Mean Value of a Function

The mean value (or average value) of a function over an interval is an elegant Further-Mathematics idea: it extends "the average of a list of numbers" to "the average of a continuously-varying quantity". It connects integration to averaging, has a clean geometric reading (the height of an equal-area rectangle), and underpins the root-mean-square value that drives electrical engineering and physics. This lesson develops the definition, the Mean Value Theorem for integrals, RMS, the parametric form, and full exam technique.

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1 Spec mapping (AQA 7367)

The mean value of a function is compulsory Pure content for Papers 1 and 2 (each 2 h, 100 marks, 33⅓%, with per-paper weighting AO1 55% / AO2 25% / AO3 20%), within the "Further calculus" strand. It is largely AO1 (apply the averaging formula and integrate correctly), with AO2 appearing whenever a question demands interpretation ("explain why the mean lies between the maximum and minimum") or a proof of the Mean Value Theorem flavour, and AO3 when set in an applied context such as average velocity or effective power. RMS is the synoptic payoff — it links to mechanics (effective force), to the normal distribution's variance in Statistics, and to A-Level Maths integration of $\sin^2$sin2 and $\cos^2$cos2 via the double-angle identities. Because the calculation is short, examiners reward precise communication and sensible checks rather than length, so this is a topic where disciplined presentation directly converts into marks.

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2 Core theory — definition and geometric meaning

The mean value of a continuous function $f$f over $[a,b]$[a,b] is

$\bar{f} = \frac{1}{b-a}\int_a^{b} f(x)\,dx.$fˉ​=b−a1​∫ab​f(x)dx.

In words: take the total signed area under the curve and share it evenly across the width $b-a$b−a. The result is the height of a rectangle, on the same base $[a,b]$[a,b], enclosing exactly the same area:

$(b-a)\,\bar{f} = \int_a^{b} f(x)\,dx = A \quad\Longrightarrow\quad \bar{f} = \frac{A}{b-a}.$(b−a)fˉ​=∫ab​f(x)dx=A⟹fˉ​=b−aA​.

This mean-value rectangle "levels out" all the bumps and dips of $y=f(x)$y=f(x) into one flat height: imagine the area under the curve as water that is then allowed to settle to a level surface — the resulting depth is $\bar f$fˉ​. Geometrically, the parts of the curve poking above the line $y=\bar f$y=fˉ​ enclose exactly as much area as the troughs falling below it, so the line splits the region into matching surplus and deficit. Because the average cannot exceed the largest value nor fall below the smallest, we always have

$\min_{[a,b]} f \ \le\ \bar{f}\ \le\ \max_{[a,b]} f,$min[a,b]​f ≤ fˉ​ ≤ max[a,b]​f,

a sanity check you should apply to every answer.

Why divide by $b-a$b−a? Compare with a discrete average $\frac{1}{n}\sum_{i=1}^{n} f(x_i)$n1​∑i=1n​f(xi​). Replacing the sum by an integral and the count $n$n by the length $b-a$b−a gives exactly $\frac{1}{b-a}\int_a^b f\,dx$b−a1​∫ab​fdx: the continuous limit of "add them up and divide by how many".

To make that precise, sample $f$f at $n$n equally-spaced points $x_i = a + i\,\delta x$xi​=a+iδx with $\delta x = \dfrac{b-a}{n}$δx=nb−a​. The ordinary average of the samples is

$\frac{1}{n}\sum_{i=1}^{n} f(x_i) = \frac{1}{n}\sum_{i=1}^{n} f(x_i) \cdot \frac{\delta x}{\delta x} = \frac{1}{n\,\delta x}\sum_{i=1}^{n} f(x_i)\,\delta x = \frac{1}{b-a}\sum_{i=1}^{n} f(x_i)\,\delta x,$n1​∑i=1n​f(xi​)=n1​∑i=1n​f(xi​)⋅δxδx​=nδx1​∑i=1n​f(xi​)δx=b−a1​∑i=1n​f(xi​)δx,

using $n\,\delta x = b-a$nδx=b−a. As $n\to\infty$n→∞ the sum $\sum f(x_i)\,\delta x$∑f(xi​)δx becomes the integral $\int_a^b f\,dx$∫ab​fdx, giving $\bar f = \dfrac{1}{b-a}\int_a^b f\,dx$fˉ​=b−a1​∫ab​fdx exactly. The definition is therefore the honest continuous version of "the average of infinitely many samples", not an arbitrary formula to be memorised. Understanding this derivation also makes the units obvious: $\bar f$fˉ​ carries the same units as $f$f itself, since dividing the integral (units of $f$f times $x$x) by the width (units of $x$x) cancels the $x$x-units.

