When two curves cross at more than two points, the top and bottom functions switch along the way, so you cannot use a single top-minus-bottom integral. You find the total area by splitting the interval at each intersection point and adding the integrals, or by integrating the absolute value of the difference of the two functions. For AP Calculus, mark every intersection point before setting your bounds.

Why This Matters for the AP Calculus Exam

Area between curves shows up across the integration applications in AP Calculus, and the "more than two points" version tests whether you really understand why order matters. When curves intersect multiple times, one function is on top in some sections and on the bottom in others. The skill being checked is recognizing that sign changes in $f(x)-g(x)$ force you to break the region into pieces.

This connects to free-response and multiple-choice work where you need to set up a correct definite integral expression and evaluate it. Writing the integral with the right limits and the right top-minus-bottom order is important for clear exam work, especially because a single wrong sign throws off the whole answer.

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Key Takeaways

Find every intersection point by solving $f(x) = g(x)$ ; these become the limits where the top and bottom functions can swap.
Sketch or graph the curves so you can see which function is on top in each section.
On each subinterval, integrate (top function) minus (bottom function) so each piece stays positive.
The total area is the sum of the integrals over each subinterval, or one integral of $|f(x)-g(x)|$ over the whole interval.
Always include units of area, such as $\text{unit}^2$ , in your final answer.

Understanding Area Between Intersecting Curves

When two curves cross at more than two points, the area between them is found by integrating the absolute value of the difference between the curves over the interval. For two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ , the area is:

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

The absolute value automatically keeps every piece of the area positive, even where the curves swap positions. The catch is that to actually evaluate this, you usually split the interval at the intersection points and replace the absolute value with the correct top-minus-bottom difference on each piece.

Pay attention to which curve is above or below in each section. That choice changes the integrand, so the intersection points are what tell you where to break the integral.

Steps for Finding the Area

Identify the points of intersection by setting the two equations equal to each other.
Graph the functions. This is optional but strongly recommended.
Decide your approach: top-minus-bottom with vertical slices ( $dx$ ), or right-minus-left with horizontal slices ( $dy$ ).
Set up the integral, splitting it at each intersection point.
Evaluate the integrals and add the results.

Worked Example

Find the area between the curves $y = x^3-2x^2$ and $y=x^2-2x$ on the interval $0\le x \le 2$ .

Step 1: Identify the points of intersection.

Set the two equations equal:

x^3−2x^2=x^2−2x

Simplify:

x^3−3x^2+2x=0

x(x−1)(x−2)=0

The curves intersect at $x=0, 1, 2$ .

Step 2: Graph the functions.

A graph helps you see which curve is on top in each section and choose your approach.

Two functions:

x^2-2x

and

x^3-2x^2

Step 3: Decide your approach.

The curves stack on top of each other, so use vertical slices and subtract bottom from top.

On $0 \le x \le 1$ , the curve $x^3−2x^2$ is the top function and $x^2−2x$ is the bottom function.

On $1\le x \le 2$ , this flips: $x^2−2x$ is now on top of $x^3−2x^2$ .

Step 4: Set up the integral, split at $x=1$ .

\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.

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

\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

\text{Area}= \tfrac{1}{2}\ \text{unit}^2

Practice Problem

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

Solution

Set the two equations equal to find the intersection points:

3x^3-x^2-10x = -x^2+2x

3x^3-12x=0

3x(x^2-4) =0

3x(x-2)(x+2)=0

The intersection points are $x = -2, 0, 2$ .

Two functions:

3x^3-x^2-10x

and

-x^2+2x

The curves stack on top of each other, so set up the integrals using vertical slices, splitting at $x=0$ .

\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

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

=\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

\text{Area} = 24 \ \text{unit}^2

How to Use This on the AP Calculus Exam

Problem Solving

Always solve $f(x) = g(x)$ first. The solutions inside your interval are exactly where you split the integral.
A quick sketch saves you from sign errors. It tells you which function is on top in each piece before you write the integral.
On each subinterval, set up (top function) minus (bottom function) so every integral comes out positive.

Common Trap

Do not integrate over the whole interval with one fixed order of functions when the curves cross in the middle. Pieces where the order flips will subtract from your total and give a wrong, too-small answer.
If you use the absolute value form $\int_a^b |f(x)-g(x)|\,dx$ , remember you still have to split it at the intersection points to evaluate it by hand.

Free Response

Show a correct integral expression with proper notation and limits before you compute. A clearly written setup makes your reasoning easy to follow.
Report your final answer with area units like $\text{unit}^2$ .

Common Misconceptions

One integral always works. When curves intersect at more than two points, a single top-minus-bottom integral over the whole interval can give a wrong answer because the top and bottom functions swap. Split at the intersection points.
The absolute value does the splitting for you. The $|f(x)-g(x)|$ form is correct, but to evaluate it without a calculator you still have to find where the difference changes sign and break the integral there.
Order does not matter inside the integral. It does. If you subtract in the wrong order on a subinterval, that piece becomes negative and shrinks your total area. Keep it top minus bottom on each piece.
You can skip finding intersection points if you have the interval. The given outer interval is not enough. You need the inner intersection points to know where to split, since that is where the functions trade places.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
absolute value of the difference	The absolute value of the difference between two functions, used to calculate area between curves regardless of which function is on top.
area between curves	The region enclosed between two or more curves, calculated using definite integrals.
definite integral	The integral of a function over a specific interval [a, b], representing the net signed area between the curve and the x-axis.

Frequently Asked Questions

How do you find area between curves with more than two intersections?

Find every intersection point, split the interval at those x-values, and add the area from each subinterval. On each piece, integrate top function minus bottom function so the area stays positive.

Why do you split integrals at intersection points?

You split at intersection points because the top and bottom functions can switch there. If you use one top-minus-bottom expression across the whole interval, some pieces may become negative and cancel actual area.

Can you use absolute value for area between curves?

Yes. The total area can be written as the definite integral of the absolute value of the difference between the functions. When evaluating by hand, you often split the integral so you can remove the absolute value correctly.

How do you know which function is on top?

Use a graph, test point, or sign chart for the difference between the functions on each subinterval. The function with the larger y-value on that interval is the top function.

When should you use horizontal slices for area between curves?

Use horizontal slices when right-minus-left with respect to y gives a simpler setup than top-minus-bottom with respect to x. This is especially useful when the region is easier to describe using y-values.

What is a common AP Calculus mistake with multiple intersections?

A common mistake is using only the first and last intersection as bounds. If curves intersect inside the interval, the integrand may change sign, so you must split at every relevant intersection or use absolute value.