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 integrals to infinity as limits, stating convergence or divergence.
- Handle integrands that blow up at an endpoint the same careful way.
WHAT YOU PROBABLY THINK
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 and the result is the height of the rectangle with the same area, the mean value of the function:
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, the integral converges; if it grows without bound, it diverges. The opening claim fails precisely when the tail shrinks fast enough.
WORKED EXAMPLE
A finite tail
Evaluate ∫ 1/x² dx from 1 to ∞, or show it diverges.
∫ from 1 to t: [−1/x] = 1 − 1/t.
As t → ∞, 1/t → 0, so the integral converges to 1.
Compare 1/x: [ln x] from 1 to t is ln t, which grows for ever. Divergent, despite the curves looking near-identical on a sketch.
TRY IT UNSEEN
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.
THE EXAM BIT
- Mean value is integral over width; forgetting the divide is the standard slip.
- Write improper integrals with a limit letter t and the words 'as t tends to'; jumping straight to ∞ loses marks.
- Say 'converges to' or 'diverges' explicitly; a bare value without the verdict drops a mark.
- 1/x is the boundary case: powers below it diverge on [1, ∞), powers above converge.
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
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Mean value: the integral divided by the interval width, the level line of equal area.
Improper integrals: integrate to a letter, then take the limit and name the verdict.
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.
CHECK YOUR PROGRESS
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- Compute the mean value of a function as the integral divided by the width.
- Evaluate integrals to infinity as limits, stating convergence or divergence.
- Handle integrands that blow up at an endpoint the same careful way.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.