Area Between Two Curves

The area trapped between two curves is an integral of vertical gaps: slice the region into thin strips, and each strip's height is top minus bottom. Add them all up and you get \(A = \int_a^b (\text{top} - \text{bottom})\,dx\).

By the end you'll be able to find where two curves enclose a region, set up the top-minus-bottom (or right-minus-left) integral, and evaluate it to get the exact area.

Predict: on [0, 1], which curve is on top — the line y = x or the parabola y = x²? Drag the vertical strip and read top − bottom to check.

The two curves cross at x = 0 and x = 1 and enclose the shaded region between them. The strip's height at position x is the vertical gap x − x². Raise the strip count to stack thin rectangles across the region and watch their total close in on the exact area, A = 1/6. Then tick Explore outside [0, 1] and drag the strip past the crossings — the curves swap and the parabola takes over on top, which is exactly why the integral stops at the crossing points.

Strip at x = 0.50 · top: line y = x − bottom: parabola y = x² · height = 0.500 − 0.250 = 0.250

8 strips: total ≈ 0.1680 · exact area A = 1/6 ≈ 0.1667

y = x and y = x² — shaded where they bound a region; strip height = top − bottom
y = x y = x² strip height (top − bottom)

The area between two curves is the integral of top minus bottom over the interval where they bound the region — or right minus left in \(y\), when the region is described sideways.

Formal

If \(f(x) \ge g(x)\) on \([a, b]\), the area between them is \[A = \int_a^b \big(f(x) - g(x)\big)\,dx.\] Because you always subtract the lower curve from the upper one, the integrand is non-negative and the result is a true positive area — even when part or all of the region sits below the x-axis. The recipe: sketch both curves, solve \(f(x) = g(x)\) to find the crossing points (those are your limits \(a\) and \(b\)), test a sample point to see which curve is on top, then integrate the difference. If the curves swap order inside the interval, split the integral at each crossing so every piece is top minus bottom.

Visual

Picture the shaded region in the demo above sliced into thin vertical strips. Each strip is nearly a rectangle: width \(dx\), height equal to the vertical gap between the top curve and the bottom curve at that x. The integral is what the Riemann sum of those rectangle areas becomes as the strips get infinitely thin — you watched the strip total march toward 1/6 as n grew. This generalizes plain area under a curve: there, the "bottom" is just the x-axis, \(g(x) = 0\).

Applied

Two cars leave a stoplight, one with velocity curve \(v_1(t)\) that starts faster and one with \(v_2(t)\) that catches up. The area between the velocity curves up to the moment they cross is exactly the lead — in meters — the first car built up: at every instant the gap \(v_1 - v_2\) is how fast the lead grows, and integrating that gap accumulates it. Same idea for revenue-rate vs. cost-rate curves: the area between them over a time window is the total profit earned in it.

When the region lies sideways: integrate in y

Some regions have a clean left and right boundary instead of a clean top and bottom — say, bounded by \(x = y^2\) on the left and \(x = 4\) on the right. Rather than splitting the region into multiple x-integrals, describe the curves as x in terms of y and integrate horizontal strips, right minus left: \[A = \int_c^d \big(\text{right}(y) - \text{left}(y)\big)\,dy.\] One warning: don't mix variables. If the integrand is in \(y\), the limits must be y-values and the differential must be \(dy\).

Worked example

Find the area enclosed by \(y = x\) and \(y = x^2\). Crossings: \(x = x^2\) gives \(x = 0\) and \(x = 1\). Top: at \(x = 0.5\), the line gives \(0.5\) and the parabola \(0.25\), so \(y = x\) is on top. Integrate: \[A = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \mathbf{\frac{1}{6}}.\]

Your turn

Find the area enclosed by \(y = 2x\) and \(y = x^2\). Crossings: \(2x = x^2\) gives \(x = 0\) and \(x = 2\). Top: at \(x = 1\), the line gives \(2\) and the parabola \(1\), so \(y = 2x\) is on top. Integrate: \[A = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = \;?\] Finish the evaluation: \(4 - \) ____ \( = \) ____

Reveal the answer

\(A = \left(4 - \dfrac{8}{3}\right) - 0 = \mathbf{\dfrac{4}{3}}\). Positive, as an area must be — if you'd gotten a negative number, that's the signal you subtracted top from bottom instead.

More info — why top minus bottom is just two areas subtracted

You already know that definite integrals split over differences: \(\int_a^b (f - g)\,dx = \int_a^b f\,dx - \int_a^b g\,dx\). So the area between the curves is literally the area under the top curve with the area under the bottom curve carved away. When both curves sit above the x-axis this picture is exact; the remarkable part is that the formula keeps working when curves dip below the axis, because the two signed-area errors cancel in the subtraction. The OpenStax chapter in Dive deeper below walks through this derivation with figures.

Check your understanding

Question 1 of 4

Find the area of the region enclosed by \(y = x^2\) and \(y = x^3\).

Question 2 of 4

Curves \(f\) and \(g\) cross at \(x = a\), \(x = m\), and \(x = b\) (with \(a < m < b\)); \(f\) is on top on \((a, m)\) and \(g\) is on top on \((m, b)\). Which setup gives the total enclosed area?

Question 3 of 4

The region bounded by \(x = y^2\) and \(x = 4\) is easiest to capture with which single integral?

Question 4 of 4

Compute the area enclosed by \(y = x\) and \(y = x^2\) by evaluating \(\int_0^1 (x - x^2)\,dx\) with the antiderivative \(F(x) = \frac{x^2}{2} - \frac{x^3}{3}\).

Recap

  • Area between curves: \(A = \int_a^b (\text{top} - \text{bottom})\,dx\) — always subtract the lower curve from the upper one, so the answer is a true positive area.
  • The limits \(a\) and \(b\) come from solving \(f(x) = g(x)\); they mark where the curves bound the region.
  • Check which curve is on top with a sample point; if the curves swap order, split the integral at each crossing.
  • For regions described sideways, integrate in \(y\): \(A = \int_c^d (\text{right} - \text{left})\,dy\) — and never mix \(dx\) with y-limits or vice versa.

Dive deeper

Sources

  • Area Between Two Curves