such problems! We hope you feel more comfortable with the process. 😊

Remember to carefully consider the intervals and points of intersection to calculate the area enclosed by the curves accurately. Happy learning!

Key Terms to Review (5)

Area Between Two Curves

: The area between two curves refers to the region enclosed by two curves on a coordinate plane. It is calculated by finding the difference between their respective integrals over a given interval.

Horizontal Slices

: Horizontal slices refer to dividing a shape or graph into thin, parallel strips that are perpendicular to the x-axis. These slices help in finding the area under a curve or calculating volumes of solids.

Integrate

: Integrating refers to finding the area under a curve by summing up infinitely many infinitesimal rectangles. It is the reverse process of differentiation.

Intersection Points

: Intersection points are the points where two or more curves intersect on a graph.

Vertical Slices

: Vertical slices refer to thin vertical strips that divide up an area or volume for calculation purposes.

8.6 Finding the Area Between Curves That Intersect at More Than Two Points

2 min readfebruary 15, 2024

Welcome back to AP Calculus with Fiveable! In the last two study guides, we discussed how to find the area between curves expressed as functions of xx and as functions of yy. Today, we’re going to apply what we know from those key topics to curves that intersect at more than two points! 🧠

📏 Understanding Finding Area Between Curves

In calculus, finding the area between curves intersecting at more than two points is crucial. This technique involves calculating the definite integral of the absolute value of the difference between the two curves over a given interval. The formula for the area between two curves y=f(x)y=f(x) and y=g(x)y=g(x) from x=ax=a to x=bx=b is given by:

Area =abf(x)g(x)dx Area \ = \int_{a}^{b} |f(x)-g(x)| \,dx

Any negative regions are guaranteed to be considered by the absolute value. We can calculate the area between the curves inside the given interval using this approach.

🪜 Finding Area Between Intersecting Curves Steps

Here are some steps you can take when approaching questions asking you to solve for the area between two curves that intersect at more than two points:

  1. 👀 Identify the points of intersection by setting the two equations equal to each other.
  2. 📈 Optional, but highly recommend: graph the functions.
  3. 🤔 Identify what approach to take (top-bottom with vertical slices, or right-left with horizontal slices).
  4. 🔢 Set up the integral with different intervals based on intersection points.
  5. 💯 Evaluate the integrals to find the area.

Let’s go step by step and walk through an example!

🤓 Area Between Curves Walkthrough

Find the area between the curves y=x32x2y = x^3-2x^2 and y=x22xy=x^2-2x between the interval 0x20\le x \le 2.

👀 Step 1) Identify the points of intersection.

To find the points of intersection, we set our two equations equal to each other, like so:

x32x2=x22xx^3−2x^2=x^2−2x

Now simplify to find the points! 🕵️

x33x2+2x=0x^3−3x^2+2x=0
x(x1)(x2)=0x(x−1)(x−2)=0

We now know that these two functions intersect at x=0,1,2x=0, 1, 2.

📈 Step 2) Graph the functions.

This helps you visualize what is going on and identify which approach to take in the next step.

Untitled

Image Created with Desmos

🤔 Step 3) Identify what approach to take

Taking a look at the graph, the curves are on top of one another. Therefore, we can use vertical slices and subtract the bottom function from the top function.

In the interval 0x10 \le x \le1, we have x32x2x^3−2x^2 as our f(x)f(x) in the blue and x22xx^2−2x as our g(x)g(x) in the red.

Vice versa is true in the interval 1x21\le x \le2, since the red curve in the graph above is on top of the blue curve.

🔢 Step 4) Set up the integral with different intervals

So we set up our integral accordingly!

Area=01((x32x2)(x22x))dx+    12((x22x)(x32x2))dx\text{Area}=\int_{0}^{1}((x^3−2x^2)−(x^2−2x))dx+\ \ \ \ \\\int_{1}^{2}((x^2-2x)−(x^3−2x^2))dx

💯 Step 5) Evaluate the integrals to find the area.

Area=01((x33x2+2x)dx+    12(x3+3x22x)dx\text{Area}=\int_{0}^{1}((x^3−3x^2+2x)dx+\ \ \ \ \\\int_{1}^{2}(-x^3+3x^2-2x)dx
Area=(14x4x3+x2)01+(14x4+x3x2)12\text{Area}=(\tfrac{1}{4}x^4−x^3+x^2)\Big|_0^1+(−\tfrac{1}{4}x^4+x^3-x^2)\Big|_1^2
Area=12 unit2\text{Area}= \tfrac{1}{2}\ unit^2

Great work! 🙏


👩‍🏫 Area Between Curves Practice

Your turn to try a question!

Find the area between 3x2x210x3x^2-x^2-10x and x2+2x-x^2+2x between the interval 2x2-2 \le x \le 2.

✏️ Solution to Practice Problem

First, set the two equations equal to find the points of intersection. ⬇️

3x3x210x=x2+2x3x^3-x^2-10x = -x^2+2x
3x312x=03x^3-12x=0
3x(x24)=03x(x^2-4) =0
3x(x2)(x+2)=03x(x-2)(x+2)=0

\therefore our points of intersection are -2, 0, and 2.

Anddd…here’s the graph of these two functions, so you can determine which approach to take.

Untitled

Image Created with Desmos

Since the functions are on top of each other, we can set up the integrals with vertical slices in mind.

Area=20((3x3x210x)(x2+2x))dx+    02((x2+2x)(3x3x210x))dx\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
Area=20(3x312x)dx+    02(3x3+12x)dx\text{Area}=\int_{-2}^{0}(3x^3−12x)dx+\ \ \ \ \\\\\int_{0}^{2}(-3x^3+12x)dx
=34x4122x220+34x4+122x202=24=\tfrac{3}{4}x^4−\tfrac{12}{2}x^2\Big|_{-2}^{0}+ \tfrac{-3}{4}x^4+\tfrac{12}{2}x^2\Big|_0^2 = 24
Area=24 unit2\text{Area} = 24 \ \text{unit}^2

Great work!


🕺Closing

This walkthrough and practice problems aimed to demystify the process, providing step-by-step guidance on how to approach such problems! We hope you feel more comfortable with the process. 😊

Remember to carefully consider the intervals and points of intersection to calculate the area enclosed by the curves accurately. Happy learning!

Key Terms to Review (5)

Area Between Two Curves

: The area between two curves refers to the region enclosed by two curves on a coordinate plane. It is calculated by finding the difference between their respective integrals over a given interval.

Horizontal Slices

: Horizontal slices refer to dividing a shape or graph into thin, parallel strips that are perpendicular to the x-axis. These slices help in finding the area under a curve or calculating volumes of solids.

Integrate

: Integrating refers to finding the area under a curve by summing up infinitely many infinitesimal rectangles. It is the reverse process of differentiation.

Intersection Points

: Intersection points are the points where two or more curves intersect on a graph.

Vertical Slices

: Vertical slices refer to thin vertical strips that divide up an area or volume for calculation purposes.


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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.