Did you know that the exact area between a quadratic curve and the straight chord joining two points on it depends only on the coefficient of x² and the horizontal separation of the points? If f(x) = ax² + bx + c, with a ≠ 0, and the chord joins the curve at x = p and x = q where p < q, then the area between the curve and that chord is |a|(q − p)³/6.
For example, the parabola f(x) = x² − 4x + 3 meets its chord endpoints at (0,3) and (2,−1). The chord is y = 3 − 2x. Between x = 0 and 2 it lies above the curve, so the area is ∫₀²[(3 − 2x) − (x² − 4x + 3)]dx = 4/3 square units. This guide explains the chord equation, why the difference factorises, how to use integration correctly and what changes for downward-opening or shifted parabolas.
A chord is a straight segment connecting two curve points
Choose two distinct points A(p,f(p)) and B(q,f(q)) on a parabola, with p < q. The straight line through these points is the secant line. Its segment between the endpoints is the chord in the geometric sense used here.
The curved piece of the parabola and the straight segment enclose a finite planar region. Its area is found by integrating the vertical separation between the straight line and the curve from p to q.
The exact shape depends on the parabola’s opening direction, but the enclosed area is always non-negative. If the curve opens upward, the chord lies above it between the endpoints. If it opens downward, the curve lies above the chord.
Begin with the actual endpoint coordinates and identify which graph is higher before writing an area integral. A negative signed integral is not a negative geometric area.
Read the endpoint coordinates from the function
Let f(x) = x² − 4x + 3. The endpoints occur at x = 0 and x = 2.
Evaluate the function: f(0) = 3 and f(2) = 4 − 8 + 3 = −1. Thus A = (0,3) and B = (2,−1).
A student might be tempted to assume that the chord lies on the x-axis because the problem uses the word chord. That is false here. Its endpoints have different heights, so the chord is a sloping straight segment.
The correct line must pass through both supplied coordinates. Calculating these values explicitly prevents the common error of integrating between a parabola and an imagined horizontal boundary instead of the given secant.
Find the chord’s gradient and equation
The gradient of the line through A(0,3) and B(2,−1) is m = [−1 − 3]/(2 − 0) = −4/2 = −2.
Because it passes through (0,3), the equation is ℓ(x) = 3 − 2x.
Check the other endpoint: ℓ(2) = 3 − 4 = −1, matching f(2). The line is therefore the chord supporting line required.
The intercept and slope of this line are different from those of the parabola’s tangent at the endpoints. A chord is defined by two points; a tangent is defined by the local gradient at one contact input. Do not mix the two constructions.
Subtract the graphs before integrating
The chord is ℓ(x) = 3 − 2x, while the quadratic is f(x) = x² − 4x + 3.
Subtract curve from chord: ℓ(x) − f(x) = 3 − 2x − (x² − 4x + 3).
Simplify carefully: the constants cancel, giving −x² + 2x = x(2 − x). This is non-negative on 0 ≤ x ≤ 2.
Therefore the area is ∫₀² x(2 − x) dx. The sign check establishes that the chosen order, upper minus lower, is correct across the entire interval and no region splitting is needed.
Integrate the vertical separation
The required integral is ∫₀² (2x − x²) dx. An antiderivative is x² − x³/3.
Evaluate at the upper limit x = 2 to obtain 4 − 8/3 = 12/3 − 8/3 = 4/3. At the lower limit x = 0, the antiderivative is zero.
Thus the geometric area between the parabola and the chord is exactly 4/3 square units.
The answer is positive and uses area units. The calculation was made easier by subtracting the curves first, which cancelled several terms. This is generally preferable to integrating the line and quadratic separately and only then subtracting two longer expressions.
Check the result at the midpoint
The midpoint of the horizontal interval [0,2] occurs at x = 1. There the chord has height ℓ(1) = 1, while the curve has height f(1) = 1 − 4 + 3 = 0.
The vertical gap is therefore 1. It is zero at each endpoint and rises to a maximum of one at the midpoint.
The region resembles an arch-shaped strip with base width 2 and maximum vertical gap 1. Its area 4/3 is less than the bounding rectangle’s area 2, as it should be.
This is a useful reasonableness check, though the exact fraction still comes from integration or a proven equivalent formula. A rough sketch alone cannot justify 4/3 precisely.
