Area Between Two Curves: Step-by-Step

The area between two curves is the size of the flat region trapped between the graphs of two functions over an interval. If the curve y = f(x) sits above the curve y = g(x) from x = a to x = b, then the area between them is the integral of their vertical gap, f(x) − g(x), from a to b.

This matters because most real shapes have curved sides that no rectangle formula can handle. The area between a parabola and a straight line or the gap between two paths on a map are all area-between-curves problems. The familiar area under a single curve is just a special case where the second curve is the x-axis, g(x) = 0.

Key Concepts

• Upper and lower function: the curve with the larger y-value is the top function f and the smaller is the bottom function g; mixing them up flips the sign.
• Intersection points: solve f(x) = g(x) to find where the curves meet; these solutions usually become the limits a and b of integration.
• Vertical strip: a thin slice of width dx and height (top − bottom); integration adds them all up.
• Horizontal strips: when the curves are given as x = h(y), slice horizontally and integrate (rightmost − leftmost) with respect to y.
• Signed vs total area: an integral returns signed area. If the curves cross inside [a, b], split the integral at the crossing point so every piece stays top-minus-bottom.

Formulas

• Vertical strips: Area = ∫(f(x) − g(x)) dx from x = a to x = b, where f(x) ≥ g(x) on [a, b]. Answer is in square units.
• Curves that cross: if they cross at x = c inside [a, b], compute integrate each piece separately, always subtracting smaller from larger.
• Horizontal strips: Area = ∫(h(y) − k(y)) dy from y = c to y = d, with x = h(y) the rightmost curve.

Worked Example

Find the area enclosed by the line y = x and the parabola y = x².

1. Find intersections: set x = x², giving x = 0 and x = 1. These are the limits.
2. Decide top and bottom: at x = 1/2 the line gives 0.5 and the parabola 0.25, so the line is on top: f(x) = x, g(x) = x².
3. Set up the integral: Area = ∫₀¹ (x − x²) dx.
4. Integrate: the antiderivative is x²/2 − x³/3. Evaluating from 0 to 1: (1/2 − 1/3) − (0 − 0) = 3/6 − 2/6 = 1/6 ≈ 0.1667 square units.

Common Misconceptions

• Subtracting in any order is fine. It is not: always use top minus bottom (or right minus left). Reversing them gives a negative area, which is wrong for a geometric region.
• The limits are whatever the question prints. When the question asks for the enclosed region, the limits come from solving f(x) = g(x), not from numbers handed to you.
• One integral always suffices. If the curves cross in the middle of the interval, a single integral lets positive and negative pieces cancel. Split at every crossing point.

Real-World Applications

• Physics: the area between two velocity–time graphs gives relative displacement, such as one car pulling ahead of another.
• Economics: consumer surplus and producer surplus are areas between demand and supply curves and the market price line.
• Engineering: cross-sectional areas of machine parts bounded by two curved profiles, used to estimate material volume and weight.

The area between two curves is the integral of the gap between them, always top minus bottom, with limits taken from the intersection points and the integral split wherever the curves cross. Sketch first to see which curve is on top, and every enclosed-region problem reduces to one careful integral.