Area between curves intersecting more than twice — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Identifying all intersection points of two curves, splitting the integration interval by crossing points, distinguishing total area vs net signed area, applying the definite integral formula for total area between curves intersecting more than twice.

You should already know: How to evaluate definite integrals of elementary functions. How to solve for roots of polynomials and transcendental functions. How to find area between two curves that intersect twice.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Area between curves intersecting more than twice?

When finding the area between two curves, the basic single-integral formula ${\int }_a^b(f(x)-g(x))dx$ only works if one function is greater than or equal to the other across the entire interval $[a,b]$. When two curves intersect more than twice, their vertical order (which is on top) swaps across the interval, so the basic formula no longer gives the correct total area. This topic is part of Unit 8 (Applications of Integration) in the AP Calculus Course and Exam Description (CED), which counts for 10-15% of the total AP exam score. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections: it commonly makes up a 3-4 point subpart of an FRQ, or 1-2 standalone MCQs per exam. Total area between multiple-intersection curves measures the entire non-negative area of all bounded regions between the two curves over an interval, unlike net signed area which allows positive and negative regions to cancel out.

2. Finding All Intersection Points to Split the Integration Interval

The first and most critical step in solving any area problem for multiple-intersection curves is identifying all intersection points between the two curves within your interval of interest. Intersections are the only points where the upper function can swap places with the lower function, so we use these points to split the full interval into smaller subintervals. By the Intermediate Value Theorem, the order of the two functions cannot change between two consecutive intersections, so one function will always be upper across the entire subinterval.

We first order the intersection points from smallest to largest: $x_0<x_1<x_2<...<x_n$. The general formula for total area becomes: $$A=\sum _{i=1}^n{\int }_{x_{i-1}}^{x_i}\left(u_i(x)-l_i(x)\right)dx$$ where $u_i(x)$ is the upper function and $l_i(x)$ is the lower function on the $i$-th subinterval. On each subinterval, we confirm the order by testing a point inside the subinterval to see which function is larger.

Worked Example

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

1. Set $f(x)=g(x)$ to find intersections: $x^3-4x=x⟹x^3-5x=0⟹x(x^2-5)=0$.
2. Solve for all roots: $x=-\sqrt{5}\approx -2.236$, $x=0$, $x=\sqrt{5}\approx 2.236$. We have 3 intersection points, meaning the curves cross twice, resulting in 2 subintervals.
3. Order intersections left to right: $x_0=-\sqrt{5}$, $x_1=0$, $x_2=\sqrt{5}$. The subintervals are $[-\sqrt{5},0]$ and $[0,\sqrt{5}]$.
4. Test the order of functions on each subinterval: for $x=-1\in [-\sqrt{5},0]$, $f(-1)=3$ and $g(-1)=-1$, so $f(x)>g(x)$ here. For $x=1\in [0,\sqrt{5}]$, $f(1)=-3$ and $g(1)=1$, so $g(x)>f(x)$ here.

Exam tip: Always factor $f(x)-g(x)$ completely to avoid missing roots; for polynomials of degree higher than 2, check for common factors first to reduce the degree and find all solutions.

3. Calculating Total Area vs Net Signed Area

A core skill tested on the AP exam for this topic is distinguishing between total area and net signed area for curves that intersect multiple times. Net signed area calculates the net difference between regions where $f(x)>g(x)$ (positive) and regions where $g(x)>f(x)$ (negative), so it can be computed as a single integral ${\int }_a^b(f(x)-g(x))dx$. By contrast, total area is always non-negative: it adds the area of every bounded region between the curves, regardless of which function is on top. When curves intersect two or fewer times, net area and total area have the same magnitude, but when they intersect three or more times, they are almost always different. AP exam questions almost always ask for total area if they just say "area"; net area is only requested if explicitly labeled.

