Finding the Area Between Curves That Intersect at More Than Two Points - AP Calculus AB Study Guide

The Core Idea: Finding the Area Between Curves That Intersect at More Than Two Points

When finding the area between two curves, the standard approach is to integrate the "top" function minus the "bottom" function over a given interval. However, this method relies on one function being consistently above the other. This topic addresses the scenario where the functions intersect multiple times, causing the roles of the "top" and "bottom" curves to switch within the region of interest.

To find the total area in such cases, the region must be partitioned into two or more sub-regions at each point of intersection. A separate definite integral is then set up for each sub-region, and the results are summed. An alternative strategy involves changing the perspective of integration. Instead of using vertical rectangles (integrating with respect to $x$), we can use horizontal rectangles and integrate with respect to $y$. This approach is often more convenient if the functions are more easily expressed as $x$ in terms of $y$, or if it simplifies a multi-integral problem into a single integral of the "right" function minus the "left" function.

Key Formulas

The fundamental formulas for finding the area between curves are adapted for these more complex regions.

1. Partitioning the Region (Integrating with respect to $x$)

   If the curves $y=f(x)$ and $y=g(x)$ intersect at $x=a$, $x=c$, and $x=b$, and the "top" function changes at $x=c$, the total area is the sum of the areas of the sub-regions:

   $$\text{Area}={\int }_a^c(\text{Top Function}-\text{Bottom Function})dx+{\int }_c^b(\text{New Top Function}-\text{New Bottom Function})dx$$

2. Integrating with respect to $y$

   If the curves are expressed as $x=f(y)$ and $x=g(y)$, and $f(y)$ is consistently to the right of $g(y)$ for $y$ in the interval $[c,d]$, the area is given by:

   $$\text{Area}={\int }_c^d(f(y)-g(y))dy={\int }_c^d(\text{Right Function}-\text{Left Function})dy$$

Understanding When to Split the Integral or Integrate with Respect to y

The choice of method depends on the geometry of the region and the form of the given equations.

• When to Split the Integral (Integrate with respect to $x$): You must split the integral if the function that defines the upper boundary of the region changes at any point between the overall integration limits. For example, if $f(x)$ is the top function on $[a,c]Formula[2]x^3-4x=-3xFormula[3]x^3-x=0Formula[4]x(x^2-1)=0Formula[5]x(x-1)(x+1)=0Formula[6]\text{Area}={\int }_{-1}^0(f(x)-g(x))dx+{\int }_0^1(g(x)-f(x))dxFormula[7]\text{Area}={\int }_{-1}^0((x^3-4x)-(-3x))dx+{\int }_0^1((-3x)-(x^3-4x))dxFormula[8]\text{Area}={\int }_{-1}^0(x^3-x)dx+{\int }_0^1(-x^3+x)dxFormula[9]{\int }_{-1}^0(x^3-x)dx={\left[\frac{x^4}{4}-\frac{x^2}{2}\right]}_{-1}^0=(0)-\left(\frac{(-1{)}^4}{4}-\frac{(-1{)}^2}{2}\right)=-\left(\frac{1}{4}-\frac{1}{2}\right)=\frac{1}{4}Formula[10]{\int }_0^1(-x^3+x)dx={\left[-\frac{x^4}{4}+\frac{x^2}{2}\right]}_0^1=\left(-\frac{1}{4}+\frac{1}{2}\right)-(0)=\frac{1}{4}Formula[11]\text{Total Area}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}Formula[12]y^2-5=4-2y^{2}Formula[13]3y^2=9Formula[14]y^2=3Formula[15]\text{Area}={\int }_{-\sqrt{3}}^{\sqrt{3}}(\text{Right Function}-\text{Left Function})dyFormula[16]\text{Area}={\int }_{-\sqrt{3}}^{\sqrt{3}}((4-2y^2)-(y^2-5))dyFormula[17]\text{Area}={\int }_{-\sqrt{3}}^{\sqrt{3}}(9-3y^2)dyFormula[18]{\int }_{-\sqrt{3}}^{\sqrt{3}}(9-3y^2)dy={\left[9y-y^3\right]}_{-\sqrt{3}}^{\sqrt{3}}Formula[19]=(9\sqrt{3}-(\sqrt{3}{)}^3)-(9(-\sqrt{3})-(-\sqrt{3}{)}^3)Formula[20]=(9\sqrt{3}-3\sqrt{3})-(-9\sqrt{3}-(-3\sqrt{3}))Formula[21]=(6\sqrt{3})-(-9\sqrt{3}+3\sqrt{3})Formula[22]=6\sqrt{3}-(-6\sqrt{3})=12\sqrt{3}$`

