Area Between Two Curves
Simple Explanation
The area of the region trapped between two curves f(x) and g(x), where f(x)β₯g(x) on [a,b], is found by integrating their difference: β«βα΅[f(x)βg(x)]dx β subtracting away the area that would otherwise be double-counted underneath the lower curve.
Why Do We Need It?
Many real regions of interest are bounded above and below by two different curves, not by a curve and the x-axis β this generalizes the area-under-a-curve idea to handle that far more common situation.
See It
A shaded lens-shaped region trapped between a straight rising line above and a parabola below, meeting at two points
Formula
Area Between Two Curves
Area = β«βα΅ [f(x) β g(x)] dx (where f(x) β₯ g(x) on [a,b])
The area of the region trapped between two curves equals the integral of the "top" function minus the "bottom" function, over the interval where the top function stays on top.
- f(x)
- β the upper (greater-valued) function on [a,b]
- g(x)
- β the lower (lesser-valued) function on [a,b]
- a, b
- β the x-values where the two curves intersect (or the given boundaries)
When to use it: Whenever the area of a region enclosed between two curves is needed β first find where they intersect, then integrate their difference.
Worked Example
Find the area between two curves
Find the area of the region bounded by f(x) = x+2 and g(x) = xΒ².
Why Does This Work?
β«[f(x)βg(x)]dx can be split into β«f(x)dx β β«g(x)dx β the area under the top curve, minus the area under the bottom curve, both measured down to the x-axis β and whatever lies below the x-axis or below g(x) cancels out exactly, leaving only the region genuinely trapped between the two curves.
Real-Life Example
Material saved by a redesigned part
An engineer redesigns a machined part, changing its outline from one curve to another, and needs to know exactly how much material the new design saves.
The saved cross-sectional area is exactly the area between the old outline curve and the new outline curve β computed with this same technique.
Practice
Find the area between f(x) = x and g(x) = xΒ², from x = 0 to x = 1.
HardCommon mistake
Integrating g(x)βf(x) instead of f(x)βg(x) (subtracting in the wrong order) β this produces a negative value instead of the true, positive area; always confirm which curve is on top first.
Quick Review
- Area = β«βα΅ [f(x)βg(x)]dx, where f(x) is the upper curve on [a,b].
- First find the intersection points β these are usually the integration bounds.
- Test a point inside the interval to confirm which curve is genuinely on top.