Derive the general chord equation without solving for its slope
Let f(x) = ax² + bx + c and choose endpoint inputs p and q, with p < q. Denote by ℓ(x) the straight line joining (p,f(p)) and (q,f(q)).
Both f and ℓ agree at p and q. Their difference f(x) − ℓ(x) is a quadratic whose leading coefficient is a, because ℓ is linear.
This difference has roots p and q, so it must equal a(x − p)(x − q). Therefore f(x) − ℓ(x) = a(x − p)(x − q).
This factorisation is the central shortcut. It avoids calculating the chord slope in every new example and immediately reveals the sign and shape of the vertical gap.
Why the quadratic–chord difference has exactly that form
Subtracting a linear function from ax² + bx + c leaves a quadratic with the same x² coefficient a.
At x = p and x = q, the curve and its chord have equal y-values, so their difference is zero at both inputs.
A nonzero quadratic with leading coefficient a and distinct roots p and q has the factor form a(x − p)(x − q). No other quadratic with that leading coefficient and those roots is possible.
This is a useful link between coordinate geometry, polynomial factorisation and integration. Instead of treating the chord as an unrelated line, it is recognised as precisely the linear part left after removing the product associated with the two contact inputs.
The opening direction decides the order of subtraction
For p < x < q, the factor x − p is positive while x − q is negative. Thus (x − p)(x − q) is negative in the interior.
If a > 0, then f(x) − ℓ(x) < 0, so the upward-opening parabola lies below its chord and ℓ − f is the positive gap.
If a < 0, then f(x) − ℓ(x) > 0, so the downward-opening parabola lies above the chord and f − ℓ is the positive gap.
This decision can be made without plotting individual y-values. It follows directly from the sign of the leading coefficient and the fact that the chord joins two inputs on the same quadratic.
Derive the universal area formula
Because the gap is positive in the appropriate subtraction order, the area is |a|∫_p^q (x − p)(q − x) dx.
Let L = q − p > 0 and introduce u = x − p. Then x − p = u and q − x = L − u, with u ranging from 0 to L.
The integral becomes |a|∫₀ᴸ u(L − u) du. Integrate Lu − u² to obtain [Lu²/2 − u³/3] from 0 to L.
The result is |a|[L³/2 − L³/3] = |a|L³/6. Therefore A = |a|(q − p)³/6. The coefficients b and c do not enter this final area.
Check the general formula against the worked example
For f(x) = x² − 4x + 3, the leading coefficient is a = 1. The chord endpoints have inputs p = 0 and q = 2.
Thus q − p = 2, and the area formula gives A = |1|·2³/6 = 8/6 = 4/3.
This agrees exactly with the direct integral of x(2 − x). The two methods use the same geometry; the general formula simply packages a calculation that recurs for every quadratic.
A student should understand the derivation before relying on the formula in a mixed-topic problem. If the upper boundary is not the actual chord or the curve is not quadratic, the shortcut may not apply.
A downward-opening example with unequal endpoint heights
Let g(x) = −2x² + 3x + 1 and choose p = −1, q = 2. Evaluate g(−1) = −2 − 3 + 1 = −4 and g(2) = −8 + 6 + 1 = −1.
The chord joins (−1,−4) and (2,−1). Its gradient is [−1 − (−4)]/[2 − (−1)] = 3/3 = 1, so ℓ(x) = x − 3.
Subtract line from curve: g(x) − ℓ(x) = −2x² + 2x + 4 = −2(x + 1)(x − 2).
Between −1 and 2 this is positive, consistent with the downward-opening parabola lying above its chord. Integrating gives the correct enclosed area.
Calculate the downward-opening area by the shortcut
For g(x) = −2x² + 3x + 1, the leading coefficient has magnitude |a| = 2.
The chord inputs are −1 and 2, so their separation is L = 3. The area formula gives A = 2·3³/6 = 54/6 = 9 square units.
A direct integral of −2x² + 2x + 4 from −1 to 2 also gives 9, confirming the result.
This example shows that the shortcut uses the magnitude of a. Using a = −2 without an absolute value would produce −9, which is a signed integral in the opposite subtraction order rather than a valid geometric area.
The linear and constant coefficients do not affect the area
Consider two quadratics with the same leading coefficient a but different b and c. If both chords use the same horizontal endpoint inputs p and q, their enclosed areas are equal.
