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

The Challenge

When two curves intersect more than twice, they carve out multiple bounded regions. Each region needs its own integral, because the functions switch positions (whichever one is on top changes).

Key principle: at each intersection point, figure out which function is on top for that region, then set up a separate integral.

Example 1: Three Intersection Points

Find the area between

f(x) = x^3 + x^2

and

g(x) = x^2 + 4x

Step 1: set the functions equal

x^3 + x^2 = x^2 + 4x

Step 2: move everything to one side. The

x^2

terms cancel:

x^3 - 4x = 0

Step 3: factor

x(x^2 - 4) = 0

Step 4: solve with the zero product property

x = 0 \quad \text{or} \quad x^2 - 4 = 0 \;\Rightarrow\; x = \pm 2

Intersection points:

x = -2, 0, 2

. Label them

a = -2

b = 0

c = 2

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

Setting Up the Integrals

Three intersection points create two regions, so we need two integrals.

Step 1: which function is on top in each region?

Region 1 (

x = -2

x = 0

). Test

x = -1

: - Cubic:

f(-1) = -1 + 1 = 0

- Parabola:

g(-1) = 1 - 4 = -3

Since

0 > -3

, the cubic is on top.

Region 2 (

x = 0

x = 2

). Test

x = 1

: - Cubic:

f(1) = 1 + 1 = 2

- Parabola:

g(1) = 1 + 4 = 5

Since

5 > 2

, the parabola is on top.

Step 2: set up an integral for each region

\text{Total Area} = \int_{-2}^{0} [(x^3 + x^2) - (x^2 + 4x)]\,dx + \int_{0}^{2} [(x^2 + 4x) - (x^3 + x^2)]\,dx

Notice that the cubic comes first in the first integral (it’s on top) and the parabola comes first in the second one. The subtraction order flips.

Step 3: simplify (optional)

= \int_{-2}^{0} (x^3 - 4x)\,dx + \int_{0}^{2} (4x - x^3)\,dx

Step 4: evaluate

= \left[\frac{x^4}{4} - 2x^2\right]_{-2}^{0} + \left[2x^2 - \frac{x^4}{4}\right]_{0}^{2} = (0 - (-4)) + (4 - 0) = 8

Total area:

8

square units

Calculator Shortcut: The Absolute Value Method

On a calculator, you can do the whole thing with one integral:

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

where

a

is the leftmost intersection and

c

is the rightmost. The absolute value automatically makes every region positive, no matter which function is on top.

Warning:

\left|\int_{a}^{c} [f(x) - g(x)]\,dx\right| \neq \text{Total Area}

The absolute value has to be inside the integral.

Correct: $\int_{a}^{c} |f(x) - g(x)|\,dx$

Wrong: $\left|\int_{a}^{c} [f(x) - g(x)]\,dx\right|$

In Example 1, the wrong version gives $\left|\int_{-2}^{2} (x^3 - 4x)\,dx\right| = 0$ , because the two regions cancel. The true area is 8.

When to use it: calculator-active questions where you just need a number quickly.

When not to: no-calculator questions, where you have to evaluate each piece by hand anyway. On free response,

\int_{a}^{c} |f(x) - g(x)|\,dx

is a correct setup, but writing the separate integrals shows which function is on top in each region.

Example 2: Three Regions

Find the area bounded by

y = 1

y = 2 - \frac{1}{2}x^2

, from

x = -2

x = 2

Step 1: find the intersections

1 = 2 - \frac{1}{2}x^2 \quad \Rightarrow \quad \frac{1}{2}x^2 = 1 \quad \Rightarrow \quad x^2 = 2 \quad \Rightarrow \quad x = \pm\sqrt{2}

So the curves cross at

x \approx \pm 1.414

Boundaries:

a = -2

b = -\sqrt{2}

c = \sqrt{2}

d = 2

Step 2: which function is on top in each region?

Region 1 (

[-2, -\sqrt{2}]

). Test

x = -1.5

: - Line:

y = 1

- Parabola:

y = 2 - \frac{1}{2}(2.25) = 0.875

Line is on top.

Region 2 (

[-\sqrt{2}, \sqrt{2}]

). Test

x = 0

: - Line:

y = 1

- Parabola:

y = 2

Parabola is on top.

Region 3 (

[\sqrt{2}, 2]

). Test

x = 1.5

: - Line:

y = 1

- Parabola:

y = 0.875

Line is on top.

