Module 5 · Lesson 5.4

Area Between Curves

Find the gap between two curves by stacking two separate accumulations and subtracting one from the other.

Intuition

You already know how to find the area under one curve. What if you want the area trapped between two curves instead — the strip that sits above one and below the other? There's no new idea needed: run the accumulation once for the top curve, run it again for the bottom curve, and subtract.

Core concept

At each thin vertical slice, the strip's height is simply the top curve's value minus the bottom curve's value, $f(x) - g(x)$. Summing those slices from $a$ to $b$ gives the total trapped area. That's exactly the difference between two ordinary definite integrals — one accumulation stacked against another and subtracted.

revenue above cost
Revenue & costR(x)=4x, C(x)=x²
Interval0 to 3
Difference function4x − x²
Antiderivative2x² − x³/3
Evaluate at 3 minus at 0(18 − 9) − 0
AnswerProfit area = 9

Interactive Lab

Slide the upper bound to grow the shaded region between the revenue curve (top) and the cost curve (bottom). Watch each accumulation total climb, and see the profit area track their difference.

R(x) = 4x C(x) = x²
x = 3.0 revenue area = 18.00 cost area = 9.00 profit area = 9.00

Key Idea

$$ \text{Area} = \int_a^b \big[f(x) - g(x)\big]\,dx $$

This holds whenever $f(x) \ge g(x)$ across the whole interval — the top curve never dips below the bottom one. Where that stops being true, the interval needs to be split at the crossing point, since the "top" and "bottom" curves swap roles there.

$f(x)$The upper curve on the interval
$g(x)$The lower curve on the interval
$a, b$The left and right bounds where the trapped region begins and ends
$f(x)-g(x)$The height of the thin vertical strip at each point

Think Further

  1. Revenue R(x) = 4x and cost C(x) = x² cross again at x = 4. Predict what happens to the shaded "profit area" region as the slider passes x = 4, and why.
  2. If a company's cost curve briefly rises above its revenue curve for part of an interval, how should the area-between-curves setup change to still represent a meaningful profit total?
  3. Two accumulation totals, revenue area and cost area, both grow as x increases in the worked example. Why can their difference (profit area) still shrink even while both individual totals are growing?
Show suggested answers
  1. Past x = 4, the cost curve rises above the revenue curve, so C(x) > R(x). The region there represents a loss, not a profit, so simply continuing $\int (R-C)\,dx$ would start subtracting area rather than adding it.
  2. The interval should be split at the crossing point, integrating $R(x)-C(x)$ where revenue leads and $C(x)-R(x)$ where cost leads, then combining the pieces — otherwise a loss region would silently cancel out a profit region.
  3. Because the difference only shrinks when the cost curve is growing faster than the revenue curve at that stretch — the gap between the two accumulation rates, not their absolute sizes, determines whether the profit area is widening or narrowing.