Mean value and accumulation

What you will be able to do

• find the mean value of a function on an interval, and the point where the function reaches it;
• find the total change of a quantity from its rate;
• tell displacement from total distance when the velocity changes sign.

Where this sits. The mean value of a function is not in the Indian syllabus (NCERT Class 12 and JEE Main). Do not mix it up with the mean value theorem for derivatives of Chapter 5. AP Calculus AB and BC include the average value of a function. Some A-Level Further Maths specifications include it, and A-Level Maths and IB use integration of velocity for displacement and distance. Check your own syllabus for the exact list.

The average of a list and the average of a function

The average of $n$ numbers is their sum divided by $n$. A graph has a height at every point, so it has too many heights to add one by one. We take thin strips instead. The sum of the strip areas is the integral. Dividing by the width $b - a$ gives the average height.

Read it as a picture. The area under the curve equals the area of a rectangle of the same width, whose height is the mean value.

A first example

Find the mean value of $f(x) = x^2$ on $[0, 3]$. The integral is $\dfrac{3^3}{3} = 9$, and the width is $3$, so the mean value is $3$.

Notice what the answer is not. It is not $f$ at the middle, which is $f(1.5) = 2.25$. It is not the average of the end heights either, which is $\dfrac{0 + 9}{2} = 4.5$. The curve is not a straight line, so those shortcuts fail.

The mean value theorem for integrals

If $f$ is continuous on $[a, b]$, its mean value lies between its smallest and its largest value. A continuous function takes every value between those two, by the Intermediate Value Theorem. So there is a point $c$ in $[a, b]$ with $f(c) = f_{\text{mean}}$. For $x^2$ on $[0, 3]$ this point is $c = \sqrt{3}$, because $c^2 = 3$.

Mean values in real life

• The mean of a velocity over a time is the mean velocity, in metres per second.
• The mean of a power $p(t)$ over a time is the mean power. A heater with $p(t) = P\sin^2(\omega t)$ has mean power $\dfrac{P}{2}$, because the mean of $\sin^2$ over a full period is $\dfrac{1}{2}$.
• The mean of a temperature over a day is the average temperature of the day.

Accumulation: a total from a rate

A rate tells how much of a quantity arrives in one unit of time. In a short time $dt$, the amount that arrives is $r(t)\,dt$. Adding these small amounts gives the net change theorem:

The unit of the answer is the unit of the rate times the unit of time. Litres per minute times minutes gives litres. Amperes times seconds gives coulombs. Watts times seconds gives joules. Always bring the time unit of the rate and the time unit of the limits to match.

Displacement and distance

Let a cart move on a line with velocity $v(t)$. The integral of $v$ is the displacement, the change of position. If $v$ is negative for a while, that part counts backwards. The distance travelled counts every metre, forwards or backwards, so it is the integral of the speed $|v(t)|$. Find the times where $v = 0$ and split the integral there.

A worked example

A cart moves with $v(t) = t^2 - 5t + 4$ metres per second for $0 \le t \le 5$ seconds. Find the displacement and the distance.

1. Where is $v = 0$? $t^2 - 5t + 4 = (t - 1)(t - 4)$, so $t = 1$ and $t = 4$. The velocity is positive on $[0, 1]$, negative on $[1, 4]$ and positive on $[4, 5]$.
2. An antiderivative. $F(t) = \dfrac{t^3}{3} - \dfrac{5t^2}{2} + 4t$. Then $F(1) = \dfrac{11}{6}$, $F(4) = -\dfrac{8}{3}$ and $F(5) = -\dfrac{5}{6}$, all in metres.
3. Displacement. $F(5) - F(0) = -\dfrac{5}{6}$ m. The cart ends a little behind its start.
4. Distance. Take the size of each piece: $\dfrac{11}{6} + \left|-\dfrac{8}{3} - \dfrac{11}{6}\right| + \left(-\dfrac{5}{6} + \dfrac{8}{3}\right) = \dfrac{11}{6} + \dfrac{9}{2} + \dfrac{11}{6} = \dfrac{49}{6}$ m, about $8.17$ m.

Practise with new numbers

Try it in Python

The first program finds the mean value of $x^2$ on $[0, 3]$ by averaging more and more heights. The second one adds up $v\,dt$ and $|v|\,dt$ for the cart of the worked example.

Common mistakes

• Forgetting to divide by $b - a$. Without it you have the integral, the total, not the mean value.
• Averaging the end values. This works only when the graph is a straight line.
• Using displacement for distance. When $v$ changes sign, the integral of $v$ is smaller than the distance.
• Not splitting at the zeros of $v$. The integral of $|v|$ must be done piece by piece.
• Mixing time units. A rate per minute needs a time in minutes.

Remember these

• Mean value: $\dfrac{1}{b - a}\displaystyle\int_a^b f(x)\,dx$. It is the height of the rectangle with the same area.
• A continuous function reaches its mean value at some point $c$.
• Total change $=\displaystyle\int_a^b r(t)\,dt$, with units of the rate times the unit of time.
• Displacement is $\displaystyle\int v\,dt$. Distance is $\displaystyle\int |v|\,dt$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the mean value of $f(x) = 2x + 1$ on $[1, 5]$. At which $x$ does $f$ reach it?

2. Find the mean value of $\sin^2 x$ on $[0, \pi]$.

3. A tank fills at the rate $r(t) = 4t$ litres per minute. How many litres arrive in the first $5$ minutes?

4. A cart has velocity $v(t) = 3 - t$ metres per second for $0 \le t \le 5$ seconds. Find the displacement and the distance.

5. Find the mean value of $e^x$ on $[0, \ln 4]$.

References

• OpenStax, Calculus Volume 1, Chapter 5, section 5.3 (the mean value theorem for integrals and the average value of a function).
• OpenStax, Calculus Volume 1, Chapter 5, section 5.4 (the net change theorem).