Signed area and integrals from geometry
When a graph dips below the x-axis, how should the region trapped there count, and how can an integral be evaluated exactly from triangles, rectangles and circles alone?
Where you are 3 in The definite integral
Why this exists
Riemann sums approximate the area under a graph with rectangles, and the definite integral is the exact number those sums approach. Where the graph runs below the x-axis, a rectangle’s height f(x_i^*) is negative, so its term f(x_i^*)\,\Delta x is negative too. The integral therefore counts area above the axis as positive and area below as negative. That reading gives the integral a meaning (net change, such as a displacement) and gives its exact value without any limit whenever the region is made of shapes with known areas.
Definition
Let f be continuous on [a, b] with a < b. The region between the graph of y = f(x) and the x-axis, for a \le x \le b, splits into pieces lying above the axis and pieces lying below it. The definite integral
is the signed area (or net area) of that region: the total area of the pieces above the axis minus the total area of the pieces below it. The function f is the integrand, and a and b are the lower and upper limits of integration.
The definition of the integral as a limit of Riemann sums is in The definite integral as a limit of Riemann sums. This lesson uses the area reading, which that limit produces; it is taken here as the working meaning.
The total area between the graph and the axis counts every piece as positive. Flipping the pieces below the axis up turns f into |f|, so the total area is \int_a^b |f(x)|\,dx.
When the limits are reversed or equal, the integral is defined by two conventions:
Intuition
Picture walking along the x-axis from a to b and collecting the area between you and the graph: a piece above the axis is a gain, a piece below it a loss. In Figure 1, the hump A_1 and the last region A_3 are gains and the valley A_2 is a loss, so
If f is a velocity, a piece above the axis is distance covered forward and a piece below is distance covered backward; the signed area is the net change in position, the total area the distance traveled (Position, displacement and distance from velocity).
Walking from b back to a reverses every step, so every gain becomes a loss. That is the content of the reversal convention.
Results
Signed area. For f continuous on [a, b], \int_a^b f(x)\,dx equals the area above the axis minus the area below. The total area is \int_a^b |f(x)|\,dx, and the two agree exactly when f(x) \ge 0 on [a, b].
Constant integrand. For a constant c and a < b, \int_a^b c\,dx = c(b - a): a rectangle above the axis when c > 0, below it when c < 0.
Half disk. For r > 0, the graph of \displaystyle y = \sqrt{r^2 - x^2} is the upper half of the circle x^2 + y^2 = r^2, so
Reversed and equal limits. \int_b^a f(x)\,dx = -\int_a^b f(x)\,dx and \int_a^a f(x)\,dx = 0.
Symmetric intervals. Let f be continuous on [-a, a]. If f is odd, meaning f(-x) = -f(x), then \int_{-a}^{a} f(x)\,dx = 0. If f is even, meaning f(-x) = f(x), then \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx.
Further rules, such as splitting an interval into parts, are in Properties of definite integrals.
Why it is true
Signed area from the sums. In a Riemann sum \sum f(x_i^*)\,\Delta x with a < b, each \displaystyle \Delta x = \frac{b - a}{n} is positive. A term with f(x_i^*) > 0 is the area of a rectangle standing above the axis; a term with f(x_i^*) < 0 is the negative of the area of a rectangle hanging below it. As the rectangles narrow, those above fill the regions above the axis and those below fill the regions below, so the limit is (area above) minus (area below). That the limit really equals these areas is stated here without proof; The definite integral as a limit of Riemann sums takes it up.
Why reversing flips the sign. Run the same sums from b to a: the width becomes \displaystyle \Delta x = \frac{a - b}{n} = -\frac{b - a}{n} while the sample heights are the same, so every term changes sign and so does the limit. The convention is also the only choice that keeps additivity true in every order: \int_a^b f(x)\,dx + \int_b^a f(x)\,dx should equal \int_a^a f(x)\,dx, the integral over no width, which is 0.
Half disk. For y \ge 0, the equation \displaystyle y = \sqrt{r^2 - x^2} is equivalent to x^2 + y^2 = r^2, the circle of radius r centered at the origin. The region under it on [-r, r] is half of a disk of area \pi r^2 (Area of a circle).
