Calculate geometric area
A region that grows
The line \(y = 4x\) and the curve \(y = x^3\) meet at \(x = 0\) and at \(x = 2\), and the line is the upper curve.
Drag the right edge b to the right. The teal region grows, and the readout shows its area.
The area up to \(b\) is the integral of the strip length \(4x - x^3\) from 0 to \(b\), which equals \(2b^2 - \tfrac{b^4}{4}\).
At \(b = 1\) the area is 1.75, and at \(b = 2\) the region is complete with area 4. The result is in square units, because each strip is a length times a width.
Any two curves
The curve \(f\) is the upper curve and \(g\) is the lower curve. Let \(H\) be an antiderivative of the strip length \(f - g\), so that \(H' = f - g\).
Move the right edge \(b\). The area from \(a\) to \(b\) is \(H(b) - H(a)\), whatever the curves are.
Show the strips. Every strip has a positive length, so the area is positive.
Notation
Integrate the upper curve minus the lower curve. The integrand is the function inside the integral, and \(H\) is an antiderivative of it. The area is \(H\) at the right limit minus \(H\) at the left limit.
\[\begin{aligned} \cb{A} &= \cb{\int_a^b} \big(f(x) - g(x)\big)\,dx \\ &= H(b) - H(a) \end{aligned}\]The result is positive, because the upper curve is above the lower curve on the whole interval. If the result is negative, check the order of the curves.
This needs \(f\) and \(g\) continuous on \([a, b]\), with \(f(x) \ge g(x)\) there.
Report the area in square units. If \(x\) and \(y\) are in meters, the area is in square meters.
\[\cb{A}\ \text{square units}\]If the curves cross inside the interval, split the region at the crossing. Compute each piece with its own upper and lower curve, and add the areas.
When it fails
A single integral from \(a\) to \(b\) needs \(f\) above \(g\) on the whole interval. Push the right limit \(b\) past the point where the curves meet.
Past that point \(g\) is on top, so \(f(x) - g(x)\) is negative. The red part counts as negative in the integral, so the integral is smaller than the area.
Example 1: a line and a parabola
Find the area of the region between \(y = x + 2\) and \(y = x^2\).
The curves meet where \(x^2 = x + 2\), so \(x^2 - x - 2 = (x + 1)(x - 2) = 0\) and the limits are \(-1\) and 2. At \(x = 0\) the line is 2 high and the parabola is 0 high, so the line is the upper curve.
\[\cb{\int_{-1}^{2}} \big(x + 2 - x^2\big)\,dx\]An antiderivative of the integrand is \(H(x) = \frac{x^2}{2} + 2x - \frac{x^3}{3}\). Evaluate it at the two limits.
\[\begin{aligned} H(2) &= 2 + 4 - \tfrac{8}{3} = \tfrac{10}{3} \\ H(-1) &= \tfrac{1}{2} - 2 + \tfrac{1}{3} = -\tfrac{7}{6} \end{aligned}\]Subtract the value at the left limit from the value at the right limit.
\[\tfrac{10}{3} - \left(-\tfrac{7}{6}\right) = \tfrac{27}{6} = \tfrac{9}{2}\]The result is positive, so the order of the curves was correct.
\[\cb{\tfrac{9}{2}\ \text{square units}}\]
Example 2: a curve that crosses the axis
Find the area of the region between \(y = x^3 - 4x\) and the x-axis, for \(-2 \le x \le 2\).
The curve meets the x-axis where \(x^3 - 4x = 0\), at \(x = -2, 0, 2\). At \(x = -1\) it is 3 high, so it is above the axis from \(-2\) to 0. At \(x = 1\) it is \(-3\) high, so it is below the axis from 0 to 2.
\[x^3 - 4x = x(x - 2)(x + 2)\]A tempting wrong move: Integrate the curve from −2 to 2, and report the result.
\[\int_{-2}^{2} (x^3 - 4x)\,dx = 0\]The red part below the axis counts as negative, so it cancels the teal part above the axis and the integral is 0. But the region has two pieces, each with a positive area.
Split at \(x = 0\). On the first piece the curve is the upper curve, and on the second piece the x-axis is the upper curve.
\[\begin{aligned} &\int_{-2}^{0} (x^3 - 4x)\,dx \\ &\quad + \int_{0}^{2} \big(0 - (x^3 - 4x)\big)\,dx \end{aligned}\]An antiderivative of \(x^3 - 4x\) is \(H(x) = \frac{x^4}{4} - 2x^2\), with \(H(-2) = -4\), \(H(0) = 0\) and \(H(2) = -4\). The minus sign in the second integral turns \(H(2) - H(0)\) into \(H(0) - H(2)\). Each piece is 4, so the area is 8.
\[\begin{aligned} H(0) - H(-2) &= 0 - (-4) = 4 \\ H(0) - H(2) &= 0 - (-4) = 4 \end{aligned}\]
Practice
Find the area of the region enclosed by \(y = \sqrt{x}\) and \(y = x^2\).
Move the right edge to 1. Which curve is the upper curve between the two meeting points?
Solution
The curves meet at \(x = 0\) and \(x = 1\). At \(x = 0.25\) the square root is 0.5 high and the parabola is 0.0625 high, so the square root is the upper curve.
\[\cb{\int_0^1} \big(\sqrt{x} - x^2\big)\,dx\]An antiderivative is \(H(x) = \frac{2}{3}x^{3/2} - \frac{x^3}{3}\). Evaluate it at the two limits, and subtract.
\[H(1) - H(0) = \left(\tfrac{2}{3} - \tfrac{1}{3}\right) - 0\]The area is one third of a square unit.
\[\cb{\tfrac{1}{3}\ \text{square unit}}\]
Questions
The graph shows the curve \(y = x^2 - 3x\). From \(x = 0\) to \(x = 3\) it lies below the x-axis, and that part is red. After \(x = 3\) it lies above the axis.
What is the area of the region between the curve and the x-axis from \(x = 0\) to \(x = 3\)?
- \(\tfrac{9}{2}\)
- \(-\tfrac{9}{2}\)
- 9
Answer
\(\tfrac{9}{2}\)
What is the area of the region between the curve and the x-axis from \(x = 0\) to \(x = 4\)?
- \(\tfrac{19}{3}\)
- \(-\tfrac{8}{3}\)
- \(\tfrac{8}{3}\)
- \(\tfrac{11}{6}\)
Answer
\(\tfrac{19}{3}\)
What is the area of the region between \(y = e^x\) and \(y = e^{-x}\) from \(x = 0\) to \(x = 1\)?
- \(e + \tfrac{1}{e} - 2\)
- \(e - \tfrac{1}{e}\)
- \(2 - e - \tfrac{1}{e}\)
- \(e - 1\)
Answer
\(e + \tfrac{1}{e} - 2\)
Summary
You subtract the lower curve from the upper curve, integrate between the limits, and report a positive result in square units.
Each strip has a positive length, so the integral adds only positive areas.
You use it to find the area of a region between two curves, and you split the region where the curves cross.
This is one of four skills in topic 8.4 of Which & Why: AP Calculus.