Area between curves intersecting more than twice

AP Calculus BC· AP Calculus BC CED — Applications of Integration· 14 min read

1. Core Concepts: When the Basic Area Formula Fails★★☆☆☆⏱ 3 min

The single integral formula $ int_a^b (f(x) - g(x)) dx $o n l y g i v escor r ec tt o t a l a r e ai f o n e f u n c t i o ni s g r e a t er t han or e q u a l t o t h eo t h er a cr oss t h ee n t i r e in t er v a l$ [a,b]$. When curves intersect more than twice, the upper/lower order swaps, so the basic formula will not work.

📘 Definition

Total Area

$A_{total}$

The sum of all non-negative bounded regions between two curves over a given interval. The AP exam almost always asks for total area when it says 'area'.

Example:

For curves intersecting 3 times, total area adds the area of two separate regions between the curves.

By contrast, net signed area allows positive and negative regions to cancel out, and can be calculated with a single integral regardless of intersections. It is only requested if explicitly labeled on the AP exam.

✓ Quick check

An AP exam question asks for 'the area of the regions bounded between the two curves'. What is requested?

- Net signed area
- Total area
- Cannot be determined without more information
Reveal answer
1 —
AP exam convention: 'area' of bounded regions always means non-negative total area.

Exam tip:

Always confirm if the question explicitly asks for net area; over 90% of the time, it will request total area.

2. Finding Intersections and Splitting the Integration Interval★★☆☆☆⏱ 4 min

The most critical step in solving any multiple-intersection area problem is identifying all intersection points between the two curves within the interval of interest. Intersections are the only points where the upper/lower order can swap, so we use these points to split the full interval into smaller subintervals.

After ordering intersections from smallest to largest $x_{0} < x_{1} < ... < x_{n}$ , the general formula for total area integrating with respect to $x$ is:

A = \sum_{i=1}^n \int_{x_{i-1}}^{x_i} \left( u_i(x) - l_i(x) \right) dx

where $u_{i} (x)$ is the upper function and $l_{i} (x)$ is the lower function on the $i$ -th subinterval. Always confirm the order by testing a point inside each subinterval.

📐 Worked Example

Find all intersection points of $f (x) = x^{3} - 4 x$ and $g (x) = x$ , and split the interval from the leftmost to rightmost intersection into valid subintervals.

1
Set $f (x) = g (x)$ to solve for intersections:
$x^3 - 4x = x \implies x^3 - 5x = 0 \implies x(x^2 - 5) = 0$
2
Solve for all roots, giving 3 intersection points:
$x = -\sqrt{5} \approx -2.236, \quad x = 0, \quad x = \sqrt{5} \approx 2.236$
3
$[- sqrt{5}, 0] $an d$ [0, \sqrt{5}]$
4
Test function order on each subinterval: For $x = - 1 \in [- 5, 0]$ , $f (- 1) = 3$ and $g (- 1) = - 1$ , so $f (x) > g (x)$ . For $x = 1 \in [0, 5]$ , $f (1) = - 3$ and $g (1) = 1$ , so $g (x) > f (x)$ .

Exam tip:

Always factor $f (x) - g (x)$ completely to avoid missing roots; for higher-degree polynomials, check for common factors first.

3. Calculating Total Area vs Net Signed Area★★★☆☆⏱ 4 min

A core AP exam skill is distinguishing between total area and net signed area for curves that intersect multiple times. Net signed area can be calculated with one integral, but total area requires splitting the interval at every intersection.

📐 Worked Example

Given $f (x) = x^{3} - 4 x$ and $g (x) = x$ between $x = - 5$ and $x = 5$ , calculate (a) net signed area, (b) total area between the curves.

1
Part (a): Net Signed Area: Calculate directly with a single integral:
$\int_{-\sqrt{5}}^{\sqrt{5}} (f(x) - g(x)) dx = \int_{-\sqrt{5}}^{\sqrt{5}} (x^3 - 5x) dx$
2
The integrand is an odd function integrated over a symmetric interval around 0, so net signed area equals:
$0$
3
Part (b): Total Area: Use the subintervals and order from the previous example to write the split integral:
$A = \int_{-\sqrt{5}}^0 (x^3 - 5x) dx + \int_0^{\sqrt{5}} (5x - x^3) dx$
4
Evaluate the first integral:
$\left[\frac{x^4}{4} - \frac{5x^2}{2}\right]_{-\sqrt{5}}^0 = 0 - \left(\frac{25}{4} - \frac{25}{2}\right) = \frac{25}{4}$
5
Evaluate the second integral:
$\left[\frac{5x^2}{2} - \frac{x^4}{4}\right]_0^{\sqrt{5}} = \left(\frac{25}{2} - \frac{25}{4}\right) - 0 = \frac{25}{4}$
6
Sum the areas to get total area:
$A = \frac{25}{4} + \frac{25}{4} = \frac{25}{2} = 12.5$

✓ Quick check

What is the total area between $f (x) = sin x$ and $g (x) = cos x$ over the interval $[0, 2 π]$ ?

- 0
- 2\sqrt{2}
- 4\sqrt{2}
- 4
Reveal answer
2 —
Option A is the net area, which is incorrect. Splitting into 3 subintervals and integrating upper minus lower gives $42$ .

Exam tip:

If the question says 'the area of the regions bounded between the two curves', it always asks for total area, not net area. Only use a single integral for net area if explicitly requested.