Odd and even functions (a geometric argument). If f(-x) = -f(x), the point (x, y) is on the graph exactly when (-x, -y) is, so the graph is unchanged by a half turn about the origin. That half turn carries the region between the graph and the axis on [0, a] onto the region on [-a, 0]. It preserves area and swaps above with below. Each piece on one side is matched by a piece of the same area on the other side with the opposite sign, so the signed areas cancel (Figure 5). For an even function, reflection in the y-axis carries the region on [0, a] onto the region on [-a, 0] without changing above or below, so the two halves have equal signed area.
Worked examples
Rectangles and a trapezoid
Evaluate (a) \int_1^5 \left(\tfrac{x}{2} + 1\right)dx and (b) \int_{-1}^{3} (-2)\,dx.
Solution(a) The graph of y = \tfrac{x}{2} + 1 is a line with heights \tfrac{3}{2} at x = 1 and \tfrac{7}{2} at x = 5, both positive, so the region is a trapezoid above the axis with parallel sides \tfrac{3}{2} and \tfrac{7}{2} and width 4 (Area of parallelograms and trapezoids):
(b) The graph of y = -2 is a horizontal line below the axis. The region is a rectangle of width 3 - (-1) = 4 and height 2, area 8, lying below the axis, so it counts negative: \int_{-1}^{3} (-2)\,dx = -8. This is c(b - a) with c = -2.
Net area and total area
For \int_0^6 (x - 2)\,dx, find the signed area and the total area between the graph and the axis.
SolutionThe line y = x - 2 crosses the axis at x = 2, so the region splits there (Figure 2). On [0, 2] the line is below the axis and cuts off a right triangle with legs 2 and 2 (the height at x = 0 is |0 - 2| = 2). Its area is \tfrac{1}{2} \cdot 2 \cdot 2 = 2 by half base times height. On [2, 6] the line is above the axis, with a triangle of base 4 and height 6 - 2 = 4, area 8.
The signed area counts the lower triangle negative:
The total area counts both positive: \int_0^6 |x - 2|\,dx = 8 + 2 = 10.
Semicircles and quarter circles
Evaluate (a) \displaystyle \int_{-5}^{5} \sqrt{25 - x^2}\,dx, (b) \displaystyle \int_0^5 \left(-\sqrt{25 - x^2}\right)dx, (c) \displaystyle \int_{-2}^{2} \left(3 + \sqrt{4 - x^2}\right)dx.
Solution(a) The graph is the upper half of the circle x^2 + y^2 = 25, radius 5, and the interval runs across the whole diameter. The half disk has area \tfrac{1}{2}\pi \cdot 5^2, so the integral is \dfrac{25\pi}{2}.
(b) The graph of \displaystyle y = -\sqrt{25 - x^2} is the lower half of the same circle. On [0, 5] it bounds a quarter disk below the axis, area \tfrac{1}{4}\pi \cdot 5^2, so the integral is -\dfrac{25\pi}{4}.
(c) Every height is 3 more than the height of the upper half of the circle of radius 2, so the region is a rectangle of width 4 and height 3 with that half disk set on top (Figure 3). Its area is 4 \cdot 3 + \tfrac{1}{2}\pi \cdot 2^2:
Integrals from a graph
The function g is graphed in Figure 4. Find (a) \int_{-4}^{0} g(x)\,dx, (b) \int_0^4 g(x)\,dx, (c) \int_{-4}^{6} g(x)\,dx, (d) the total area between the graph and the axis on [-4, 6], (e) \int_6^0 g(x)\,dx.
Read the pieces off the figure. On [-4, -2] the graph is level at height 2: a rectangle above the axis, area 2 \cdot 2 = 4. On [-2, 0] it falls from 2 to 0: a triangle above, area \tfrac{1}{2} \cdot 2 \cdot 2 = 2. On [0, 4] it is the lower half of the circle of radius 2 centered at (2, 0): a half disk below, area \tfrac{1}{2}\pi \cdot 2^2 = 2\pi. On [4, 6] it rises from 0 to 3: a triangle above, area \tfrac{1}{2} \cdot 2 \cdot 3 = 3.