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3 Worked examples with M1/A1 mark scheme

Worked Example 1 — a polynomial

Find the mean value of $f(x)=x^2$f(x)=x2 over $[0,3]$[0,3].

$\int_0^{3} x^2\,dx = \left[\frac{x^3}{3}\right]_0^{3} = \frac{27}{3} = 9 \qquad \text{(M1 integrate, A1 value 9)}$∫03​x2dx=[3x3​]03​=327​=9(M1 integrate, A1 value 9)

$\bar{f} = \frac{1}{3-0}\times 9 = 3 \qquad \text{(M1 apply } \tfrac{1}{b-a}, \text{ A1 value 3)}$fˉ​=3−01​×9=3(M1 apply b−a1​, A1 value 3)

The mean value is $\boxed{3}$3​; since $0 \le 3 \le 9 = f(3)$0≤3≤9=f(3), it sits between the min and max as expected.

Worked Example 2 — a trigonometric function

Find the mean value of $\sin x$sinx over $[0,\pi]$[0,π].

$\int_0^{\pi} \sin x\,dx = \big[ -\cos x \big]_0^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2$∫0π​sinxdx=[−cosx]0π​=−cosπ+cos0=1+1=2

$\bar{f} = \frac{1}{\pi}\times 2 = \frac{2}{\pi} \approx 0.637.$fˉ​=π1​×2=π2​≈0.637.

This is approximately $0.637$0.637. The peak of $\sin x$sinx on $[0,\pi]$[0,π] is $1$1 and the minimum is $0$0, and indeed $0 < \tfrac{2}{\pi} < 1$0<π2​<1, so the answer passes the min–max sanity check comfortably.

Worked Example 3 — an exponential

Find the mean value of $e^{x}$ex over $[0,1]$[0,1].

$\int_0^{1} e^{x}\,dx = \big[ e^{x} \big]_0^{1} = e - 1, \qquad \bar{f} = \frac{e-1}{1-0} = e - 1 \approx 1.718.$∫01​exdx=[ex]01​=e−1,fˉ​=1−0e−1​=e−1≈1.718.

Worked Example 3b — a rational function

Find the mean value of $f(x) = \dfrac{1}{x}$f(x)=x1​ over $[1,e]$[1,e].

$\int_1^{e} \frac{1}{x}\,dx = \big[ \ln x \big]_1^{e} = \ln e - \ln 1 = 1, \qquad \bar{f} = \frac{1}{e-1}\times 1 = \frac{1}{e-1} \approx 0.582.$∫1e​x1​dx=[lnx]1e​=lne−ln1=1,fˉ​=e−11​×1=e−11​≈0.582.

Notice the answer lies between the function's values at the ends, $f(e)=\tfrac1e\approx 0.368$f(e)=e1​≈0.368 and $f(1)=1$f(1)=1 — the usual sanity check holds.

Worked Example 3c — a curve dipping below the axis

Find the mean value of $f(x) = \cos x$f(x)=cosx over $[0,\pi]$[0,π]. Because $\cos x$cosx is positive on $[0,\tfrac{\pi}{2})$[0,2π​) and negative on $(\tfrac{\pi}{2},\pi]$(2π​,π], expect substantial cancellation:

$\int_0^{\pi} \cos x\,dx = \big[ \sin x \big]_0^{\pi} = \sin\pi - \sin 0 = 0, \qquad \bar{f} = \frac{1}{\pi}\times 0 = 0.$∫0π​cosxdx=[sinx]0π​=sinπ−sin0=0,fˉ​=π1​×0=0.