Worked Example

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

1. Net signed area: Calculate directly as ${\int }_{-\sqrt{5}}^{\sqrt{5}}(f(x)-g(x))dx={\int }_{-\sqrt{5}}^{\sqrt{5}}(x^3-5x)dx$. The integrand is an odd function, so integrating over a symmetric interval around 0 gives a net area of $0$.
2. Total area: Use the subintervals and function order from the previous example: $f(x)\ge g(x)$ on $[-\sqrt{5},0]$, $g(x)\ge f(x)$ on $[0,\sqrt{5}]$. Write the total area as: $$A={\int }_{-\sqrt{5}}^0(x^3-5x)dx+{\int }_0^{\sqrt{5}}(5x-x^3)dx$$
3. Evaluate the integrals: First integral: ${\left[\frac{x^4}{4}-\frac{5x^2}{2}\right]}_{-\sqrt{5}}^0=0-\left(\frac{25}{4}-\frac{25}{2}\right)=\frac{25}{4}$. Second integral: ${\left[\frac{5x^2}{2}-\frac{x^4}{4}\right]}_0^{\sqrt{5}}=\left(\frac{25}{2}-\frac{25}{4}\right)-0=\frac{25}{4}$.
4. Sum the areas: $A=\frac{25}{4}+\frac{25}{4}=\frac{25}{2}=12.5$.

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 the single integral for net area if explicitly requested.

4. Integrating with Respect to $y$ for Multiple Horizontal Crossings

When curves are given as functions of $y$ ($x=f(y)$ and $x=g(y)$), or when integrating with respect to $y$ simplifies the calculation, the same core logic applies but we adjust the formula for the variable of integration. We first solve for all $y$-values where $f(y)=g(y)$, order them from lowest to highest, split the interval into subintervals between consecutive intersections, then check which function gives the rightmost (larger) $x$-value on each subinterval. The formula for total area becomes: $$A=\sum _{i=1}^n{\int }_{y_{i-1}}^{y_i}\left(r_i(y)-l_i(y)\right)dy$$ where $r_i(y)$ is the rightmost $x$-value and $l_i(y)$ is the leftmost $x$-value on the $i$-th subinterval. This is most commonly used for horizontally opening parabolas or other curves that fail the vertical line test as functions of $x$.

Worked Example

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

1. Find intersections: set $y^3-2y=y⟹y^3-3y=0⟹y(y-\sqrt{3})(y+\sqrt{3})=0$, so intersections at $y=-\sqrt{3},0,\sqrt{3}$.
2. Split into subintervals $[-\sqrt{3},0]$ and $[0,\sqrt{3}]$, test which function is rightmost: for $y=-1\in [-\sqrt{3},0]$, $x=(-1{)}^3-2(-1)=1$ and $x=-1$, so $y^3-2y$ is rightmost. For $y=1\in [0,\sqrt{3}]$, $x=1-2=-1$ and $x=1$, so $y$ is rightmost.
3. Write the total area: $$A={\int }_{-\sqrt{3}}^0\left((y^3-2y)-y\right)dy+{\int }_0^{\sqrt{3}}\left(y-(y^3-2y)\right)dy={\int }_{-\sqrt{3}}^0(y^3-3y)dy+{\int }_0^{\sqrt{3}}(3y-y^3)dy$$
4. Evaluate: First integral: ${\left[\frac{y^4}{4}-\frac{3y^2}{2}\right]}_{-\sqrt{3}}^0=0-\left(\frac{9}{4}-\frac{9}{2}\right)=\frac{9}{4}$. Second integral: ${\left[\frac{3y^2}{2}-\frac{y^4}{4}\right]}_0^{\sqrt{3}}=\frac{9}{2}-\frac{9}{4}=\frac{9}{4}$. Sum gives $A=\frac{9}{2}=4.5$.

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 (and how to avoid them)

• Wrong move: Stopping at two intersection points when $f(x)-g(x)$ has higher degree, missing one or more roots inside the interval of integration. 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 moving to integration.
• Wrong move: Using a single integral of $f(x)-g(x)$ over the entire interval, calculating net area instead of the 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. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

What is the total area between $f(x)=\sin x$ and $g(x)=\cos x$ over the interval $[0,2\pi ]$? A) $0$ B) $2\sqrt{2}$ C) $4\sqrt{2}$ D) $4$

Worked Solution: First, find all intersection points on $[0,2\pi ]$ by solving $\sin x=\cos x$, which gives solutions at $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}$, resulting in three subintervals. Check the order: $\cos x\ge \sin x$ on $[0,\frac{\pi }{4}]$ and $[\frac{5\pi }{4},2\pi ]$, while $\sin x\ge \cos x$ on $[\frac{\pi }{4},\frac{5\pi }{4}]$. Evaluate the sum of integrals of upper minus lower, which simplifies to $(\sqrt{2}-1)+2\sqrt{2}+(1+\sqrt{2})=4\sqrt{2}$. Option A is the net area, which is incorrect. The correct answer is C.

