Difficulty: Intermediate | Prerequisites: Definite integrals, u-substitution, graphing polynomials and trig functions, basic geometry (area of a triangle).
Once you can evaluate definite integrals, the natural next question is: what area does a curve actually enclose? This topic takes the definite integral and gives it a geometric meaning, letting you compute the exact area between two curves, between a curve and an axis, or inside a region bounded by several functions. It sits squarely in the middle of Integral Calculus, after substitution and before volumes of revolution. You should be comfortable setting up and evaluating definite integrals before working through these problems.
To find the area between two curves, integrate the top function minus the bottom function over the relevant interval. If the curves cross, split the integral at the crossing points and take absolute values. You can also integrate with respect to y when horizontal slicing is simpler.
Area between curves
The total region enclosed between two functions f(x) and g(x) on an interval [a, b], computed as ∫ₐᵇ |f(x) - g(x)| dx. Think of it as stacking up infinitely thin vertical slices, each as tall as the gap between the two curves.
Upper and lower functions
On a given interval, the "upper" function is the one with greater y-values (closer to the top of the graph), and the "lower" function is the one beneath it. The area formula is ∫(upper - lower) dx. If they swap positions partway through, you need to split the integral.
Intersection points
The x-values (or y-values) where two curves meet. These are found by setting f(x) = g(x) and solving. They determine the limits of integration and, when curves cross, where to split the integral.
Integrating with respect to y
Instead of vertical slices (dx), you use horizontal slices (dy). The area becomes ∫(right function - left function) dy. This is useful when the region is easier to describe as a function of y, for example when the boundary is x = 6y - y².
Signed area vs geometric area
A definite integral ∫ₐᵇ f(x) dx gives signed area: positive above the x-axis, negative below. Geometric area is always positive, so when a curve dips below the axis you must account for the sign change, either by splitting the integral or taking absolute values.
If f(x) ≥ g(x) on [a, b], then:
Area = ∫ₐᵇ [f(x) - g(x)] dx
The integral of (top minus bottom) gives the area enclosed between the two curves over that interval.
Before setting up the integral, check which function has greater values on the interval. Evaluate both at a test point, or sketch the graphs. For parabolas, note whether they open upward or downward.
Find the area enclosed by f(x) = 5 + x - x² and g(x) = x + 1 on [0, 1].
Check: f(0) = 5 ≥ 1 = g(0), and f(1) = 5 ≥ 2 = g(1). So f(x) ≥ g(x) on [0, 1].
f(x) - g(x) = (5 + x - x²) - (x + 1) = 4 - x².
Area = ∫₀¹ (4 - x²) dx = [4x - x³/3] from 0 to 1 = 4 - 1/3 = 11/3 sq. units.
Find the area bounded by f(x) = x² - 3x and g(x) = 6x - 2x² on [0, 3].
Both cross the x-axis at x = 0 and x = 3. The parabola g opens downward and sits above f on [0, 3].
g(x) - f(x) = (6x - 2x²) - (x² - 3x) = 9x - 3x².
Area = ∫₀³ (9x - 3x²) dx = [9x²/2 - x³] from 0 to 3 = 81/2 - 27 = 27/2 sq. units.
Find the area between f(x) = (1/2)cos x and g(x) = (1/2)sin x on [0, π].
The curves cross at x = π/4 (where cos x = sin x). On [0, π/4], cos x ≥ sin x. On [π/4, π], sin x ≥ cos x (or more precisely, cos x - sin x ≤ 0).
Area = (1/2) ∫₀^{π/4} (cos x - sin x) dx - (1/2) ∫_{π/4}^π (cos x - sin x) dx.
Using [sin x + cos x] as the antiderivative:
A₁ = (1/2)[sin x + cos x] from 0 to π/4 = (1/2)(√2 - 1).
A₂ = (1/2)[sin x + cos x] from π/4 to π = -(1/2)(1 + √2).
Area = A₁ - A₂ = √2.
The integral ∫₀⁴ [(x + 1) + (1/2)x] dx represents the area between f(x) = x + 1 and g(x) = -(1/2)x on [0, 4]. The integrand is f(x) - g(x) = (x + 1) - (-(1/2)x) = (x + 1) + (1/2)x. The region is between the line y = x + 1 (above) and the line y = -(1/2)x (below), from x = 0 to x = 4.
