Premed · Premed · Calculus 1
Lecture 22: Applications — Areas Between Curves
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- Set up and evaluate integrals for the area between two curves
- Determine which function is on top over a given interval
- Handle regions where curves cross (split into sub-intervals)
- Compute areas by integrating with respect to y when appropriate
- Apply these techniques to practical problems
Lecture Content
I. Area Between Two Curves — Basic Setup
Given two continuous functions with f(x) >= g(x) on [a, b], the area of the region between them is A = integral from a to b of [f(x) - g(x)] dx. Here f(x) is the upper curve (top boundary) and g(x) is the lower curve (bottom boundary). This formula works even when both functions take negative values, because only their difference matters. The familiar area between a curve and the x-axis is the special case where g(x) = 0.
II. Identifying the Top and Bottom Curves
Before integrating, it is essential to determine which function is larger on the interval in question. A practical method is to evaluate both functions at a test point c in the interval: if f(c) > g(c), then f is on top in that sub-interval. If the curves cross within [a, b], the roles of "top" and "bottom" switch at the crossing points. These crossing points are found by solving f(x) = g(x).
III. Regions Where Curves Cross
When f and g cross at points within [a, b], the integral must be split at each crossing point. The general formula is A = integral from a to b of |f(x) - g(x)| dx, which in practice means splitting into sub-intervals where one function dominates the other.
For example, to find the area between y = x and y = x^2 on [0, 2], first find the intersections: x = x^2 gives x = 0 and x = 1. On [0, 1], x >= x^2, so the integrand is x - x^2. On [1, 2], x^2 >= x, so the integrand is x^2 - x. The total area is A = integral from 0 to 1 of (x - x^2) dx + integral from 1 to 2 of (x^2 - x) dx.
<image>A graph showing y = x and y = x^2 on [0, 2]. The curves intersect at (0, 0) and (1, 1). The region between the curves on [0, 1] (where x > x^2) is shaded blue. The region between the curves on [1, 2] (where x^2 > x) is shaded green. Each region is labeled with its integral formula. Title: "Area between curves that cross: split at the intersection."</image>
IV. Worked Examples
To find the area between y = x^2 and y = x + 2, solve for intersections: x^2 = x + 2 gives x^2 - x - 2 = 0, so (x - 2)(x + 1) = 0 and x = -1, 2. Testing at x = 0 shows 2 > 0, so x + 2 is on top throughout [-1, 2]. The area is A = integral from -1 to 2 of [(x + 2) - x^2] dx = [x^2/2 + 2x - x^3/3] from -1 to 2 = (2 + 4 - 8/3) - (1/2 - 2 + 1/3) = 10/3 + 7/6 = 9/2.
For the area enclosed by y = sin x and y = cos x from x = 0 to x = 2pi, the intersections occur at x = pi/4 and x = 5pi/4. Determining which function is on top in each sub-interval requires checking test points, and the computation must be split into three intervals: [0, pi/4], [pi/4, 5pi/4], and [5pi/4, 2*pi].
V. Integrating with Respect to y
Sometimes it is easier to integrate horizontally, with respect to y. If the region is bounded by x = f(y) on the right and x = g(y) on the left, for c <= y <= d, then A = integral from c to d of [f(y) - g(y)] dy. This approach is especially useful when the boundary curves are given as functions of y, or when integrating with respect to x would require splitting into multiple integrals.
For example, to find the area between y^2 = x and x = 2y + 3, express both as functions of y: the right boundary is x = 2y + 3 and the left boundary is x = y^2. The intersections come from y^2 = 2y + 3, giving y^2 - 2y - 3 = 0 and (y - 3)(y + 1) = 0, so y = -1 and y = 3. The area is A = integral from -1 to 3 of [(2y + 3) - y^2] dy = [y^2 + 3y - y^3/3] from -1 to 3 = (9 + 9 - 9) - (1 - 3 + 1/3) = 9 + 5/3 = 32/3.
<image>Panel A: The region between y^2 = x (a rightward-opening parabola) and x = 2y + 3 (a line). The region is shaded, with horizontal strips indicating integration with respect to y. The intersections at y = -1 and y = 3 are marked. Panel B: The same region shown with vertical strips, illustrating that integrating with respect to x would require two separate integrals (different formulas above and below the vertex). The horizontal approach (Panel A) is clearly simpler. Title: "Choosing the right variable of integration."</image>
VI. Area Between Curves — General Principles
Several general principles guide the computation. Always sketch the region first. Find all intersection points to determine the limits of integration. Determine which curve is "outer" (top or right) on each sub-interval. The integrand is always (outer) - (inner), which gives a positive quantity. Choose to integrate with respect to x or y depending on which approach is simpler. As a sanity check, estimate whether the area is reasonable by comparing to the area of a bounding rectangle.
VII. Applications
The technique of computing areas between curves has applications across many fields. In economics, the area between supply and demand curves gives consumer and producer surplus. In engineering, it can determine the cross-sectional area of irregular shapes. In experimental science, the area between a measured curve and a theoretical prediction quantifies the discrepancy. These two-dimensional techniques also extend naturally to three-dimensional applications involving volumes, which are the subject of the next lecture.
<image>An economics-themed graph showing a supply curve (upward-sloping) and a demand curve (downward-sloping) intersecting at equilibrium point (Q, P). The area between the demand curve and the horizontal price line P (above equilibrium) is shaded blue and labeled "Consumer Surplus." The area between the supply curve and the horizontal price line P (below equilibrium) is shaded green and labeled "Producer Surplus." Title: "Areas between curves in economics: consumer and producer surplus."</image>
