Gradients of Curves & Areas Under Graphs

SPEC A15 • ALGEBRA

Higher tier only. Estimate the gradient at a point with a tangent, and the area under a curve with the trapezium rule. Two 3–4 mark questions, one lesson.

Straight Lines Won't Work Here

Up to now, gradient has been rise ÷ run on a straight line, and area has come from standard shapes — triangles, rectangles, trapeziums.

On a curve, both methods break down. The gradient changes at every point, and the shape under the curve is not a standard region.

The Higher-Tier Toolkit

Gradient at a single point — draw a tangent, then use rise over run on that straight line.

Area under a curve — split into vertical strips and use the trapezium rule.

Both give estimates, not exact values. Accuracy depends entirely on how you draw the tangent and how many strips you use.

Key Terms

Tangent

A straight line that just touches a curve at a single point, without crossing it at that point. The gradient of the tangent gives the gradient of the curve at that point.

Chord

A straight line joining two points on the curve. A chord gives an average gradient between those two points — not the gradient at a single point. Using a chord when a tangent is needed is the most common mistake on this topic.

Ordinate

A y-value on the curve at a specific x. The trapezium rule needs a list of ordinates — y₀, y₁, y₂, …, yₙ — one for each vertical strip boundary.

Gradient at a Point: Draw a Tangent

To estimate the gradient of a curve at x = 2, place a straight edge so it just grazes the curve at that point — not cutting through it. Extend the line a reasonable distance either side.

The tangent (light blue) touches the curve at P(2, 2) and extends well either side. Pick two clear points on the tangent — here (1, 1) and (4, 4) — and use rise ÷ run.

Compute the Gradient of the Tangent

Once the tangent is drawn, pick two points on the tangent — ideally with readable integer coordinates so there is no guessing.

gradient = (y₂ − y₁) ÷ (x₂ − x₁)

From the diagram on the previous slide, the two chosen points on the tangent are (1, 1) and (4, 4):

Calculation

gradient = (4 − 1) ÷ (4 − 1) = 3 ÷ 3 = 1

So the gradient of the curve at x = 2 is estimated as 1. Mark schemes accept a range of values around the true gradient — the exact range is stated in each question's mark scheme — because drawing a perfectly accurate tangent by hand is impossible.

Exam Tip

Always show the two points you used on the tangent and the subtraction. A written gradient with no working can still score method marks if the tangent is visible, but both together is worth full marks.

The Trap: Tangent vs Chord

The single most common error on this topic is drawing a chord instead of a tangent — typically by joining two marked points on the curve.

Weak student

Joins two points already plotted on the curve (e.g. at x = 1 and x = 3) and reads that as the gradient at x = 2. This is a chord, not a tangent.

Gives the average gradient between two points — not the gradient at one point.

Strong student

Places a transparent ruler so it just grazes the curve at x = 2 without crossing it, then draws the tangent and picks two clear points on that line.

Gives the gradient at exactly that point.

Examiner Alert

"Chord instead of tangent" scores zero. Mark schemes explicitly state: "Do not accept — chord instead of tangent." Your line must touch the curve at one point only. Extend it well beyond the point so you can read coordinates easily.

Area Under a Curve

For straight-line graphs (like a velocity-time graph made of triangles and trapeziums), area comes from standard shape formulae. For a curved boundary, no single formula works.

Strategy

Split the region under the curve into vertical strips of equal width, and treat each strip as a trapezium. Sum their areas to estimate the total.

This is the trapezium rule. More strips → closer to the true area. Examiners will tell you how many strips to use.

Why it works

Each strip has a straight top (the chord connecting two points on the curve) rather than the actual curve. The straight top makes the strip a trapezium — and trapezium area is easy: ½(a + b) × h.

Trapezium Rule: Visual

Four strips of width h = 1 approximate the area under y = x² + 1 from x = 0 to x = 4. Each strip's top is a chord — making a trapezium. For this concave-up curve, each chord lies above the actual curve, so the total is an over-estimate.