Using Your Calculator

For problems involving complex functions or multiple intersections, a graphing calculator is an essential tool.

To find the area of a region that must be partitioned:

1. Graph Functions: Enter the top/bottom functions into Y1 and Y2.

2. Find Intersections: Use the CALC menu (2nd + TRACE) and select $5:intersect$. Move the cursor near each intersection point and press ENTER three times. Find all three (or more) intersection points, e.g., $x=a$, $x=b$, and $x=c$. It is highly recommended to store these values into variables ($STO->$, $ALPHA$, A, etc.) to maintain precision.

3. Identify Top/Bottom: Look at the graph to see which function (Y1 or Y2) is on top for each interval ($[a,b]$ and $[b,c]$).

4. Calculate Integrals: Use the fnInt command (MATH -> 9: fnInt) from the home screen to calculate the area of each sub-region and add them together. The syntax will look like this:

   fnInt(Y1-Y2, X, a, b) + fnInt(Y2-Y1, X, b, c)

Note on Integrating with Respect to $y$: Standard graphing calculators are designed to graph and integrate functions of $x$. To use a calculator for a $dy$ integral, you can either solve the equations for $y$ (which may be complicated) or simply replace all instances of $y$ in your integrand with $x$ and use your $y$-bounds as the $x$-bounds in fnInt. For example, to calculate $\int_{-\sqrt{3}}^{\sqrt{3}} (9 - 3y^2) \, dy, you would enter fnInt(9-3X^2, X, -√(3), √(3))`.

AP Exam Quick Hit

Common Question Types

• Multiple Intersections (Calculator Active): You will be given two functions, often transcendental (e.g., `f(x) = e^{x/2}and $g(x) = \sin(x)+2), and asked to find the area of the region enclosed by them. You must use your calculator to find all intersection points and then set up the sum of two or more definite integrals to find the total area.

• Integrating with Respect to $y$ (Non-Calculator): You will be given functions that are difficult to work with in terms of $x$, such as $x=y^2$ and $x=2-y$. The prompt will ask for the area of the enclosed region, and the most direct solution is to set up and evaluate a single integral with respect to $y$.

Common Mistakes

• Ignoring Middle Intersections: Finding only the two outermost intersection points and setting up a single integral (${\int }_a^b(f(x)-g(x))dx$). This is incorrect because the integral will compute a net area, where one sub-region's area is subtracted from the other, leading to a wrong answer.

• Incorrect Integrand Order: Forgetting to switch the order of subtraction ($top-bottom$) when the functions cross. For example, using $(f-g)$ for all integrals when it should be $(f-g)$$ for the first and $(g-f)$ for the second.

• Setup Errors for $dy$ Integrals: When integrating with respect to $y$, students may mistakenly use $x$-values for the bounds of integration or forget to write the differential as $dy$. The entire setup—integrand, bounds, and differential—must be consistent with the variable of integration.

• Using Absolute Value in Setup: On a Free Response Question, writing ${\int }_a^b|f(x)-g(x)|dx$ as your setup is not sufficient. You must show the explicit setup with separate integrals for each sub-region, clearly identifying the correct $(top-bottom)$ order for each.