The mean is exactly $0$0: the positive area on the first half is cancelled by the equal negative area on the second. This is a vivid reminder that the mean value is a signed average — and a perfect motivation for the root-mean-square, which never suffers this cancellation.

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4 Specimen-style exam question

(Specimen-style — not a real past paper.)

(a) Find the mean value of $f(x) = \sin^2 x$f(x)=sin2x over one full period $[0, 2\pi]$[0,2π]. (b) Hence write down the mean value of $\cos^2 x$cos2x over the same interval, justifying your answer.

(a) Use the double-angle identity $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$sin2x=21​(1−cos2x):

$\int_0^{2\pi} \sin^2 x\,dx = \int_0^{2\pi} \frac{1 - \cos 2x}{2}\,dx = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_0^{2\pi} = \frac{2\pi}{2} - 0 = \pi.$∫02π​sin2xdx=∫02π​21−cos2x​dx=[2x​−4sin2x​]02π​=22π​−0=π.

$\bar{f} = \frac{1}{2\pi}\times \pi = \frac{1}{2}.$fˉ​=2π1​×π=21​.

(b) Since $\sin^2 x + \cos^2 x = 1$sin2x+cos2x=1, the two means must sum to $1$1; by symmetry they are equal, so the mean of $\cos^2 x$cos2x is also $\tfrac12$21​. (Equivalently $\cos^2 x = \tfrac12(1+\cos 2x)$cos2x=21​(1+cos2x) integrates to $\pi$π likewise.)

(c) Hence write down the RMS value of $\sin x$sinx over $[0,2\pi]$[0,2π]. The RMS is the square root of the mean of the square, and the mean of $\sin^2 x$sin2x is $\tfrac12$21​ from part (a), so $f_{\text{rms}} = \sqrt{\tfrac12} = \tfrac{1}{\sqrt2}$frms​=21​​=2​1​. The word "Hence" signals exactly this reuse of the part-(a) result — no fresh integration is required, and re-deriving it from scratch would waste time and risk error.

Key result. The mean of $\sin^2$sin2 or $\cos^2$cos2 over any whole number of periods is always $\tfrac12$21​ — the single most-used fact in RMS calculations. Commit it to memory together with its consequence $f_{\text{rms}} = \tfrac{A}{\sqrt2}$frms​=2​A​ for a sinusoid of amplitude $A$A.

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5 The Mean Value Theorem for integrals, and a linear shortcut

Mean Value Theorem (for integrals). If $f$f is continuous on $[a,b]$[a,b], there exists at least one $c \in (a,b)$c∈(a,b) with $\displaystyle f(c) = \frac{1}{b-a}\int_a^{b} f(x)\,dx$f(c)=b−a1​∫ab​f(x)dx.

Why it is true (proof sketch). Let $m$m and $M$M be the minimum and maximum of the continuous function $f$f on $[a,b]$[a,b]. Then $m \le f(x) \le M$m≤f(x)≤M for all $x$x, and integrating preserves the inequalities:

$m(b-a) \le \int_a^{b} f(x)\,dx \le M(b-a) \quad\Longrightarrow\quad m \le \bar f \le M.$m(b−a)≤∫ab​f(x)dx≤M(b−a)⟹m≤fˉ​≤M.

So the average $\bar f$fˉ​ is a value between the minimum and maximum of $f$f. By the Intermediate Value Theorem, a continuous function takes every value between its minimum and maximum, so there is some $c\in[a,b]$c∈[a,b] with $f(c) = \bar f$f(c)=fˉ​. That is, the function genuinely attains its average somewhere. For $f(x)=x^2$f(x)=x2 on $[0,3]$[0,3] we found $\bar f = 3$fˉ​=3; solving $c^2 = 3$c2=3 gives $c = \sqrt 3 \approx 1.732 \in (0,3)$c=3​≈1.732∈(0,3), confirming the theorem. Note that $c$c is not generally the midpoint — here the midpoint is $1.5$1.5, but the mean is attained at $\sqrt3 \approx 1.73$3​≈1.73.

Worked Example 4 — linear function and a useful shortcut

Find the mean value of $f(x) = 2x + 1$f(x)=2x+1 over $[0,4]$[0,4].