For example, f₁(x) = x² and f₂(x) = x² − 4x + 3 have different heights and slopes. Their chords between x = 0 and x = 2 are different lines, yet each encloses area 4/3 with the corresponding parabola.
Adding a linear function bx + c to a parabola also adds that same linear function to the chord through its selected endpoints. The vertical difference cancels it.
This invariance is the reason the universal formula contains only |a| and the horizontal separation. It is not an accidental cancellation in one numerical example.
Why horizontal translation does not change the area
Suppose the parabola is shifted right by three units and the two chord inputs are also shifted right by three units.
The coefficient a remains unchanged, and the horizontal separation q − p stays the same. Thus |a|(q − p)³/6 is unchanged.
This is consistent with geometry: a horizontal translation moves the enclosed region but does not stretch it or change its area.
The actual equations of the curve and chord may look more complicated after a shift. The universal formula reveals the invariant feature without requiring fresh integration of every translated expression.
Why the area scales with the cube of the width
The formula A = |a|L³/6 is cubic in the chord’s horizontal width L.
If the width doubles while a stays the same, the enclosed area is multiplied by 2³ = 8. If the width triples, the area is multiplied by 3³ = 27.
This is because widening the interval increases not just its horizontal extent but also the maximum vertical separation between the parabola and chord, which grows as L².
The result is a useful model of nonlinear scaling. A twice-as-wide chord interval generally encloses far more than twice the area when the quadratic curvature remains fixed.
The maximum vertical gap occurs halfway between the endpoints
The magnitude of the vertical difference is |a|(x − p)(q − x) on p ≤ x ≤ q.
This is a downward-opening quadratic in x, with roots p and q. Its maximum occurs at the midpoint m = (p + q)/2.
At that input, both factors have value L/2, so the maximum gap is H = |a|(L/2)² = |a|L²/4.
For the opening example L = 2 and a = 1, giving H = 1. This agrees with the direct y-value comparison at x = 1. The result connects the area problem with the vertex and symmetry of an ordinary quadratic.
The area is two-thirds of base times maximum gap
Combine A = |a|L³/6 with H = |a|L²/4. Solving the second equation gives |a|L² = 4H.
Substitute into the area formula: A = (4HL)/6 = (2/3)LH.
Thus the area between a quadratic arc and its chord is two-thirds of the rectangle with base L and height equal to the maximum vertical gap.
This is not a universal area formula for any curved segment. It works here because the gap itself is a parabola with zeros at both endpoints. It offers a second useful route to the same exact result once H and L are known.
Recover curvature from a known area
Suppose a quadratic’s chord endpoints have horizontal separation L = 4 and the enclosed area is 32/3 square units. Find the magnitude of its leading coefficient.
Use A = |a|L³/6. Thus 32/3 = |a|·64/6 = (32/3)|a|.
Therefore |a| = 1. The area alone determines the magnitude of a, not whether the parabola opens upward or downward.
If the problem also says the chord lies above the curve, then a > 0, so a = 1. If the curve lies above the chord, then a < 0, giving a = −1. This illustrates why geometric orientation supplies information that an unsigned area cannot.
Recover endpoint separation from curvature and area
Suppose |a| = 2 and the enclosed area is 9 square units. The formula gives 9 = 2L³/6 = L³/3.
Hence L³ = 27 and L = 3, taking the positive real cube root because L is a distance.
This matches the downward-opening example with endpoints −1 and 2. No individual endpoint needs to be known to recover their separation.
The area equation determines only q − p. It does not determine the interval’s midpoint or actual location on the x-axis. Additional information is needed if the question asks for both endpoint coordinates.
An arbitrary secant line is not always the chord being requested
A problem may supply a straight line that intersects a parabola at two points, without explicitly giving p and q. In that case, solve f(x) = ℓ(x) to find the intersection inputs first.
Once those two distinct real inputs are known, they become p and q in the universal area formula.
If the line is tangent to the parabola, the intersection inputs coincide and L = 0. There is no positive-area enclosed segment between two distinct crossings.
If the line does not intersect the parabola in real points, there is no real chord of the described kind. The formula must not be used with imaginary endpoints or an arbitrary integration interval.
A secant example with an unknown crossing pair
Consider the upward parabola y = x² and the line y = 2x + 3. Solve their intersections by x² = 2x + 3.
Rearrange to x² − 2x − 3 = (x − 3)(x + 1) = 0. Thus the crossing inputs are p = −1 and q = 3, separated by L = 4.
