The AP® exams love questions about the area between 2 graphs. These problems appear on both the AB and BC tests because they blend algebra, geometry, and integral calculus. After reading, students will know how to set up definite integrals, handle curve crossings by splitting integrals, and double-check answers with graphing technology. Mastering these techniques turns a scary topic into a predictable, step-by-step routine.

Concept Basics: What Does “Area Between 2 Graphs” Mean?

Think of two curves as slices of bread. The space trapped between them is like the filling of a sandwich. A calculus “knife” cuts that sandwich into thin strips and adds all the tiny areas together.

Vertical Slices (Top minus Bottom)

Most textbook examples use vertical strips that run parallel to the y-axis. Each strip has width \Delta x and height\text{height}= \text{top function} - \text{bottom function}.

Adding every strip from x=a to x=b leads to

\displaystyle \text{Area} = \int_{a}^{b} \bigl[f(x)-g(x)\bigr]dx.

Horizontal Slices (Right minus Left)

However, if vertical strips overlap or fail to cover the region, horizontal strips parallel to the x-axis save the day. Each horizontal slice has width \text{width}= \text{right function} - \text{left function} and thickness \Delta y. Therefore, the area is

\displaystyle \text{Area} = \int_{c}^{d} \bigl[r(y)-\ell(y)\bigr]dy.

Mini Example A

Find the area bounded by y = 3 and y = x on x=0 to x=3.

1. Top minus bottom: f(x)=3, g(x)=x.
2. Therefore, \displaystyle \text{Area}= \int_{0}^{3} (3 - x)dx.
3. Compute:
   • = \bigl[3x - \tfrac{1}{2}x^{2}\bigr]_{0}^{3}= (9 - 4.5) - 0 = 4.5 square units.

Setting Up the Definite Integral

First, locate the intersection points. This step sets the limits of integration.

• Algebraic method: solve f(x)=g(x).
• Calculator method: use “Intersect” on a TI-84 or tap curves in Desmos.

Next, decide on vertical or horizontal slices. For vertical strips, ensure the top curve stays above the bottom curve between the limits. Units also matter; area is always positive.

Example 1

Find the area between y = x^{2} and y = 2x on x\in[0,2].

1. Intersection: solve x^{2} = 2x \implies x(x-2)=0. Thus x=0,2.
2. Order check: on (0,2), 2x > x^{2}, so 2x is on top.
3. Set up integral: \displaystyle \text{Area}= \int_{0}^{2} \bigl(2x - x^{2}\bigr)dx.
4. Evaluate:
   • = \Bigl[x^{2} - \tfrac{1}{3}x^{3}\Bigr]_{0}^{2}= \bigl(4 - \tfrac{8}{3}\bigr)= \tfrac{4}{3} square units.

A quick sketch shows a parabola opening up and a straight line above it from 0 to 2, confirming the setup.

Splitting Integrals & Absolute Value

Curves often swap positions inside the interval. Therefore, a single “top minus bottom” integrand may turn negative, which ruins the area. Two flexible strategies fix this problem:

1. Splitting integrals at each intersection point and recomputing “top minus bottom” in every sub-interval.
2. Using a single integral of the absolute value: \displaystyle \text{Area}= \int_{a}^{b} \bigl|f(x)-g(x)\bigr|dx.

Example 2

Region between y = x^{3}-4x and y = -x on x\in[-2,2].

Step 1: Intersections

Solve x^{3}-4x = -x \implies x^{3}-3x = 0 \implies x(x^{2}-3)=0. Thus x=-\sqrt{3},0,\sqrt{3}\approx -1.732,0,1.732.

Method A: Splitting Integrals

Top function check:

• On [-2,-\sqrt{3}], -x is above.
• On [-\sqrt{3},0], x^{3}-4x is above.

Symmetry shows the same flip on the right.

Therefore, \displaystyle \text{Area}= \int_{-2}^{-\sqrt{3}} \bigl(-x - (x^{3}-4x)\bigr)dx + \int_{-\sqrt{3}}^{0} \bigl((x^{3}-4x) - (-x)\bigr)dx + \text{mirror}.

Because the region is symmetric, double the area of these integrals. Final numeric answer: 5 square units.

Method B: Absolute Value in One Shot

\displaystyle \text{Area}= \int_{-2}^{2} \bigl|x^{3}-3x\bigr|dx.

The numeric result matches Method A, proving both paths are equivalent.

Horizontal Distances: When dx Won’t Work

Occasionally, vertical strips overlap awkwardly—for instance, sideways parabolas or functions given as x in terms of y. In such cases, horizontal strips and dy integration shine.

Example 3

Find the area enclosed by x = y^{2} (a right-opening parabola) and x = y + 2.

1. Intersection in y: solve y^{2} = y + 2 \implies y^{2}-y-2=0. Roots: y=-1,2.
2. Right minus left: for y\in[-1,2], the line x=y+2 sits right of the parabola x=y^{2}.
3. Integral: \displaystyle \text{Area}= \int_{-1}^{2} \bigl[(y+2) - y^{2}\bigr]dy.

4. Evaluate:
   • = \Bigl[\tfrac{1}{2}y^{2} + 2y - \tfrac{1}{3}y^{3}\Bigr]_{-1}^{2}= \bigl(2 + 4 - \tfrac{8}{3}\bigr) - \bigl(\tfrac{1}{2} + (-2) + \tfrac{1}{3}\bigr)= \tfrac{9}{2} square units.

Graphing & Technology Checks

Technology reduces algebraic slips. Desmos or a TI-84 can:

• Plot both curves quickly.
• Identify intersection points with “Calc → Intersect.”
• Provide a numeric value for the definite integral under “Math → fnInt” (TI) or shaded area (Desmos).

However, remember to show all analytic steps; the AP® rubric awards points for setup, not just the final decimal.

Quick Reference Chart: Key Vocabulary

Term	Definition
Definite integral	A precise “add-them-all-up” command that returns area or net change.
Intersection points	Where two graphs meet; set f(x)=g(x) to find them.
Top function	Curve with bigger y value over an interval.
Bottom function	Curve with smaller y value in that same band.
Splitting integrals	Dividing one big integral into smaller ones when curves switch places.
Absolute value method	One integral that forces the height to stay positive by using \bigl|f-g\bigr|.
Vertical slices	Thin strips parallel to the y-axis; integrate with dx.
Horizontal slices	Strips parallel to the x-axis; integrate with dy.

Summary & Key Takeaways

Sketch the region, then identify limits by solving intersection equations. Next, choose vertical or horizontal slices. If curves trade positions, split the interval or add an absolute value. Finally, integrate and verify on a graphing utility. With this checklist, students can tackle any area between 2 graphs problem and feel ready for related topics such as volumes of revolution and average value.

Sharpen Your Skills for AP® Calculus AB-BC

Are you preparing for the AP® Calculus exam? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world math problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now!

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