A stepbystep exploration of how to find the region bounded by two or more functions. In calculus, the definite integral gives the signed area under a single curve. Realworld problems, however, often involve two (or more) curves that intersect, creating a strip whose area we need. Examples include calculating the difference in revenue between two pricing models, the volume of a tank formed by two rotating profiles, or the probability that a random point falls inside a specific region. Mastering the technique of subtracting one integral from another unlocks a whole family of applications. If the graphs of functions f(x) and g(x) enclose a region on an interval [a, b], and f(x) g(x) throughout that interval, the areaA is The integral adds up vertical slices of widthdx, each slice having height equal to the difference between the top curve f and the bottom curve g. When the curves cross, the integral is split at the crossing points so that the correct top and bottom functions are used on each subinterval. When two curves intersect only at the endpoints of the interval, the region is a single strip. Example: y = x and y = x + 2 intersect at x = -1 and x = 2. A lensshaped region occurs when the curves cross twice, creating a bounded area that lies between them. In such cases the integral is taken from the left intersection to the right intersection with a single expression f(x) g(x) (provided the top curve does not change). For regions bounded by more than two curves, locate the outermost top and bottom functions on each subinterval, then subtract appropriately. Sometimes it is easier to integrate with respect toy, especially when the curves are given as x = h(y). Find the area enclosed by y = x and y = 4 x. The enclosed area equals (16/3)2 square units. Determine the area between y = sin x and y = cos x from 0 to /2. The area between the curves on the given interval is 2(21) square units. Sometimes integrating with respect to y is simpler. Find the area bounded by the curves x = y and x = 2 y. The regions area is 9/2 square units. When curves intersect at x = a and x = b (af(x) g(x) on that interval, the area is: If the top function changes, write: where each subinterval [a_i, b_i] corresponds to a region with a consistent ordering of the curves. Interested readers may explore these resources for deeper insight: Computing the area between curves combines algebra (to find intersection points) with integral calculus (to accumulate vertical or horizontal slices). The core workflowsketch, intersect, split, integrateremains the same regardless of the functions involved. With practice, the technique becomes an intuitive tool for a wide range of problems, from pure mathematics to realworld engineering. Areas Between Curves
1. Why Study Areas Between Curves?
2. The Core Idea
A = a [f(x) g(x)]dx
3. Setting Up the Integral A Checklist
4. Common Scenarios
4.1 One Intersection (Simple Strip)
4.2 Two Intersections (A Lens Shape)
4.3 Multiple Curves
5. Example 1 Polynomial Functions
A = 22 [(4 x) x]dx = 22 (4 2x)dx
(4 2x)dx = 4x (2/3)x
A = [4x (2/3)x]22 = (42 (2/3)(2)) (42 + (2/3)(2)) = 82 (4/3)(22) = 82 (8/3)2 = (24/3 8/3)2 = (16/3)2
6. Example 2 Trigonometric Functions
A = 0/4 (cos x sin x)dx + /4/2 (sin x cos x)dx
(cos x sin x)dx = sin x + cos x (sin x cos x)dx = cos x sin x
A = [sin x + cos x]/4 + [cos x sin x]/4/2 = (sin/4 + cos/4 (0 + 1)) + (cos/2 sin/2 + cos/4 + sin/4) = (2/2 + 2/2 1) + (0 1 + 2/2 + 2/2) = (2 1) + (2 1) = 22 2
7. Example 3 Using Horizontal Slices
A = 21 [(2 y) y]dy
(2 y y)dy = 2y (1/2) y (1/3) y
A = [2y (1/2) y (1/3) y]21 = (21 1 1) (2(2) 4 (8)) = (2 0.5 0.333) (4 2 + 2.666) = 1.166 (3.333) = 4.5
8. Applications in Science and Engineering
9. Common Mistakes to Avoid
10. Quick Reference Formula
A = ab [f(x) g(x)]dx
A = a_ib_i [top_i(x) bottom_i(x)]dx
11. Further Reading
12. Summary