The leading coefficient of the parabola is a = 1, so the enclosed area is L³/6 = 64/6 = 32/3 square units.
The chord line lies above the parabola on the interval because (2x + 3) − x² = −(x − 3)(x + 1) is positive between the roots. The result agrees with the geometry and requires no additional lengthy integration.
Do not confuse chord area with area under the curve
The area between y = f(x) and its chord y = ℓ(x) is ∫|ℓ(x) − f(x)| dx over the chord interval.
The area under f(x) relative to the x-axis is a different quantity. It depends on where f is positive or negative and may require splitting at x-intercepts.
In the opening example the quadratic has an output below zero at x = 2, so an ordinary “area under the graph” calculation would not even have the same upper and lower boundaries as the chord region.
Always identify the closed region described in the problem. An integral can be algebraically correct and still answer the wrong question if the chosen boundaries are incorrect.
Signed integration must become positive geometric area
Suppose an upward parabola lies below its chord. Then f − ℓ is negative between the endpoints, and ∫_p^q [f(x) − ℓ(x)] dx is negative.
The geometric area is its opposite, ∫_p^q [ℓ(x) − f(x)] dx, which is positive. For a downward parabola, the subtraction order reverses.
The universal formula uses |a| specifically to avoid presenting a negative area. The absolute value is not a cosmetic addition; it records the magnitude of the vertical separation.
A student who calculates −4/3 for the opening problem has likely integrated lower minus upper. That signed result can help diagnose the mistake, but the geometric area must be reported as +4/3.
Why the derivative gives the midpoint of the gap
The gap has the form G(x) = |a|(x − p)(q − x). Differentiate: G′(x) = |a|[(q − x) − (x − p)].
Simplify to G′(x) = |a|(p + q − 2x). Setting this to zero gives x = (p + q)/2.
The second derivative G″(x) = −2|a| is negative, confirming a maximum. This matches the elementary symmetry reasoning.
The calculus is useful but not necessary to discover the midpoint. In a time-limited solution, recognising the zeros p and q of a quadratic gap immediately gives its axis of symmetry and maximum-gap location.
A short examination solution format
First obtain the two distinct intersection inputs p and q, whether they are given or found from f(x) = ℓ(x). Ensure p < q.
Second identify the quadratic’s leading coefficient a and determine which graph is above the other between the endpoints.
Third either calculate the difference explicitly and integrate upper minus lower, or use the proven shortcut A = |a|(q − p)³/6. State positive area units.
Finally check the maximum gap or a direct integral if the answer seems surprising. Keep the shortcut tied to a genuine quadratic curve and its own chord, rather than applying it automatically to unrelated functions.
Practice questions with checked answers
Question one: f(x) = x² − 4x + 3 with chord endpoints at x = 0 and x = 2. Answer: area 4/3.
Question two: f(x) = −2x² + 3x + 1 with chord inputs −1 and 2. Answer: area 9.
Question three: y = x² and chord line y = 2x + 3. Answer: intersections at x = −1 and 3; enclosed area 32/3.
Question four: for f(x) = ax² + bx + c and chord width L, state the area. Answer: |a|L³/6.
Question five: for chord width 4 and area 32/3, find |a|. Answer: 1.
Question six: if width doubles but a stays fixed, what happens to area? Answer: it increases by a factor of eight.
Question seven: express area using maximum vertical gap H and width L. Answer: (2/3)LH.
Attempt each by explaining why the quadratic–chord difference vanishes at the two endpoints.
Syllabus scope and connected learning
Area between a quadratic and its chord connects simultaneous equations, factorisation, graph shape, integration and optimisation. The universal cubic-width formula is a useful extension that can be derived from familiar definite-integral work.
For Singapore, use the relevant SEAB syllabus documents for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2, according to the student’s school sequence.
Read companion guides How to Find a Function’s Average Value from a Definite Integral, How to Prove a Quadratic Stays Above a Line for Every x and How to Find the Maximum Area of a Rectangle Under a Parabola.
For further resources, visit the Additional Mathematics archive or the eduKate reference tutorial approach; families can enquire about suitable small-group tuition. The transferable insight is that subtracting the chord reveals a factored quadratic gap whose exact area depends on curvature and horizontal width alone. Properly taught kids shine a bright light into the future.