When a function crosses the x-axis on the interval, the definite integral gives signed area (negative below the axis). To find the total geometric area, split the integral at each root and take the absolute value of each piece.
Find the area between f(x) = 2x - x² and the x-axis on [0, 4].
f(x) = x(2 - x), so the roots are x = 0 and x = 2. The parabola opens downward: f(x) ≥ 0 on [0, 2] and f(x) ≤ 0 on [2, 4].
Area = ∫₀² (2x - x²) dx - ∫₂⁴ (2x - x²) dx.
∫₀² (2x - x²) dx = [x² - x³/3] from 0 to 2 = 4 - 8/3 = 4/3.
∫₂⁴ (2x - x²) dx = [x² - x³/3] from 2 to 4 = (16 - 64/3) - (4 - 8/3) = -20/3.
Area = 4/3 - (-20/3) = 4/3 + 20/3 = 24/3 = 8 sq. units.
Notice: you subtract the second integral because f(x) is negative on [2, 4], so the integral itself is negative. Subtracting a negative gives you the positive area.
Some regions are more naturally described using horizontal slices. Instead of expressing curves as y = f(x), you express them as x = h(y) and integrate with respect to y.
Area = ∫_c^d [h_right(y) - h_left(y)] dy
where h_right is the rightmost boundary and h_left is the leftmost boundary (often the y-axis, i.e. x = 0).
Find the area bounded by x = 6y - y² and the y-axis.
The parabola x = 6y - y² = y(6 - y) intersects the y-axis at y = 0 and y = 6. On [0, 6], the curve is to the right of the y-axis (x ≥ 0), so the y-axis is the left boundary.
Area = ∫₀⁶ (6y - y²) dy = [3y² - y³/3] from 0 to 6 = 108 - 72 = 36 sq. units.
Find the area enclosed by f(y) = y² and g(y) = 2 + y.
Set y² = 2 + y, so y² - y - 2 = (y + 1)(y - 2) = 0. The curves intersect at y = -1 and y = 2.
On [-1, 2], the line g(y) = 2 + y lies to the right of the parabola f(y) = y².
Area = ∫_{-1}^{2} [(2 + y) - y²] dy = [2y + y²/2 - y³/3] from -1 to 2.
At y = 2: 4 + 2 - 8/3 = 10/3. At y = -1: -2 + 1/2 + 1/3 = -7/6.
Area = 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 9/2 sq. units.
You can use integration to find the area of a triangle or polygon by expressing each side as a linear function and integrating.
Find the area of triangle ABC with A = (0, 0), B = (2, 6), C = (6, 2).
The three sides are lines:
AB: y = 3x (slope 6/2 = 3, through the origin)
AC: y = (1/3)x (slope 2/6 = 1/3, through the origin)
BC: slope = (2 - 6)/(6 - 2) = -1, so y - 6 = -(x - 2), giving y = -x + 8
The top boundary changes at x = 2. On [0, 2], the top is AB (y = 3x) and the bottom is AC (y = (1/3)x). On [2, 6], the top is BC (y = -x + 8) and the bottom is AC (y = (1/3)x).
Area = ∫₀² [3x - (1/3)x] dx + ∫₂⁶ [(-x + 8) - (1/3)x] dx.
∫₀² (8/3)x dx = [(4/3)x²] from 0 to 2 = 16/3.
∫₂⁶ (-4x/3 + 8) dx = [-2x²/3 + 8x] from 2 to 6 = (-24 + 48) - (-8/3 + 16) = 24 - 40/3 = 32/3.
Area = 16/3 + 32/3 = 48/3 = 16 sq. units.
When given a figure with regions A₁, A₂, A₃ defined by curves y = f(x) and y = g(x) inside a rectangle, the area of the region above f and below the top of the rectangle is:
A₂ = (rectangle area) - ∫₁⁴ f(x) dx
This works because the integral of f gives the area under f, and subtracting from the rectangle gives the area above f. Be careful: an analogous formula using g works for A₁ only if g is the lower boundary, which depends on the specific geometry.
\text{Area} = \int_a^b |f(x) - g(x)|\, dx\text{Area} = \int_c^d |h_{\text{right}}(y) - h_{\text{left}}(y)|\, dyWhen f(x) ≥ g(x) throughout [a, b], the absolute value signs are unnecessary and the formula simplifies to ∫ₐᵇ [f(x) - g(x)] dx.
