01 · Exponential curvature
Does the trapezium rule overestimate ∫₀² eˣ dx?
Hint
Compute the second derivative.
Worked solution
Yes. f″ = eˣ > 0, so chords lie above the curve.
Understand · explore · practise
Trapezium rule accuracy
Decide whether the trapezium rule overestimates or underestimates using curvature. Compare strip counts, signed and percentage error, mixed curvature and rounded ordinates.
Before you startTrapezium rule, definite integrals and second derivatives.
01 / Compare each chord with the true curve
Curvature determines the direction of the signed error.
Convex curve: T ≥ I. Concave curve: T ≤ I.
For a function with that curvature throughout the interval; I is the signed integral.
A trapezium integrates the straight chord joining two neighbouring ordinates. Compare the chord with the curve over the entire strip. Being increasing or decreasing alone does not tell you which is above.
Compare chords with the curveExplore
T = 2.75; exact integral I = 2.666667.
Signed error T − I = 0.083333; absolute error = 0.083333.
Percentage error relative to |I| = 3.125%.
The function is convex: each chord lies on or above the curve, so T ≥ I.
These are signed integrals. Overestimate means a larger numerical value, including when the true integral is negative. The exact values here are available so you can check the approximations.
02 / A convex graph lies below its chords
Nonnegative second derivative gives an overestimate.
Use four strips to estimate ∫₀² x² dx.Worked example
f″(x) = 2 > 0
The curve is convex throughout.
T = (1/4)[0 + 4 + 2(1/4 + 1 + 9/4)] = 11/4
The chord approximation lies above the curve.
I = [x³/3]₀² = 8/3
Evaluate the exact integral for comparison.
T − I = 1/12 > 0
This is an overestimate.
The animation uses two strips to make the chord gaps easier to see.
Watch: chords lie above or below
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / A concave graph lies above its chords
Nonpositive second derivative gives an underestimate.
Use four strips for ∫₀² (4 − x²) dx.Worked example
f″(x) = −2 < 0
The curve is concave throughout.
T = 8 − 11/4 = 21/4
The constant part integrates exactly; subtract the x² estimate.
I = 8 − 8/3 = 16/3
Use the exact integral.
T − I = −1/12
This is an underestimate.
02 · Logarithmic curvature
For x > 0, does the trapezium rule overestimate or underestimate ∫₁² ln x dx?
Hint
f″ = −1/x².
Worked solution
Underestimate: ln x is concave on [1,2].
03 · Linear exactness
What happens when f″ = 0 throughout an interval?
Hint
The graph is a straight line there.
Worked solution
The trapezium rule is exact, apart from numerical rounding.
04 / Increasing does not mean overestimated
Gradient and curvature answer different questions.
Compare eˣ and ln x on [1,2].Worked example
Both functions are increasing
Their first derivatives are positive.
eˣ is convex, so its trapezium estimate is above
Its second derivative is positive.
ln x is concave, so its estimate is below
Its second derivative is negative.
04 · A decreasing convex curve
Classify the trapezium estimate for ∫₁² 1/x dx.
Hint
f′ = −1/x² but f″ = 2/x³.
Worked solution
Overestimate. The function decreases, but it is convex.
05 · A decreasing concave curve
Classify the trapezium estimate for ∫₀¹ −x² dx.
Hint
f″ = −2.
Worked solution
Underestimate of the signed integral: T is more negative than I.
05 / Compare numerical values for signed integrals
More negative means smaller, not larger.
Use one strip for ∫₀¹ −x² dx.Worked example
T = (1/2)(0 − 1) = −1/2
Apply the signed rule.
I = −1/3
This is the true signed integral.
−1/2 < −1/3, so T is an underestimate
The geometric area magnitude 1/2 instead overestimates 1/3.
06 · Name the quantity
If T = −2 and I = −3, is T an overestimate of I?
Hint
Compare them on a number line.
Worked solution
Yes: −2 > −3. But |T| underestimates |I|.
07 · An error sign
If T − I = −0.04, what does that say?
Hint
The approximation is below the exact signed integral.
Worked solution
It is an underestimate by 0.04 in absolute value.
06 / Do not make a global claim from mixed curvature
Positive and negative strip errors can cancel.
Apply equal strips to f(x) = x³ − 3x on [−2,2].Worked example
f″(x) = 6x changes sign at zero
There is no single curvature sign over the full interval.
The function is odd and the nodes are symmetric
Paired weighted ordinates cancel.
