Area Between Curves
In short: Area between curves is the size of the patch that two graphs wall off between them, computed as the integral of the higher graph minus the lower one across the stretch where they bound it.
AP Calculus ABCalculusAP Calculus AB Unit 8
What is Area Between Curves?
The area between curves measures the space that two graphs fence in. Cut that space into thin upright strips; each strip is as tall as the gap between the top graph and the bottom graph, and \(dx\) wide.
If \(f(x) \ge g(x)\) on \([a, b]\), then the region has area \(\int_a^b \left[f(x) – g(x)\right]dx\). The limits are usually the \(x\)-coordinates where the graphs intersect. If the curves cross inside the interval, split the integral at the crossing so that the top function is always the one written first.
How to spot it
- The wording asks for the size of the shape trapped by two named graphs.
- A picture shows two graphs forming a closed shape, like a lens or a bowl.
- No limits are given, which means you must solve f(x) = g(x) to find them.
- The curves swap which is on top, so the region needs more than one integral.
The method
- Sketch both curves and find the intersection points by solving \(f(x) = g(x)\).
- Decide which function is on top between consecutive intersections.
- Set up \(\int (\text{top} – \text{bottom})\,dx\) with the intersection \(x\)-values as limits.
- Integrate and evaluate. The result is positive and measured in squared units, which is a quick sanity check on area between curves.
Area Between Curves: solved example
Problem
How much area do the graphs of \(y = 2x + 3\) and \(y = x^2\) wall off between them?
Solution
Intersections: \(x^2 = 2x + 3\) gives \(x^2 – 2x – 3 = (x – 3)(x + 1) = 0\), so \(x = -1\) and \(x = 3\). The line lies higher in between (at \(x = 0\): 3 versus 0).
Area \(= \int_{-1}^{3} (2x + 3 – x^2)\,dx = \left[x^2 + 3x – \frac{x^3}{3}\right]_{-1}^{3}\).
At 3: \(9 + 9 – 9 = 9\). At \(-1\): \(1 – 3 + \frac{1}{3} = -\frac{5}{3}\). The area is \(9 + \frac{5}{3} = \) \(\frac{32}{3}\) square units.
Try it
Question 1
Find the area between \(y = \sqrt{x}\) and \(y = x\) for \(0 \le x \le 1\).
Show the answer
On this interval \(\sqrt{x} \ge x\). \(\int_0^1 (\sqrt{x} – x)\,dx = \frac{2}{3} – \frac{1}{2} = \frac{1}{6}\).
Question 2
Find the total area enclosed by \(y = x\) and \(y = x^3\).
Show the answer
They meet at \(-1, 0, 1\) and swap positions at 0. By symmetry the area is \(2\int_0^1 (x – x^3)\,dx = 2\left(\frac{1}{2} – \frac{1}{4}\right) = \frac{1}{2}\).
Area Between Curves vs. Definite Integral
A single definite integral gives the area between one curve and the x-axis. The area between curves subtracts one integral from the other, or equivalently integrates a difference. Using only the top curve would count the strip all the way down to the axis, which is too much.
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Whole dictionary: A–Z · AP Calculus AB · Calculus