Questions on mean values and improper integrals

Question 1

Define the mean value of a function f on the interval [a, b], and describe what it represents geometrically.

Accepted 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.

Question 2

Find the mean value of f(x) = sin x on the interval [0, π].

Accepted 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.

Question 3

Find the mean value of f(x) = x − 3 on the interval [0, 2].

Accepted Answer

∫(x − 3) dx from 0 to 2 = [x²/2 − 3x] = (2 − 6) − 0 = −4. Dividing by the width 2 gives mean −2. M1 for integrating, A1 for −4, A1 for −2. The mean value is negative here because f is negative throughout the interval, from −3 at x = 0 to −1 at x = 2: a mean value is a signed average, not an area.

Question 4

Without evaluating it, state whether ∫ x⁻⁴ dx from 1 to ∞ converges or diverges, and give a reason.

Accepted Answer

It converges. B1 for convergence. The power 4 is above the boundary case 1, so 1/x⁴ shrinks to zero fast enough for the tail to trap a finite area. B1 for the reason. The boundary itself, p = 1, is the one power for which the antiderivative is a logarithm rather than another power of x.

Question 5

Find the exact mean value of f(x) = e^2x on [0, 1].

Accepted Answer

∫e^2x dx from 0 to 1 = [e^2x/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.

Question 6

Evaluate ∫ x^−3/2 dx from 1 to ∞, or show that it diverges.

Accepted 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.

Question 7

Evaluate ∫ 1/√x dx from 0 to 4, treating the lower limit with care.

Accepted 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.

Question 8

Both 1/x and 1/x^3/2 tend to zero as x → ∞, yet only one of ∫ from 1 to ∞ converges. Identify which, and explain what distinguishes them.

Accepted Answer

The 1/x^3/2 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.

Question 9

Evaluate ∫ xe^−2x dx from 0 to ∞, or show that it diverges.

Accepted Answer

Integrate to a limit t by parts, with u = x and dv = e^−2x dx, so v = −½e^−2x: ∫xe^−2x dx = −½xe^−2x + ½∫e^−2x dx = −½xe^−2x − ¼e^−2x. M1 for the parts split, A1 for the antiderivative. Between 0 and t this is (−½te^−2t − ¼e^−2t) − (0 − ¼) = ¼ − ½te^−2t − ¼e^−2t. As t → ∞, te^−2t → 0 by the standard limit with k = 1, and e^−2t → 0, so the integral converges to ¼. M1 for both limits, A1 for ¼.

Question 10

Evaluate ∫ 1/x² dx from 2 to ∞, or show that it diverges.

Accepted Answer

Integrate to a limit t: [−1/x] from 2 to t = −1/t + 1/2. M1 for integrating to a limit t, A1 for −1/t + 1/2. As t → ∞, 1/t → 0, so the integral converges to 1/2. M1 for letting t → ∞, A1 for 1/2.

Question 11

The temperature T °C during a 24-hour period is modelled by T(t) = 10 + 8 sin(πt/12) for 0 ≤ t ≤ 24, with t in hours from midnight. Find the mean temperature over the 24 hours, and interpret the result.

Accepted Answer

∫T(t) dt from 0 to 24 = ∫10 dt + 8∫sin(πt/12) dt, both from 0 to 24. The constant term gives 240. M1 for splitting the integral. The sine term integrates to [−(96/π)cos(πt/12)] from 0 to 24 = −(96/π)(cos 2π − cos 0) = −(96/π)(1 − 1) = 0, since t = 0 and t = 24 are exactly one full period apart. M1 for the sine term over a full period. So the total is 240, and dividing by the width 24 gives mean 10 °C. A1 for 240, A1 for 10. The mean equals the model's constant term exactly, because the sine term contributes nothing to the integral once a whole period has passed, whatever its amplitude.

Question 12

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⁺.

Accepted 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.

Question 13

For p > 1, show that ∫ x^−p dx from 1 to ∞ converges, and find its value in terms of p. Then show that the boundary case p = 1 diverges. Finally, check your general value against the case p = 3/2, which you evaluated directly earlier in this workbook.

Accepted Answer

Integrate to a limit t: ∫x^−p dx = x^1−p/(1 − p), valid since p ≠ 1. M1 for the antiderivative. Between 1 and t this is (t^1−p − 1)/(1 − p). Since p > 1, the exponent 1 − p is negative, so t^1−p → 0 as t → ∞, and the expression tends to −1/(1 − p) = 1/(p − 1). M1 for using p > 1 to justify the limit, A1 for the exact value in terms of p. At p = 1 the antiderivative above is undefined, and directly, ∫1/x dx = [ln x] from 1 to t = ln t, which grows without bound as t → ∞, so the integral diverges. B1 for evaluating the p = 1 case separately, B1 for the conclusion that it diverges. Setting p = 3/2 in the general formula gives 1/(3/2 − 1) = 2, which is exactly the value found by integrating x^−3/2 from 1 to ∞ earlier in this workbook. A1 for the check against that known value. The 1/x² integral evaluated earlier is not a check on this formula, because it runs from 2 rather than from 1: moving the lower limit changes the value, here from 1 to 1/2, while leaving convergence untouched.

Question 14

Evaluate ∫ 1/x^2/3 dx from −8 to 1, or show that it diverges.

Accepted Answer

The integrand is undefined at x = 0, which lies inside the interval, so the integral is split there into two one-sided limits. B1 for recognising the interior singularity and the need to split. The antiderivative is ∫x^−2/3 dx = 3x^1/3. M1 for the antiderivative. Right piece, from t to 1: [3x^1/3] = 3 − 3t^1/3, which tends to 3 as t → 0⁺. A1 for 3. Left piece, from −8 to s: [3x^1/3] = 3s^1/3 − 3(−8)^1/3 = 3s^1/3 + 6, using the real cube root (−8)^1/3 = −2, which tends to 6 as s → 0⁻. M1 for the left piece with the cube root of −8, A1 for 6. Both one-sided limits are finite, so the integral converges, to 3 + 6 = 9. A1 for 9.

Question 15

A tank is filled by a valve whose flow rate is modelled by R(t) = 20 − t² litres per minute, for 0 ≤ t ≤ 4. Find the mean flow rate over these 4 minutes, and explain why it is not the same as the average of the rates at t = 0 and at t = 4.

Accepted Answer

∫(20 − t²) dt from 0 to 4 = [20t − t³/3] = 80 − 64/3 = 176/3. M1 for integrating, A1 for 176/3. Dividing by the width 4 gives mean 44/3 litres per minute, about 14.7. M1 for dividing by the width, A1 for 44/3. The rate at t = 0 is 20 and at t = 4 is 20 − 16 = 4, and their average is (20 + 4)/2 = 12, which is less than 44/3. B1 for the comparison with a reason: R(t) is not linear, since the −t² term makes its graph curve above the straight line joining the two endpoint readings, so the rate spends more of the interval closer to its higher early values than a steady decline would, pulling the true mean above the naive average of the two readings.