Further Maths/Lessons/Methods in Calculus (Year 2)/The Mean Value of a Function
Core Pure · Methods in Calculus · Year 2

The Mean Value of a Function

Averaging a function over an interval is just integration in disguise — a direct extension of the mean of a finite list of numbers.

7 min read 1 worked example

The mean value of a function

Definition 3.2 — mean value of a function

The mean value of $f(x)$ on the interval $[a,b]$ is

$$\bar f=\frac1{b-a}\int_a^bf(x)\,dx$$

Geometrically, this is the constant height of a rectangle with base $[a,b]$ whose area equals the area under $y=f(x)$ on that interval — a direct extension of the mean of a finite list of numbers, replacing "sum and divide by count" with "integrate and divide by interval length."

A curve y=f(x) rising across the interval [a,b], with a rectangle of base b-a and height f-bar overlaid: where the curve dips below the rectangle's top a pink deficit region is shaded, and where the curve rises above the rectangle's top a grey excess region is shaded, illustrating that these two regions have equal area
The mean value $\bar f$ is the height of the rectangle (base $[a,b]$) whose area matches the area under the curve — the pink deficit and grey excess regions have equal area.
Example 1 · Finding a mean value
Find the mean value of $f(x)=\dfrac6{\sqrt{1+2x}}$ over the interval $[0,4]$.

$\bar f=\dfrac1{4-0}\displaystyle\int_0^4\dfrac6{\sqrt{1+2x}}\,dx=\dfrac64\int_0^4(1+2x)^{-1/2}\,dx$.

Using the reverse chain rule, $\displaystyle\int(1+2x)^{-1/2}\,dx=(1+2x)^{1/2}+c$ (the coefficient of $x$ inside, $2$, cancels with the $\frac12$ power's chain-rule factor).

$=\dfrac64\Big[(1+2x)^{1/2}\Big]_0^4=\dfrac64\big(\sqrt9-\sqrt1\big)=\dfrac64(3-1)=\dfrac64\times2=3$.

A useful shortcut

If $f(x)$ has mean value $m$ on $[a,b]$, then $f(x)+c$ has mean value $m+c$ on the same interval — adding a constant to the function shifts every value, and hence the average, by that same constant. This follows directly from linearity of the integral: $\displaystyle\int_a^b(f(x)+c)\,dx=\int_a^bf(x)\,dx+c(b-a)$, so dividing by $b-a$ adds exactly $c$ to the mean value.

Exam-style questions

Question 1 5 marks

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

$\bar f=\dfrac1{4-1}\displaystyle\int_1^4x^2\,dx=\dfrac13\Big[\dfrac{x^3}3\Big]_1^4$. (M1, A1)

$=\dfrac13\left(\dfrac{64}3-\dfrac13\right)=\dfrac13\times\dfrac{63}3=\dfrac13\times21=7$. (M1, A1, A1)

Question 2 7 marks

Find the mean value of $f(x)=\dfrac1{x\ln x}$ over the interval $[e,e^3]$.

$\bar f=\dfrac1{e^3-e}\displaystyle\int_e^{e^3}\dfrac1{x\ln x}\,dx$. Using the substitution $u=\ln x$, $du=\dfrac1x\,dx$: $\displaystyle\int\dfrac1{x\ln x}\,dx=\int\dfrac1u\,du=\ln|u|+c=\ln(\ln x)+c$. (M1, A1, A1)

Evaluating: $\Big[\ln(\ln x)\Big]_e^{e^3}=\ln(\ln e^3)-\ln(\ln e)=\ln3-\ln1=\ln3$. (M1, A1)

$\bar f=\dfrac{\ln3}{e^3-e}$. (A1)

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