(a) Both pieces on [-4, 0] are above the axis: 4 + 2 = 6.
(b) The half disk is below the axis: \int_0^4 g(x)\,dx = -2\pi.
(c) Add every piece with its sign:
(d) The total area counts the half disk positive: 9 + 2\pi \approx 15.28.
(e) First integrate in the usual direction: on [0, 6] the signed area is -2\pi + 3. Reversing the limits flips the sign:
Odd integrands over symmetric intervals
Evaluate (a) \int_{-2}^{2} x^3\,dx and (b) \int_{-1}^{1} \left(3 + x^3\right)dx.
Solution(a) f(x) = x^3 is continuous and odd, since (-x)^3 = -x^3. In Figure 5 the region above the axis on [0, 2] and the region below it on [-2, 0] are matched by a half turn about the origin, so they have the same area A and opposite signs:
(b) On [-1, 1] the heights 3 + x^3 run from 2 to 4, so the whole region lies above the axis. The graph is the graph of x^3 raised 3 units, so a half turn about the point (0, 3) maps it to itself. Compare the region with the rectangle of height 3 on [-1, 1], area 6. On [0, 1] the region pokes above the line y = 3; on [-1, 0] it falls short of that line. The half turn about (0, 3) carries the bump exactly onto the gap, so they have the same area, and the region’s area equals the rectangle’s:
Choosing a limit to hit a target
Rafaela wants every number b with \int_0^b (4 - x)\,dx = 6. Find them all.
SolutionThe line y = 4 - x is above the axis for x < 4 and below it for x > 4. Sort the cases by where b is.
If 0 \le b \le 4, the region is a trapezoid above the axis with parallel sides 4 and 4 - b and width b, area \tfrac{1}{2}(4 + 4 - b)\,b = 4b - \tfrac{b^2}{2}. Set this equal to 6:
Only b = 2 lies in this case. Check: a trapezoid with sides 4 and 2 and width 2 has area 6.
If b > 4, the triangle on [0, 4] above the axis has area \tfrac{1}{2} \cdot 4 \cdot 4 = 8, and a triangle below the axis on [4, b] with legs b - 4 subtracts \tfrac{(b - 4)^2}{2}. Then 8 - \tfrac{(b - 4)^2}{2} = 6 gives (b - 4)^2 = 4, so b = 6.
If b < 0, then \int_0^b (4 - x)\,dx = -\int_b^0 (4 - x)\,dx, and on [b, 0] the line is above the axis, so the value is negative and cannot be 6.
The answers are b = 2 and b = 6.
Boundaries
- The area reading needs the integral to exist. For a continuous f on a closed interval it always does. A function that blows up inside the interval, such as \displaystyle \frac{1}{x} on [-1, 1], has no integral in this sense, and the symmetry rule for odd functions does not apply to it: the rule assumes f is continuous on [-a, a].
- Signed area and total area are equal only when f(x) \ge 0 on the interval. Otherwise the integral is smaller, and it can be 0 or negative while the region is large.
- Geometry evaluates an integral only when the region is built from shapes with known area: rectangles, triangles, trapezoids, parts of disks, or regions related to these by symmetry. The region under y = x^2 on [0, 1] is none of these; its area needs the limit definition or the fundamental theorem of calculus, a later unit.
Where this goes next
The definite integral as a limit of Riemann sums gives the integral its exact definition, and Properties of definite integrals collects the rules the area reading makes plain. For a velocity, signed area is displacement and total area is distance (Position, displacement and distance from velocity). Letting the upper limit move, as in Rafaela’s question, makes the integral a function of b: see Accumulation and net change and Graphing an accumulation function, where it peaks where the integrand crosses from positive to negative. The fundamental theorem of calculus, a later unit, evaluates integrals that no formula from geometry can measure.
Exercises
Lines, rectangles and triangles
- Evaluate \int_{-2}^{5} 3\,dx.