The Trapezium Rule Formula

For n strips of equal width h, with ordinates y₀, y₁, …, yₙ:

Area ≈ ½h × [y₀ + yₙ + 2(y₁ + y₂ + … + yₙ₋₁)]

What each part means

½ h — half the strip width, pulled out as a common factor.

y₀ + yₙ — the two end ordinates, counted once each.

2(y₁ + … + yₙ₋₁) — every middle ordinate, counted twice (because each middle ordinate is a boundary shared between two strips).

Memory hook

"Ends once, middles twice." Mis-doubling the ends (or forgetting to double the middles) is the most common arithmetic error.

The MCE Chain for This Topic

Every trapezium-rule question is multi-step. Mark schemes split the marks across method, calculation, and evaluation — miss any one and you drop a mark.

MMethod. State what you're doing: "I will use the trapezium rule with n strips of width h. I need ordinates at x = …" — list them. Or for a tangent: "I will draw a tangent at x = … and compute gradient = (y₂−y₁)/(x₂−x₁) using two points on the tangent."

CCalculate. Compute each ordinate by substituting into the curve equation. Substitute into the trapezium rule formula — show the substitution before you simplify. Keep exact values until the final line.

EEvaluate. State the estimated area with correct units (y-axis unit × x-axis unit). If the question asks for interpretation, say what the area represents — e.g. distance, total change. State whether it's an over- or under-estimate if relevant.

Worked Example: Trapezium Rule

Ex

Higher — 3 marks

Use the trapezium rule with 4 strips of width 1 to estimate the area under y = x² + 1 between x = 0 and x = 4.

M4 strips of width h = 1 between x = 0 and x = 4. I need the 5 ordinates at x = 0, 1, 2, 3, 4, found from y = x² + 1. Then apply the trapezium rule.

Worked Example: Calculate

Step 1: Find the 5 ordinates

x = 0 → y = 0² + 1 = 1

x = 1 → y = 1² + 1 = 2

x = 2 → y = 2² + 1 = 5

x = 3 → y = 3² + 1 = 10

x = 4 → y = 4² + 1 = 17

Step 2: Substitute into the trapezium rule

Area ≈ ½ × 1 × [y₀ + y₄ + 2(y₁ + y₂ + y₃)]

= ½ × [1 + 17 + 2(2 + 5 + 10)]

= ½ × [18 + 2(17)]

= ½ × [18 + 34]

= ½ × 52 = 26

Worked Example: Evaluate

Step 3: State the answer with context

Estimated area ≈ 26 square units.

The true value (by integration, for reference only) is 76/3 ≈ 25.33. The trapezium rule gave 26 — slightly larger than the true value.

Why is it an over-estimate?

The curve y = x² + 1 is concave up — it curves upward like a cup. The straight chord joining two ordinates on a concave-up curve lies above the curve itself, so each trapezium contains a sliver of extra area that isn't really under the curve.

Rule of Thumb

Concave up → over-estimate. Concave down → under-estimate. Use more strips (or include a stated reason) to improve or justify your estimate.

Area in Context: What Does It Represent?

When a curve is on real axes (not just y vs x), the area under it has a physical meaning — and naming it correctly is worth marks.

Curve typeGradient representsArea represents

Velocity–time (v–t)Acceleration (m/s²)Distance travelled (m)

Distance–time (d–t)Speed (m/s)(Not usually meaningful)

Rate of flow – time(How fast the flow is changing)Total volume flowed

Financial: amount vs timeRate of growth (£ per year)(Total paid over time, for a rate graph)

Units

Area has units = (y-axis unit) × (x-axis unit). So the area under a v–t graph with axes in m/s and s has units m/s × s = metres — which matches "distance."

Examiner Alerts

Over- / under-estimate

For a concave-up curve (curves upward), the trapezium rule overestimates — the straight tops of the trapeziums stick up above the curve. For a concave-down curve, it underestimates. This is a common follow-up part: "Is your estimate an over- or under-estimate? Explain." — worth 1 mark and easy to drop.