Step 3: three integrals

\text{Total Area} = \int_{-2}^{-\sqrt{2}} \left[1 - \left(2 - \tfrac{1}{2}x^2\right)\right]dx + \int_{-\sqrt{2}}^{\sqrt{2}} \left[\left(2 - \tfrac{1}{2}x^2\right) - 1\right]dx + \int_{\sqrt{2}}^{2} \left[1 - \left(2 - \tfrac{1}{2}x^2\right)\right]dx

Simplified:

= \int_{-2}^{-\sqrt{2}} \left(\tfrac{1}{2}x^2 - 1\right)dx + \int_{-\sqrt{2}}^{\sqrt{2}} \left(1 - \tfrac{1}{2}x^2\right)dx + \int_{\sqrt{2}}^{2} \left(\tfrac{1}{2}x^2 - 1\right)dx

Regions 1 and 3 have the same integrand, because the same function is on top in both. That’s symmetry doing you a favor.

Storing Intersection Points on a Calculator

When the intersections are irrational or ugly decimals, store them so you don’t lose precision.

Step 1: find the intersections 1. Enter both functions in Y= 2. Press GRAPH 3. 2nd → CALC → 5:intersect 4. Select the first curve, second curve, then make a guess 5. Note the

x

-value

Step 2: store it 1. 2nd → MODE (QUIT) 2. Press STO→ 3. Press ALPHA → A (or any letter) 4. Press ENTER

Step 3: repeat, storing the other intersections as B, C, and so on.

Step 4: use the stored variables in your integrals:

\int_{A}^{B} [\text{top} - \text{bottom}]\,dx + \int_{B}^{C} [\text{top} - \text{bottom}]\,dx

Then evaluate with MATH → 9:fnInt( using A, B, C.

General Strategy

Step 1: find ALL intersection points. Set the functions equal, solve algebraically or with a calculator, and list them in order:

a, b, c, d, \ldots

Step 2: count the regions. Number of regions = (number of boundaries) − 1. The boundaries are all the intersections, plus any other endpoints the problem gives you. Each region needs its own integral.

Step 3: for each region, pick a test point inside the interval, evaluate both functions there, and whichever is larger is the “top” function.

Step 4: set up one integral per region, each of the form

\int_{x_i}^{x_{i+1}} [\text{top} - \text{bottom}]\,dx

Step 5: add them all together.

Example 3: Setup Only

Set up (but don’t evaluate) integrals for the area between

f(x) = x^3 - 3x

and

g(x) = x

from

x = -2

x = 2

Step 1: find the intersections

x^3 - 3x = x \quad \Rightarrow \quad x^3 - 4x = 0 \quad \Rightarrow \quad x(x-2)(x+2) = 0

x = -2, 0, 2

Step 2: test each region

From

x = -2

0

, test

x = -1

: -

f(-1) = -1 + 3 = 2

g(-1) = -1

f(x)

is on top.

From

x = 0

2

, test

x = 1

: -

f(1) = 1 - 3 = -2

g(1) = 1

g(x)

is on top.

Step 3: write the integrals

\boxed{\text{Area} = \int_{-2}^{0} [(x^3 - 3x) - x]\,dx + \int_{0}^{2} [x - (x^3 - 3x)]\,dx}

Common Mistakes to Avoid

Forgetting to find all intersection points. Solve completely, and check whether the functions cross more than twice

Using the same integrand for every region. Check which function is on top in EACH region separately

Putting the absolute value outside the integral. It goes inside: $\int |f-g|\,dx$ , not $|\int (f-g)\,dx|$

Not using parentheses when subtracting. Write $(x^2 + 4x) - (x^3 + x^2)$ , not $x^2 + 4x - x^3 + x^2$

Forgetting to add all the integrals together

When the Intersections Aren’t Nice Values

Non-calculator problems: leave answers in exact form. Write

\int_{-\sqrt{3}}^{\sqrt{3}}

, not

\int_{-1.732}^{1.732}

, and simplify radicals where you can.

Calculator problems: store values with STO→, reference the stored variables in your integrals, and you keep full precision.

Summary Table

Boundaries	Regions Created	Integrals Needed
2	1	1
3	2	2
4	3	3
$n$	$n-1$	$n-1$

Key Takeaways

Multiple intersections create multiple regions, and each needs its own integral

Find all intersections: set the functions equal and solve completely

Test each region. Don’t assume the same function stays on top

One integral per region: number of integrals = (boundaries) − 1, where the boundaries are the intersections plus any other given endpoints

Add the areas together for the total

Calculator shortcut: $\int_{a}^{c} |f(x) - g(x)|\,dx$ works, but doesn’t show full work

Absolute value placement: inside the integral, not outside

Store intersection values on your calculator for precision with decimals

Order matters: always subtract bottom from top (or left from right for $dy$ )

Sketch a rough graph to see which function is where

Parentheses are essential when subtracting multi-term functions

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