4. Integration With Respect to $y$ for Multiple Crossings★★★☆☆⏱ 3 min

✓ Calculator OK

When curves are given as functions of $y$ ( $x = f (y)$ and $x = g (y)$ ), or when integrating with respect to $y$ simplifies calculation, the same core logic applies, adjusted for the variable of integration. We find all intersection $y$ -values, order them, split into subintervals, then check which function gives the rightmost (larger) $x$ -value on each subinterval.

The general formula for total area integrating with respect to $y$ is:

A = \sum_{i=1}^n \int_{y_{i-1}}^{y_i} \left( r_i(y) - l_i(y) \right) dy

where $r_{i} (y)$ is the rightmost $x$ -value and $l_{i} (y)$ is the leftmost $x$ -value on the $i$ -th subinterval.

📐 Worked Example

Find the total area between $x = y^{3} - 2 y$ and $x = y$ from the lowest intersection to the highest intersection.

1
Find intersections by setting the functions equal:
$y^3 - 2y = y \implies y^3 - 3y = 0 \implies y(y - \sqrt{3})(y + \sqrt{3}) = 0$
2
Intersections are at $y = - 3, 0, 3$ . Split into subintervals and test order: For $y = - 1 \in [- 3, 0]$ , $y^{3} - 2 y = 1$ which is larger than $x = - 1$ , so $y^{3} - 2 y$ is rightmost. For $y = 1 \in [0, 3]$ , $y = 1$ is larger than $y^{3} - 2 y = - 1$ , so $y$ is rightmost.
3
Write the total area integral:
$A = \int_{-\sqrt{3}}^0 (y^3 - 3y) dy + \int_0^{\sqrt{3}} (3y - y^3) dy$
4
Evaluate both integrals and sum for total area:
$\frac{9}{4} + \frac{9}{4} = \frac{9}{2} = 4.5$

Exam tip:

Never confuse the formula when integrating with respect to $y$ : it is always (right $x$ minus left $x$ ), not (top $y$ minus bottom $y$ ) — mixing this up is a common FRQ point deduction.

5. Common Pitfalls

Wrong move:

Stopping at two intersection points when $f (x) - g (x)$ has higher degree, missing roots inside the interval.

Why:

Students often factor out one root and forget to solve the remaining polynomial, assuming only two intersections for any two curves.

Correct move:

Always count the degree of $f (x) - g (x)$ and confirm you have found all roots that fall inside your interval before integrating.

Wrong move:

Using a single integral of $f (x) - g (x)$ over the entire interval, calculating net area instead of requested total area.

Why:

Confusion between the two definitions, or forgetting that the upper function swaps after each intersection.

Correct move:

If asked for area, always split the interval at every intersection and integrate upper minus lower on each subinterval.

Wrong move:

Flipping the order of upper and lower function on a subinterval, leading to a negative contribution to total area.

Why:

Picking a test point outside the subinterval, or making an arithmetic error when evaluating function values.

Correct move:

Always pick a test point strictly inside the subinterval, write down both function values explicitly, and confirm their order before integrating.

Wrong move:

When integrating with respect to $y$ , using (top $y$ minus bottom $y$ ) instead of (right $x$ minus left $x$ ).

Why:

Muscle memory from integrating with respect to $x$ leads to mixing up the formula.

Correct move:

Remind yourself that the integrand is always (larger variable value minus smaller variable value) for the variable of integration.

Wrong move:

Forgetting to include endpoints of the interval that are also intersections, leading to an incorrect number of subintervals.

Why:

Students only solve for intersections in the open interval and ignore endpoints.

Correct move:

Always check if the endpoints of your interval are intersections, and add them to your ordered list of split points if they are.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Total Area (w.r.t $x$ )	$A = \sum_{i = 1}^{n} \int_{x_{i - 1}}^{x_{i}} (u_{i} (x) - l_{i} (x)) d x$	Ordered intersections; $u_{i}$ = upper, $l_{i}$ = lower; always positive
Total Area (w.r.t $y$ )	$A = \sum_{i = 1}^{n} \int_{y_{i - 1}}^{y_{i}} (r_{i} (y) - l_{i} (y)) d y$	Ordered intersections; $r_{i}$ = rightmost $x$ , $l_{i}$ = leftmost $x$
Absolute Value Form	$A = \int_{a}^{b} ∣ f (x) - g (x) ∣ d x$	Equivalent to split sum; evaluated by splitting at intersections
Net Signed Area	$Net Area = \int_{a}^{b} (f (x) - g (x)) d x$	Can be positive/negative/zero; cancels opposite regions
First Step	Solve $f (x) = g (x)$ for all roots in $[a, b]$	Missing any root inside the interval guarantees an incorrect answer
Test Point Check	Evaluate $f (x) - g (x)$ at one point per subinterval	If positive, $f$ is upper; if negative, $g$ is upper; confirms order

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
3-point multiple choice question
2023 · FRQ
4-point subpart of full question

Going deeper

Unit OverviewAP Calculus BC Unit 8 OverviewParent unit learning objectives

What's Next

Mastering area between multiple-intersection curves is a critical prerequisite for upcoming topics in AP Calculus BC Unit 8. When calculating volumes of solids with the washer and shell methods, you need to correctly identify inner/outer radii or bounds of integration, which relies on the same skill of splitting intervals at intersections learned here. This topic also builds the foundation for calculating arc length and surface area of curves, where similar absolute value integration techniques are used. Beyond the AP exam, this core skill of splitting intervals at crossing points is used for improper integrals and line integrals in future college calculus courses.