- Evaluate \int_0^3 2x\,dx.
- Evaluate \int_2^6 (x + 1)\,dx.
- Evaluate \int_{-6}^{2} \left(1 - \tfrac{x}{2}\right)dx.
- Find \int_0^5 (3 - x)\,dx and the total area between y = 3 - x and the axis on [0, 5].
- Find \int_{-1}^{3} (2x - 2)\,dx and the total area between the graph and the axis on [-1, 3].
- Caetano walks along a straight hallway with velocity v(t) = 4 - 2t meters per second for 0 \le t \le 3, positive meaning toward the far door. Find \int_0^3 v(t)\,dt, his change in position, and the total distance he walks.
Circles
- Evaluate \displaystyle \int_{-6}^{6} \sqrt{36 - x^2}\,dx.
- Evaluate \displaystyle \int_0^7 \sqrt{49 - x^2}\,dx.
- Evaluate \displaystyle \int_{-4}^{0} \left(-\sqrt{16 - x^2}\right)dx.
- Evaluate \displaystyle \int_1^7 \sqrt{9 - (x - 4)^2}\,dx.
- Evaluate \displaystyle \int_{-3}^{3} \left(\sqrt{9 - x^2} - 3\right)dx. Sketch the region first.
From a graph
Use the graph of h in Figure 6.
- Evaluate \int_0^4 h(x)\,dx.
- Evaluate \int_4^8 h(x)\,dx.
- Evaluate \int_0^{10} h(x)\,dx.
- Find the total area between the graph of h and the axis on [0, 10].
- Evaluate \int_{10}^{6} h(x)\,dx.
Symmetry and reversed limits
- Evaluate \int_{-\pi}^{\pi} x^2 \sin x\,dx.
- Evaluate \int_{-\pi/2}^{\pi/2} (1 + \sin x)\,dx.
- Given \int_1^4 f(x)\,dx = 7, find \int_4^1 f(x)\,dx and \int_4^4 f(x)\,dx.
- Let f be continuous and odd with \int_0^3 f(x)\,dx = 5. Find \int_{-3}^{0} f(x)\,dx, \int_{-3}^{3} f(x)\,dx and \int_0^{-3} f(x)\,dx.
- Find every b \ne 1 with \int_1^b (x - 4)\,dx = 0.
Reasoning
- Explain why \int_a^b f(x)\,dx \le \int_a^b |f(x)|\,dx for f continuous on [a, b] with a < b, and say when the two are equal.
- Leonel writes \displaystyle \int_{-1}^{1} \frac{1}{x}\,dx = 0 because \displaystyle \frac{1}{x} is odd. Explain what is wrong.
- Let f be continuous and even on [-a, a]. Explain with a reflection why \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx.
Answers
- 21. A rectangle above the axis of width 7 and height 3.
- 9. A triangle above the axis with base 3 and height 2 \cdot 3 = 6: \tfrac{1}{2} \cdot 3 \cdot 6 = 9.
- 20. A trapezoid with parallel sides 3 and 7 and width 4: \tfrac{3 + 7}{2} \cdot 4 = 20.
- 16. The height is 4 at x = -6 and 0 at x = 2, so the region is a triangle above the axis with base 8 and height 4.
- Signed area \tfrac{5}{2}, total area \tfrac{13}{2}. Above the axis on [0, 3]: a triangle with legs 3 and 3, area \tfrac{9}{2}. Below on [3, 5]: legs 2 and 2, area 2. So \tfrac{9}{2} - 2 = \tfrac{5}{2} and \tfrac{9}{2} + 2 = \tfrac{13}{2}.
- Signed area 0, total area 8. The line crosses at x = 1. Below on [-1, 1]: base 2, height 4, area 4. Above on [1, 3]: base 2, height 4, area 4. So 4 - 4 = 0 and 4 + 4 = 8.
- \int_0^3 v(t)\,dt = 3, so he ends 3 meters toward the far door; he walks 5 meters. The velocity is positive until t = 2: a triangle of area \tfrac{1}{2} \cdot 2 \cdot 4 = 4. On [2, 3] it is negative: a triangle of area \tfrac{1}{2} \cdot 1 \cdot 2 = 1. Signed 4 - 1 = 3, total 4 + 1 = 5.