$\int_0^{4} (2x+1)\,dx = \big[ x^2 + x \big]_0^{4} = 16 + 4 = 20, \qquad \bar{f} = \frac{20}{4} = 5.$∫04​(2x+1)dx=[x2+x]04​=16+4=20,fˉ​=420​=5.

For any linear $f(x)=mx+c$f(x)=mx+c, the mean over $[a,b]$[a,b] equals the value at the midpoint, $f\!\left(\tfrac{a+b}{2}\right)$f(2a+b​). Here $f(2) = 5$f(2)=5, matching — a quick check worth knowing. The reason is that a straight line's graph over $[a,b]$[a,b] has its "level surface" exactly at the midpoint height, so the surplus above the line on one side balances the deficit on the other by symmetry. This shortcut is only valid for linear functions; do not be tempted to use it on curves.

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6 Mark-scheme literacy

Examiners reward the structure as much as the answer. Write the formula $\bar f = \tfrac{1}{b-a}\int_a^b f\,dx$fˉ​=b−a1​∫ab​fdx before substituting — the $\tfrac{1}{b-a}$b−a1​ factor is itself a method mark, and omitting it is the commonest way to lose marks on this topic. "Find the mean value" and "find the average value" mean the same thing, so do not be thrown by the wording. The command "Hence" requires you to reuse the integral or result you just computed (do not restart from scratch — a fresh method may even be penalised under "Hence"). For RMS, the structure is square $\to$→ integrate $\to$→ average $\to$→ root — showing each stage protects follow-through marks if one slips. Leave answers exact (e.g. $\tfrac{2}{\pi}$π2​, $\tfrac{1}{\sqrt2}$2​1​) unless a decimal is requested, and always perform the min–max sanity check: a mean value outside the range of $f$f on the interval is definitely an error. A final-answer-only response with no working caps the marks even when the number is right.

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7 Grade-band model answers

Question: Find the mean value of $\cos x$cosx over $\left[0, \tfrac{\pi}{2}\right]$[0,2π​] and state, with reason, what it tells you.

Mid-band response. "$\int_0^{\pi/2}\cos x\,dx = [\sin x] = 1$∫0π/2​cosxdx=[sinx]=1. Mean $= 1 \div \tfrac{\pi}{2} = \tfrac{2}{\pi}$=1÷2π​=π2​."

Examiner-style commentary. Correct value and a correctly evaluated integral, but the limits are not written in the bracket and there is no interpretation — the AO2 "state what it tells you" is unaddressed, so a communication mark is lost.

Stronger response. "$\int_0^{\pi/2}\cos x\,dx = [\sin x]_0^{\pi/2} = 1 - 0 = 1$∫0π/2​cosxdx=[sinx]0π/2​=1−0=1. Mean $= \tfrac{1}{\pi/2}\times 1 = \tfrac{2}{\pi}\approx 0.64$=π/21​×1=π2​≈0.64. This lies between $0$0 and $1$1, the min and max of $\cos x$cosx on the interval."

Examiner-style commentary. Complete and well-presented; limits shown, the $\tfrac{1}{b-a}$b−a1​ factor explicit, and a sensible check. Full marks.

Top-band response. "$\displaystyle \bar f = \frac{1}{\frac{\pi}{2}-0}\int_0^{\pi/2}\cos x\,dx = \frac{2}{\pi}\big[\sin x\big]_0^{\pi/2} = \frac{2}{\pi}(1) = \frac{2}{\pi}\approx 0.637$fˉ​=2π​−01​∫0π/2​cosxdx=π2​[sinx]0π/2​=π2​(1)=π2​≈0.637. This equals the mean of $\sin x$sinx over $[0,\pi]$[0,π] — unsurprising, since $\cos x = \sin\!\left(\tfrac{\pi}{2}-x\right)$cosx=sin(2π​−x) maps one interval onto the other. The value lies strictly between the function's minimum $0$0 and maximum $1$1, as the Mean Value Theorem guarantees, and is attained at $x=\arccos\tfrac{2}{\pi}\approx 0.88$x=arccosπ2​≈0.88."