When the curves cross at a point x = p inside [a, b], split:
Area = ∫ₐᵖ [f(x) - g(x)] dx + ∫_p^b [g(x) - f(x)] dx
(swapping which function is subtracted based on which is on top in each sub-interval).
Area-between-curves calculations are used in economics to compute consumer and producer surplus (the area between supply and demand curves up to the equilibrium price). In physics, the area between a velocity curve and the time axis gives displacement, and in engineering, the cross-sectional area of irregular shapes (beams, channels, ducts) is computed by integrating boundary functions.
Students often compute ∫ₐᵇ f(x) dx when asked for the area between f and the x-axis, without checking whether f dips below zero. A negative integral does not mean negative area; it means part of the region is below the axis and you need to split the integral.
When two curves cross inside the interval, students sometimes integrate (f - g) across the whole interval without splitting. This gives net signed area, not the total geometric area.
Students sometimes confuse integrating with respect to x versus y. If the region is described by x = h(y), slicing horizontally (dy) is correct. Trying to invert h and integrate in x often creates unnecessary difficulty.
A common error with triangles is computing the area using one integral from x = 0 to x = 6, without recognising that the "top" boundary changes at x = 2. You must split at every vertex.
⚠️ Area-between-curves problems are a near-certainty on exams. Expect at least one question asking for area between two given functions, and possibly one that requires integrating with respect to y.
⚠️ Problems that ask you to identify which shaded region corresponds to a given integral test conceptual understanding, not just computation. Practise reading integrands as (top - bottom) and matching to a graph.
⚠️ When f crosses the x-axis, the exam may ask for the area between f and the axis on an interval that includes both positive and negative portions. You must split.
⚠️ The tangent-line problem (area between a parabola, its tangent at a point, and the x-axis) combines derivative skills with area computation. Know how to find the tangent line, its x-intercept, then set up the area as a difference of integrals.
True or False: ∫₀⁴ f(x) dx always gives the area between f and the x-axis on [0, 4].
Answer: False. It gives signed area. If f(x) < 0 on part of the interval, you need to split the integral.
Fill in the blank: The area between y = f(x) and y = g(x) on [a, b], where f(x) ≥ g(x), equals ∫ₐᵇ ____ dx.
Answer: [f(x) - g(x)].
True or False: When integrating with respect to y, you integrate (top - bottom) dy.
Answer: False. You integrate (right - left) dy.
Fill in the blank: To find where two curves intersect, set ____ and solve.
Answer: f(x) = g(x).
True or False: The area of a region is always positive.
Answer: True. Area is a geometric quantity and is always non-negative.
Q: Find the area between f(x) = 2x - x² and the x-axis on [0, 4].
A: f crosses the axis at x = 0 and x = 2. Area = ∫₀² (2x - x²) dx - ∫₂⁴ (2x - x²) dx = 4/3 + 20/3 = 8 sq. units.
Q: Find the area enclosed by f(x) = 5 + x - x² and g(x) = x + 1 on [0, 1].
A: f(x) - g(x) = 4 - x². Area = ∫₀¹ (4 - x²) dx = 4 - 1/3 = 11/3 sq. units.
Q: Find the area bounded by x = 6y - y² and the y-axis.
A: The curve meets the y-axis at y = 0 and y = 6. Area = ∫₀⁶ (6y - y²) dy = 108 - 72 = 36 sq. units.
Q: Find the area bounded by f(x) = x² - 3x and g(x) = 6x - 2x².
A: Both cross at x = 0 and x = 3. Area = ∫₀³ (9x - 3x²) dx = 27/2 sq. units.
Q: Use integration to find the area of triangle ABC with A = (0,0), B = (2,6), C = (6,2).
A: Split at x = 2. Area = ∫₀² (3x - x/3) dx + ∫₂⁶ (-x + 8 - x/3) dx = 16/3 + 32/3 = 16 sq. units.
Q: Find the area enclosed by f(y) = y² and g(y) = 2 + y.
A: Intersections at y = -1 and y = 2. Area = ∫_{-1}^{2} (2 + y - y²) dy = 9/2 sq. units.
Area between curves is the geometric interpretation of the definite integral, which you first encountered as a Riemann sum. This topic connects forward to volumes of revolution (disk/washer and shell methods), where you rotate these regions around an axis and compute the resulting 3D volume. It also connects to probability, where the area under a probability density function between two values gives the probability of an outcome falling in that range.
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