Maths › Further calculus › Mean values and improper integrals
Mean values and improper integrals
The average height of a curve, and integrals that dare an infinite limit. Both come down to watching what an ordinary integral does at the edges.
Builds on Areas and the limit of a sum and Geometric series.
IN THIS TOPIC
- Compute the mean value of a function as the integral divided by the width.
- Evaluate improper integrals as limits, stating convergence or divergence.
- Split an integral at an interior singularity and test each one-sided limit separately.
COMMON MISCONCEPTION
An integral with an infinite limit must give an infinite answer.
The average height of a curve
Divide the area under a curve by the width of the interval. The result is the height of the rectangle with the same area, the mean value of the function:
The booklet has no entry for it, so learn the shape. Area first, then divide by the width.
WORKED EXAMPLE
Mean of a parabola
Find the mean value of f(x) = x² on the interval [0, 3].
∫x² dx from 0 to 3 = [x³/3] = 9.
Mean = 9/(3 − 0) = 3.
The curve runs from height 0 to height 9, and the average sits at 3, well below halfway. The parabola spends most of the interval low.
Daring the edge
An improper integral has an infinite limit, or an integrand that blows up at an endpoint. Replace the awkward endpoint with a letter, integrate normally, then let the letter tend to its limit. If the answer settles on a finite value the integral converges; if it grows without bound, or oscillates, or otherwise fails to approach a finite limit, the integral diverges. An unbounded region can still enclose a finite area, provided the tail shrinks fast enough.
WORKED EXAMPLE
A finite tail
Evaluate ∫ 1/x² dx from 1 to ∞, or show it diverges.
Integrate from 1 to t: [−1/x] = 1 − 1/t.
As t → ∞, 1/t → 0, so the integral converges to 1.
Compare 1/x. Now [ln x] from 1 to t is ln t, which grows for ever, so that one diverges. The two curves look near-identical on a sketch.
INDEPENDENT PRACTICE
Blow-up at the bottom
Evaluate ∫ 1/√x dx from 0 to 1, treating the lower limit with care.
Show the working
The integrand is unbounded at 0, so integrate from t to 1 first: [2√x] = 2 − 2√t.
As t → 0⁺, 2√t → 0, so the integral converges to 2.
An infinite spike can still trap finite area. What matters is how sharply it narrows.
A hole in the middle
The singularity need not sit at an endpoint. If the integrand is undefined at a point inside the interval, split the integral there and treat each piece as its own one-sided limit. The integral converges only if both pieces do, and its value is then the sum of the two.
WORKED EXAMPLE
Splitting at an interior singularity
Evaluate ∫ 1/x2/3 dx from −1 to 1, or show it diverges.
The integrand is undefined at x = 0, inside the interval, so split there: one integral from −1 to 0 and one from 0 to 1, each improper at 0.
Right piece: ∫ from t to 1 gives [3x1/3] = 3 − 3t1/3, which tends to 3 as t → 0⁺.
Left piece: ∫ from −1 to s gives 3s1/3 + 3, which tends to 3 as s → 0⁻.
Both one-sided limits settle, so the integral converges to 6.
Skipping the split is not a shortcut but a wrong answer. Sweep 1/x² straight across from −1 to 1 and [−1/x] gives (−1) − (1) = −2: a negative answer from an integrand that is positive wherever it is defined, which is the absurdity flag. Split properly and ∫ from t to 1 of 1/x² dx = 1/t − 1, which grows without bound as t → 0⁺. One piece already diverges, so the whole integral diverges, and the tidy −2 was an artefact of integrating across a point the antiderivative cannot cross.
ASSESSMENT FOCUS
- Mean value is the integral over the width. Forgetting to divide is the standard slip.
- Write improper integrals with a limit letter and the words 'as t tends to'. Jumping straight to ∞ loses marks.
- Say 'converges to' or 'diverges' explicitly; the verdict is part of the answer.
- For ∫x−p dx on [1, ∞), convergence needs p > 1. The case p = 1 diverges, so 1/x sits just outside.
- An integrand undefined inside the interval means splitting there first. Integrating straight across can return a finite number that is simply wrong: 1/x² across [−1, 1] gives −2 that way.
CHECK YOURSELF
Find the mean value of f(x) = 1/x² on the interval [1, 2].
Show a hint
Integrate to get [−1/x], then divide by the width, which is 1.
Show the answer
∫ 1/x² dx from 1 to 2 = [−1/x] = 1 − 1/2 = 1/2. The width is 1, so the mean value is 1/2.
Mean value is the integral divided by the interval width, the level line of equal area.
Improper integrals: integrate to a letter, take the limit, then name the verdict.
A singularity inside the interval splits the integral in two, and it converges only if both halves do.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the mean values and improper integrals questions page.
CHECK YOUR PROGRESS
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- Compute the mean value of a function as the integral divided by the width.
- Evaluate improper integrals as limits, stating convergence or divergence.
- Split an integral at an interior singularity and test each one-sided limit separately.
Open the full revision checklist to see every objective in the course in one place.