- 18\pi. A half disk of radius 6: \tfrac{1}{2}\pi \cdot 36.
- \dfrac{49\pi}{4}. A quarter disk of radius 7 above the axis.
- -4\pi. A quarter disk of radius 4 below the axis: -\tfrac{1}{4}\pi \cdot 16.
- \dfrac{9\pi}{2}. The graph is the upper half of the circle of radius 3 centered at (4, 0), and [1, 7] is its full diameter.
- \dfrac{9\pi}{2} - 18 \approx -3.86. The graph is the upper half of the circle of radius 3 centered at (0, -3); it lies at or below the axis, touching it at x = 0. The region between it and the axis is the rectangle [-3, 3] \times [-3, 0], area 18, with the half disk of area \tfrac{9\pi}{2} removed. It lies below the axis, so the integral is -(18 - \tfrac{9\pi}{2}).
- 4. Rectangle 2 \cdot 2 = 4 on [0, 2], triangle +1 on [2, 3], triangle -1 on [3, 4]: 4 + 1 - 1 = 4.
- -4 - \pi. A rectangle below of area 2 \cdot 2 = 4 on [4, 6] and a quarter disk below of area \tfrac{1}{4}\pi \cdot 4 = \pi on [6, 8].
- 3 - \pi \approx -0.14. Add \int_8^{10} h(x)\,dx = \tfrac{1}{2} \cdot 2 \cdot 3 = 3 to the previous two: 4 + (-4 - \pi) + 3.
- 13 + \pi \approx 16.14. Every piece positive: 4 + 1 + 1 + 4 + \pi + 3.
- \pi - 3 \approx 0.14. On [6, 10] the signed area is -\pi + 3; reversing the limits flips the sign.
- 0. The integrand is continuous and odd: (-x)^2 \sin(-x) = -x^2 \sin x. So the regions on the two sides cancel.
- \pi. The heights 1 + \sin x run from 0 to 2, so the region is above the axis. The graph is symmetric about (0, 1), so the bump above y = 1 on the right fills the gap below it on the left, leaving the rectangle of height 1 and width \pi.
- \int_4^1 f(x)\,dx = -7 by reversing the limits, and \int_4^4 f(x)\,dx = 0.
- \int_{-3}^{0} f(x)\,dx = -5, \int_{-3}^{3} f(x)\,dx = 0, \int_0^{-3} f(x)\,dx = 5. The half turn about the origin carries the region on [0, 3] onto the one on [-3, 0] with the sign switched. The sides cancel over [-3, 3], and reversing \int_{-3}^{0} gives 5.
- b = 7. The line y = x - 4 is below the axis on [1, 4], a triangle of area \tfrac{9}{2}. To cancel it, the region above the axis on [4, b] needs area \tfrac{9}{2}, a triangle with legs b - 4: \tfrac{(b - 4)^2}{2} = \tfrac{9}{2}, so b = 7. For b < 1 the integral is -\int_b^1 (x - 4)\,dx, the negative of a region lying below the axis, so it is positive, not 0.
- Both integrals count the areas above the axis as positive. The signed area subtracts each area below the axis, while the total area adds it, so the signed area is never larger. They are equal exactly when there is no area below the axis, which for continuous f means f(x) \ge 0 for every x in [a, b].
- The symmetry rule needs f continuous on [-1, 1], and \displaystyle \frac{1}{x} is undefined at 0. Near 0 its graph climbs without bound, and the region on [0, 1] has no finite area (stated without proof). There are no two numbers to cancel, so the integral does not exist in this lesson’s sense.
- If f(-x) = f(x), the point (x, y) is on the graph exactly when (-x, y) is, so reflection in the y-axis maps the graph to itself. It carries the region between the graph and the axis on [0, a] onto the region on [-a, 0], keeping areas and keeping above above and below below. So the signed areas on the two halves are equal, and the whole is twice the right half.