Combine the region areas
Two ways to add two pieces
The curve is \(f(x) = x^2\) and the line is \(g(x) = 9\). They cross at \(x = 3\), which cuts the region between the bounds \(x = 0\) and \(x = 4\) into two pieces.
Each piece has an area, 18 on the left and \(\tfrac{10}{3}\) on the right, and an area is never negative.
The integral of \(f - g\) on a piece is a signed value, negative where \(f\) is below \(g\). On the left, \(f\) is below \(g\): the integral is \(-18\) and the piece is red. On the right, it is \(\tfrac{10}{3}\) and the piece is teal.
Press the first button to add the signed values. The sum is \(-\tfrac{44}{3}\), which is not an area. Press the second button to add the areas: both pieces turn teal, and the sum is \(\tfrac{64}{3}\), the area of the region.
Areas add, signed values cancel
The two curves cross twice between the bounds, so the region has three pieces. Drag the slider to move \(g\) up or down: the crossings move, and the pieces change shape.
The red pieces, where \(f\) is below \(g\), subtract from the signed total, which is the integral of \(f - g\) over the whole region.
Press the button to turn the red pieces teal. Each piece then counts as its area, and however the curves sit, the area of the region is the sum of the piece areas.
Notation
Cut the region at every crossing, so one curve stays on top on each piece. The area of a piece is the integral of top minus bottom, which is never negative.
\[\cb{A_i} = \int \left(\text{top} - \text{bottom}\right) dx \ge 0\]The area of the region is the sum of the piece areas.
\[\cb{A} = \cb{A_1} + \cb{A_2} + \cdots\]A single integral of \(f - g\) over the whole interval is different, because the pieces where \(f\) is below \(g\), shown red, subtract.
\[\begin{aligned} \int_a^b (f - g)\,dx &= \cb{\text{teal areas}} \\ &\quad - \cr{\text{red areas}} \end{aligned}\]This equals the area only when \(f\) is on top on the whole interval.
When it fails
For \(f(x) = x^2\) and \(g(x) = 9\) from 0 to 4, the integral of \(f - g\) is \(-\tfrac{44}{3}\). A negative number cannot be an area.
Press the button to replace the sum by this integral. If a teal piece and a red piece have equal areas, the integral is 0, even though the region has area.
Example 1: A parabola and a line
Find the area between \(f(x) = 2x\) and \(g(x) = x^2\) from \(x = 1\) to \(x = 3\).
Set the formulas equal. The solutions are \(x = 0\) and \(x = 2\). Only \(x = 2\) lies inside the bounds, and it cuts the region into two pieces.
\[2x = x^2 \;\Longrightarrow\; x(x - 2) = 0\]On the first piece, test \(x = 1.5\): \(2x = 3\) is above \(x^2 = 2.25\), so the line is on top.
\[\begin{aligned} \cb{A_1} &= \int_1^2 \left(2x - x^2\right) dx \\ &= \Big[x^2 - \tfrac{x^3}{3}\Big]_1^2 = \tfrac23 \end{aligned}\]On the second piece, test \(x = 2.5\): the parabola is on top, so the piece turns red and its area is parabola minus line.
\[\begin{aligned} \cb{A_2} &= \int_2^3 \left(x^2 - 2x\right) dx \\ &= \Big[\tfrac{x^3}{3} - x^2\Big]_2^3 = \tfrac43 \end{aligned}\]Add the two areas. The red piece counts as \(\tfrac43\), not \(-\tfrac43\).
\[\cb{A} = \tfrac23 + \tfrac43 = 2\]
Example 2: A cosine and the x-axis
Find the area between \(y = \cos x\) and the x-axis from \(x = 0\) to \(x = \tfrac{3\pi}{2}\).
The curve crosses the x-axis at \(x = \tfrac{\pi}{2}\), inside the bounds, which cuts the region into two pieces. Its next zero, \(x = \tfrac{3\pi}{2}\), is the right bound itself.
Integrate \(\cos x\) on each piece. The first integral is 1. The second is \(-2\) because the curve is below the axis there, so that piece is red.
\[\begin{aligned} \int_0^{\pi/2} \cos x\,dx &= 1 \\ \int_{\pi/2}^{3\pi/2} \cos x\,dx &= -2 \end{aligned}\]A tempting wrong move: Add the two integrals.
\[1 + (-2) = -1\]The total is negative, but the figure shows a region with area. The value \(-2\) is the negative of the red piece's area, so adding it subtracts that piece.
On the second piece the x-axis is on top, so its area is \(\int_{\pi/2}^{3\pi/2} (0 - \cos x)\,dx = 2\). Add the areas.
\[\cb{A} = 1 + 2 = 3\]
Practice
Find the area between \(f(x) = x^3\) and \(g(x) = x^2\) from \(x = 0\) to \(x = 2\).
The top end of the bar changes curve at the crossing. Add the piece areas, not the signed integrals.
Solution
Set the formulas equal. The solution \(x = 0\) is the left bound, so \(x = 1\) is the only cut, and it makes two pieces.
\[x^3 = x^2 \;\Longrightarrow\; x^2(x - 1) = 0\]Test \(x = 0.5\): \(x^2\) is above \(x^3\), so the first piece turns red. Test \(x = 1.5\): \(x^3\) is above \(x^2\). Find each piece's area as top minus bottom.
\[\begin{aligned} \cb{A_1} &= \int_0^1 \left(x^2 - x^3\right) dx = \tfrac{1}{12} \\ \cb{A_2} &= \int_1^2 \left(x^3 - x^2\right) dx = \tfrac{17}{12} \end{aligned}\]Add the areas.
\[\cb{A} = \tfrac1{12} + \tfrac{17}{12} = \tfrac32\]One signed integral from 0 to 2 gives \(\tfrac{17}{12} - \tfrac1{12} = \tfrac43\), not the area, because the first piece subtracts.
Questions
Questions 1 and 2 use \(f(x) = x\) and \(g(x) = 4 - x\) from \(x = 0\) to \(x = 5\).
What is the area of the left piece?
- 4
- −4
- 8
- 2
Answer
4
The right piece has area 9. What is the area of the whole region?
- 13
- 5
- 9
- 36
Answer
13
The curves \(f(x) = x^2\) and \(g(x) = 3x\) cross at \(x = 3\). What is the area between them from \(x = 0\) to \(x = 4\)?
- \(\tfrac{19}{3}\)
- \(-\tfrac83\)
- \(\tfrac83\)
- \(\tfrac92\)
Answer
\(\tfrac{19}{3}\)
Summary
The area between two curves is the sum of the piece areas, each found as top minus bottom.
Where \(f\) is below \(g\), the integral of \(f - g\) is negative, so one integral across a crossing lets pieces cancel.
When the curves cross between the bounds, split there and add the areas.
This is one of four skills in topic 8.6 of Which & Why: AP Calculus.