Question 2 (Free Response)

Let $f(x)=x^4-5x^2$ and $g(x)=-4$. (a) Find all $x$-values where $f(x)=g(x)$. (b) Find the total area of the regions bounded between the graphs of $f(x)$ and $g(x)$. (c) What is the net signed area between $f(x)$ and $g(x)$ over the interval from the leftmost intersection to the rightmost intersection?

Worked Solution: (a) Set $x^4-5x^2=-4⟹x^4-5x^2+4=0⟹(x^2-1)(x^2-4)=0$. All solutions are $x=-2,x=-1,x=1,x=2$. (b) Ordered intersections are $-2<-1<1<2$, giving three subintervals. Testing confirms $g(x)>f(x)$ on $[-2,-1]$ and $[1,2]$, and $f(x)>g(x)$ on $[-1,1]$. Total area is: $$A={\int }_{-2}^{-1}(-4-x^4+5x^2)dx+{\int }_{-1}^1(x^4-5x^2+4)dx+{\int }_1^2(-4-x^4+5x^2)dx$$ Using even function symmetry to simplify evaluation gives $A=\frac{44}{15}+\frac{76}{15}=8$. The total bounded area is $\boxed{8}$. (c) Net signed area is ${\int }_{-2}^2(f(x)-g(x))dx={\int }_{-2}^2(x^4-5x^2+4)dx=\frac{32}{15}\approx 2.13$. The net signed area is $\boxed{\frac{32}{15}}$.

Question 3 (Application / Real-World Style)

A city's water demand over a 4-day period is modeled by $D(t)=10+4\sin \left(\frac{\pi t}{2}\right)$ million gallons per day, and the city's water supply over the same period is modeled by $S(t)=10+4\cos \left(\frac{\pi t}{2}\right)$ million gallons per day, for $0\le t\le 4$ days. The total volume of water that needs to be moved into or out of storage to balance the system is equal to the total area between the two curves. Calculate this total volume.

Worked Solution: First, find intersections on $[0,4]$: set $\sin \left(\frac{\pi t}{2}\right)=\cos \left(\frac{\pi t}{2}\right)⟹\tan \left(\frac{\pi t}{2}\right)=1$, so solutions at $t=0.5$ and $t=2.5$. Split into three subintervals: $S(t)>D(t)$ on $[0,0.5]$ and $[2.5,4]$, $D(t)>S(t)$ on $[0.5,2.5]$. Total area is: $$A={\int }_0^{0.5}4\left(\cos \left(\frac{\pi t}{2}\right)-\sin \left(\frac{\pi t}{2}\right)\right)dt+{\int }_{0.5}^{2.5}4\left(\sin \left(\frac{\pi t}{2}\right)-\cos \left(\frac{\pi t}{2}\right)\right)dt+{\int }_{2.5}^{4}4\left(\cos \left(\frac{\pi t}{2}\right)-\sin \left(\frac{\pi t}{2}\right)\right)dt$$ Evaluating the integral gives $A=\frac{32\sqrt{2}}{\pi }\approx 14.4$ million gallons. In context, the city needs approximately 14.4 million gallons of total storage capacity to cover daily imbalances between water supply and demand over the 4-day period.

7. Quick Reference Cheatsheet

8. What's Next

Mastering area between multiple-intersection curves is a critical prerequisite for calculating volumes of solids with the washer and shell methods, the next major topic in Unit 8 Applications of Integration. When finding volumes of solids with holes or cross-sections between multiple curves, you need to correctly identify inner/outer radii or upper/lower bounds, which requires 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 later in Unit 8, where you will use similar absolute value integration techniques. Even beyond AP Calculus, this core skill of splitting intervals at crossing points is used for improper integrals and line integrals in future calculus courses.

• Area between curves (two intersections)
• Volume of revolution: washer method
• Volume of revolution: shell method
• Arc length of planar curves