Nov 2024 Paper 2H Q21(c) asked exactly this — the expected answer was "underestimate" with a valid reason involving the area under the graph. Many students gave the reason but forgot to name the direction (over/under), or vice versa.

"Middles twice, ends once"

The most common arithmetic slip: doubling all ordinates, or forgetting to double any middle ordinate. Re-read the formula before you substitute — and write out 2(y₁ + y₂ + y₃) explicitly before evaluating.

Memorise the formula

The trapezium rule is not given on the AQA formulae sheet at GCSE. Learn it word-perfect: ½h × [ends + 2×middles]. Writing the formula before you substitute gets you the first method mark even if the arithmetic goes wrong later.

Practice Question 1

Tangent gradient. A curve passes through the points (0, 0), (1, 1), (2, 4) and (3, 9). A tangent to the curve is drawn at the point (2, 4). Two clearly-marked points on the tangent are (1, 0) and (4, 12).

(a) Estimate the gradient of the curve at x = 2. [2 marks]

(b) State the units of this gradient if x is time in seconds and y is distance in metres. [1 mark]

M

Use the two points given on the tangent: (1, 0) and (4, 12). Apply gradient = (y₂ − y₁) / (x₂ − x₁).

C

gradient = (12 − 0) / (4 − 1) = 12 / 3 = 4.

E

(a) Estimated gradient at x = 2 is 4. (b) On a distance–time graph, gradient = speed, so the units are metres per second (m/s).

Practice Question 2

Trapezium rule. A curve has equation y = x² between x = 0 and x = 4. Use the trapezium rule with 4 strips of equal width to estimate the area under the curve between x = 0 and x = 4. [3 marks]

M

4 strips over an interval of length 4, so width h = 1. Need ordinates at x = 0, 1, 2, 3, 4.

C

y₀ = 0, y₁ = 1, y₂ = 4, y₃ = 9, y₄ = 16. Area ≈ ½ × 1 × [0 + 16 + 2(1 + 4 + 9)] = ½ × [16 + 28] = ½ × 44 = 22.

E

Estimated area ≈ 22 square units. (Exact area is 64/3 ≈ 21.33 — so the trapezium rule gives a slight over-estimate because y = x² is concave up.)

Practice Question 3

Interpretation in context. The speed of a car, v in m/s, is plotted against time t in seconds. A student uses the trapezium rule to estimate the area under the v-t curve between t = 0 and t = 10 and gets 72.

(a) State what the area represents, including units. [2 marks]

(b) The v-t curve is concave up throughout this interval. State, with a reason, whether 72 is an over- or under-estimate of the true distance. [2 marks]

M

On a velocity–time graph, area = distance. Units = (y-axis unit) × (x-axis unit) = m/s × s = m.

C

(a) The area represents the distance travelled by the car between t = 0 and t = 10, in metres. So estimated distance ≈ 72 m.

E

(b) 72 is an over-estimate. Because the curve is concave up, the straight tops of the trapeziums lie above the curve, so each trapezium contains extra area not actually under the curve.

Summary

Gradient of a Curve at a Point

Draw a tangent — a line that just touches the curve at that one point.

Pick two clear points on the tangent with readable coordinates.

Gradient = (y₂ − y₁) / (x₂ − x₁).

Never use a chord — it gives average gradient, not gradient at a point.

Trapezium Rule

Area ≈ ½h × [y₀ + yₙ + 2(y₁ + … + yₙ₋₁)]

Ends once, middles twice.

Concave up → over-estimate . Concave down → under-estimate .

More strips → better estimate.

Area Means…

v-t graph → distance

Rate-of-flow graph → total volume

Units: y-unit × x-unit

Avoid the Traps

Chord ≠ tangent

Don't forget to double middles

State units in the final answer

Top Exam Tip

On any trapezium-rule question, write the list of ordinates in a column before substituting into the formula. Examiners see the method clearly, and you're far less likely to miscount which ones get doubled.