Practise › Questions › Mean values and improper integrals
Mean values and improper integrals questions
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.
7 original questions · 26 marks · the mean values and improper integrals notes · Further calculus
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Define the mean value of a function f on the interval [a, b], and describe what it represents geometrically.
Worked answer
The integral of f from a to b divided by b − a. It is the height of the rectangle on the same base that has the same area as the region under the curve, the level the curve would settle to if its area were poured flat. B1 for the definition, B1 for the geometric description.Find the mean value of f(x) = sin x on the interval [0, π].
Worked answer
∫sin x dx from 0 to π = [−cos x] = 2. Dividing by the width π gives mean 2/π ≈ 0.637. M1 for integrating, A1 for 2, A1 for 2/π. Less than the maximum 1, as the curve spends its time rising to the peak and falling away.Find the exact mean value of f(x) = e2x on [0, 1].
Worked answer
∫e2x dx from 0 to 1 = [e2x/2] = (e² − 1)/2. The width is 1, so the mean is (e² − 1)/2 ≈ 3.19. M1 for integrating, A1 for the antiderivative, M1 for dividing by the width, A1 for the exact mean. The mean sits well below the endpoint value e² ≈ 7.39 because an exponential spends most of any interval far below its final value.Evaluate ∫ x−3/2 dx from 1 to ∞, or show that it diverges.
Worked answer
Integrate to a limit t: [−2x−1/2] from 1 to t = 2 − 2/√t. As t → ∞ the second term vanishes, so the integral converges to 2. M1 for integrating to a limit t, A1 for 2 − 2/√t, M1 for letting t → ∞, A1 for 2. The power 3/2 is above the boundary case 1, which is what buys convergence.Evaluate ∫ 1/√x dx from 0 to 4, treating the lower limit with care.
Worked answer
The integrand is unbounded at 0, so integrate from t to 4: [2√x] = 4 − 2√t. As t → 0⁺ this tends to 4, so the integral converges to 4. M1 for replacing the lower limit by t, A1 for 4 − 2√t, M1 for the limit as t → 0, A1 for 4. The spike at zero is infinitely tall but narrow enough to hold finite area.Both 1/x and 1/x3/2 tend to zero as x → ∞, yet only one of ∫ from 1 to ∞ converges. Identify which, and explain what distinguishes them.
Worked answer
The 1/x3/2 integral converges, to 2. The 1/x integral gives ln t, which grows without bound. B1 for naming the convergent integral, B1 for the divergence of ln t, B1 for the explanation in terms of rate. Tending to zero is not enough on its own, since what settles the question is how fast. Powers of x below −1 shrink quickly enough to trap finite area, and 1/x sits exactly on the divergent side of the boundary.Show that ∫ ln x dx from 0 to 1 converges, and find its exact value. You may assume that t ln t → 0 as t → 0+.
Worked answer
ln x is unbounded below as x → 0+, so the integral is improper at the lower limit and must be handled as a limit. Integrating by parts with u = ln x and dv = dx gives ∫ln x dx = x ln x − x. Between t and 1 this is (0 − 1) − (t ln t − t) = −1 − t ln t + t. As t → 0+ the term t ln t tends to 0 by the given result and t tends to 0, so the limit exists and the integral converges to −1. B1 for recognising the integral is improper at 0, M1 for integration by parts, A1 for x ln x − x, M1 for evaluating between t and 1, A1 for −1 − t ln t + t, A1 for the value −1. The value is negative because ln x lies below the axis throughout 0 < x < 1. Writing [x ln x − x] from 0 to 1 straight out is worth no marks, since x ln x is undefined at 0 and the limit is exactly what needs showing.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise mean values and improper integrals one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.