MISSION · Calculus
The Area Between Two Curves
Top minus bottom, integrated across where they overlap
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LEVEL 2 · LEARN
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The idea in one sentence
Integrate the vertical distance between the two curves — upper minus lower — across the interval where they enclose a region.
\[A = \int_a^b \left[f(x) - g(x)\right]dx\]Where it comes from
A single integral adds up rectangles of height \(f(x)\) standing on the axis. For a region between two curves, the rectangles stand on the lower curve instead, so each one has height \(f(x) - g(x)\). Nothing else changes.
That’s also why the axis stops mattering. Shift both curves up by 10 and each rectangle’s height is unchanged, so the area is the same — and the formula agrees, because the 10s cancel in the subtraction. A region partly below the axis needs no special treatment.
A worked example
\[y = x \quad \text{and} \quad y = x^2\]Find where they meet. \(x = x^2\) gives \(x(x-1) = 0\), so \(x = 0\) and \(x = 1\). These are the limits, and finding them is usually where the real work is.
Decide which is on top. Between 0 and 1, test \(x = 0.5\): the line gives 0.5, the parabola gives 0.25. The line is above.
Integrate the difference.
\[A = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\]When they cross in the middle
If the curves swap places inside the interval, one integral won’t do it — the part where you subtracted the wrong way comes out negative and cancels real area.
Split at every crossing point and integrate each piece with its own top and bottom, then add the results. Sketching first isn’t decoration; it’s how you find out that a split is needed.
Integrating with respect to y
Some regions are described much more easily sideways. If a region is bounded left and right by curves, write everything as \(x\) in terms of \(y\) and integrate:
\[A = \int_c^d \left[x_{\text{right}} - x_{\text{left}}\right]dy\]A region between \(y^2 = x\) and \(y = x - 2\) needs two integrals in \(x\), because the lower boundary changes halfway. In \(y\) it takes one. Choosing the variable well can halve the work.
Volumes are the same idea
Rotate that gap around an axis and each rectangle becomes a washer — a disc with a hole:
\[V = \pi\int_a^b \left[f(x)^2 - g(x)^2\right]dx\]Note the squares are subtracted, not the difference squared. The two are not the same, and mixing them up is the standard error in volumes of revolution.
Common mix-ups
- Getting top and bottom backwards, which gives a negative area. A negative answer is a sign flip, not a real result.
- Not solving for the intersections, and using limits the question never gave.
- Missing a crossing inside the interval, so parts of the region cancel.
- Writing \([f(x) - g(x)]^2\) in the washer formula instead of \(f^2 - g^2\).
Check yourself
Where do \(y = x\) and \(y = x^2\) meet, and what area do they enclose?
They meet at x = 0 and x = 1. The line is on top there, so the area is \(\int_0^1 (x - x^2)dx = 1/2 - 1/3 = 1/6\).
LEVEL 3 · REFLECT
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