Both the trapezium estimate and the exact signed integral are zero
The cancellation makes the signed result exact here.
The chords still differ from the curve on individual strips
Exact total does not mean exact geometry or zero geometric area.
08 · An inflection inside the interval
Can f″ changing sign tell you by itself whether the total trapezium estimate is above or below?
Hint
Different strips can contribute opposite errors.
Worked solution
No. Analyse the pieces or calculate a comparison.
09 · Zero denominator
Can you quote percentage error relative to an exact integral of zero?
Hint
The usual formula divides by |I|.
Worked solution
No. Report absolute error; relative percentage error is undefined.
07 / More strips usually improve a smooth approximation
Do not turn a useful pattern into an unconditional guarantee.
Compare n = 2, 4 and 8 for ∫₀² x² dx.Worked example
Exact I = 8/3
Use this to measure actual error.
T₂ = 3; T₄ = 11/4; T₈ = 43/16
Halve the strip width each time.
Errors: 1/3, 1/12, 1/48
Here each doubling divides the error by four.
The estimates approach I from above
This agrees with convexity.
For sufficiently smooth functions, trapezium error commonly scales like h² as the mesh becomes fine. Mixed-curvature cancellation and rounded data can spoil simple monotonic improvement. Do not promise that every larger strip count always gives a closer result.
10 · A further quadratic refinement
For the same quadratic, predict the error with 16 strips.
Hint
The exact error pattern here divides by four.
Worked solution
1/192.
08 / State what your error is relative to
Use the exact magnitude in the denominator.
Find percentage error for the four-strip x² estimate.Worked example
Absolute error = |11/4 − 8/3| = 1/12
Discard the sign only for the absolute error.
Percentage error = [(1/12)/(8/3)] × 100
Use |I|, not the approximation, in this convention.
3.125%
The signed error is positive, so this is an overestimate.
11 · Percentage error from given values
If I = 5 and T = 5.1, find the percentage error relative to |I|.
Hint
Use 0.1/5 × 100.
Worked solution
2%.
12 · Negative exact value
If I = −4 and T = −4.2, find absolute and percentage errors.
Hint
Use |T − I| and |I|.
Worked solution
Absolute error 0.2; percentage error 5%. T underestimates the signed integral.
09 / Separate curve approximation from rounded data
More strips cannot undo inaccurate ordinates.
All stored ordinates differ from their true values by at most ε.Worked example
The weights sum to 2n
Endpoints contribute 2; interiors contribute 2(n − 1).
The maximum propagated change is (h/2)(2n)ε
Use absolute values for a worst-case bound.
Since nh = b − a, the bound is (b − a)ε
This bounds ordinate-rounding error only, not the trapezium approximation error.
13 · A rounding bound
On an interval of width 3, each ordinate is rounded to the nearest 0.001. Bound the change in T caused by rounding.
Hint
Each ordinate error is at most 0.0005.
Worked solution
At most 3 × 0.0005 = 0.0015, ignoring additional calculator arithmetic rounding.
10 / Give a reason, not just a numerical guess
A sketch supports your explanation; curvature substantiates it.
A useful answer names the chord position and connects it to the integral. For example: “f″ is positive on the interval, so the curve is convex and each chord lies above it; therefore the trapezium rule overestimates the signed integral.” If curvature changes, explain why a single global conclusion does not follow.
14 · Repair a justification
“The curve rises, so the estimate is too high.” What is missing?
Hint
A rising graph may be concave or convex.
Worked solution
You need the curvature or a direct chord comparison over the interval.
15 · Exactness by cancellation
Does a zero total error prove every strip is exact?
Hint
Opposite errors can cancel.
Worked solution
No. The odd cubic example has exact total signed integral despite curved gaps on individual strips.
11 / Check curvature, quantity and precision
Keep signed error, absolute error and percentage error distinct.
- Convex chords lie above; concave chords lie below.
- Compare signed values when classifying over- and underestimates.
- Increasing or decreasing alone is insufficient.
- Mixed curvature can cause cancellation.
- Refinement helps under suitable conditions; check rather than assume.
- Use |I| for percentage error and reject I = 0.
- Keep full working precision and separate data error from approximation error.
16 · Final classification
For f(x) = e⁻ˣ on [0,2], classify the trapezium estimate.
Hint
Differentiate twice.
Worked solution
f″ = e⁻ˣ > 0. The function is convex, so the trapezium estimate is an overestimate.
Section 1 of 11 · Compare each chord with the true curve