Applications of integration including the average value of a function on a closed interval, total change from a rate of change function, and kinematics (displacement and distance from velocity)

What this dot point is asking

VCAA wants you to use the definite integral in three application contexts: the average value of a function on a closed interval, the total change of a quantity from a given rate of change, and kinematics (displacement and distance from velocity). The dot point bridges the abstract calculus of Unit 3 to the modelling Paper 2 SAC and exam.

Average value of a function

The average value of $f$ on $[a, b]$ is:

Geometrically, $\bar f$ is the height of the rectangle on $[a, b]$ whose area equals $\int_{a}^{b} f(x) \, dx$. The rectangle has width $(b - a)$ and area equal to the integral, so its height is the integral divided by the width.

Procedure

1. Compute the definite integral $\int_{a}^{b} f(x) \, dx$.
2. Divide by the interval length $b - a$.

Why it matters

In modelling contexts, average value is the natural "typical level" of a continuous quantity over time. The average temperature over a day, the average flow rate, the average drug concentration. All are integrals divided by interval length.

Example

Average value of $f(x) = \sin(x)$ on $[0, \pi]$.

$\bar f = \frac{1}{\pi - 0} \int_{0}^{\pi} \sin(x) \, dx = \frac{1}{\pi} [-\cos(x)]_{0}^{\pi} = \frac{1}{\pi} ((-(-1)) - (-1)) = \frac{1}{\pi} \cdot 2 = \frac{2}{\pi}$.

Average value approximately $0.637$.

Total change from a rate of change

If $r(t) = \frac{dQ}{dt}$ is the rate of change of a quantity $Q$ with respect to time, the total change over $[t_1, t_2]$ is:

This is the fundamental theorem of calculus applied to a rate function.

Procedure

1. Identify the rate function $r(t)$.
2. Compute the definite integral.
3. Interpret the result in the context's units.

Worked contexts

Water flowing into / out of a container

Rate in litres per minute integrated over time gives volume in litres.

Drug concentration in blood

Rate of change of concentration integrated gives concentration change.

Population growth

Rate of change of population integrated gives population change.

Energy flow

Power (rate of energy transfer) integrated gives energy.

Example

Water flows into a tank at $r(t) = 3 + 0.5 t$ litres per minute, $0 \leq t \leq 10$ minutes. Volume added in the first 10 minutes:

$V = \int_{0}^{10} (3 + 0.5 t) \, dt = [3 t + 0.25 t^2]_{0}^{10} = 30 + 25 = 55$ litres.

Kinematics: displacement and distance from velocity

The velocity $v(t)$ of a particle is the rate of change of its position $x(t)$, so $v(t) = \frac{dx}{dt}$.

Displacement (signed)

The displacement of the particle on $[t_1, t_2]$ is:

Displacement is signed: positive if the net motion is in the positive direction, negative if in the negative direction, zero if the particle returns to its starting point.

Distance travelled (unsigned)

The total distance travelled is the integral of speed (always non-negative):

If the velocity changes sign on the interval, the distance is not the same as the displacement. Compute distance by splitting the interval at the zeros of $v$ and summing the absolute values of each piece.

Procedure for distance

1. Find the zeros of $v(t)$ on $[t_1, t_2]$. These are the times when the particle changes direction.
2. Split the interval at each zero.
3. Compute $\lvert \int v \, dt \rvert$ on each sub-interval and sum.

Example

A particle has velocity $v(t) = 2t - 6$ for $0 \leq t \leq 5$ seconds.

Displacement. $\int_{0}^{5} (2t - 6) \, dt = [t^2 - 6t]_{0}^{5} = (25 - 30) - 0 = -5$.

Displacement is $-5$ metres (5 metres in the negative direction net).

Distance. $v(t) = 0$ at $t = 3$. Split at $t = 3$.

On $[0, 3]$: $v < 0$ for $t < 3$. $\int_{0}^{3} (2t - 6) \, dt = [t^2 - 6t]_{0}^{3} = 9 - 18 = -9$. Distance on this piece: $|-9| = 9$.

On $[3, 5]$: $v > 0$ for $t > 3$. $\int_{3}^{5} (2t - 6) \, dt = (25 - 30) - (9 - 18) = -5 - (-9) = 4$. Distance: $4$.

Total distance: $9 + 4 = 13$ metres.

Note: displacement $-5$ differs from distance $13$ because the particle changed direction at $t = 3$.

Position from velocity (with initial condition)

If $v(t) = \frac{dx}{dt}$ and $x(0) = x_0$ is the initial position, then:

Use this to find the position function from the velocity function and an initial condition.

Average rate of change

The average rate of change of a quantity over an interval is:

This is the average value of the rate function. The same formula applies to velocity (average velocity = displacement / time) and speed (average speed = distance / time).

Examples in context

Example 1. Average temperature over a morning. A room's temperature is $T(t) = 18 + 2t - 0.25t^2$ degrees, $t$ hours after 6 am, for $0 \le t \le 4$. Its average value is $\bar T = \frac{1}{4}\int_0^4 (18 + 2t - 0.25t^2)\,dt = \frac{1}{4}\left[18t + t^2 - \frac{t^3}{12}\right]_0^4 = \frac{1}{4}\left(72 + 16 - \frac{64}{12}\right) = \frac{1}{4}(82.67) \approx 20.7^\circ$C.

Example 2. Distance from a velocity that reverses. A trolley has velocity $v(t) = 3t - 12$ m/s for $0 \le t \le 6$ s. The displacement is $\int_0^6(3t - 12)\,dt = \left[\frac{3t^2}{2} - 12t\right]_0^6 = 54 - 72 = -18$ m. But $v = 0$ at $t = 4$: distance is $\left|\int_0^4 v\,dt\right| + \int_4^6 v\,dt = |-24| + 6 = 30$ m, larger than the displacement because the trolley reversed direction.

Try this

Q1. Find the average value of $f(x) = 4x$ on $[0, 5]$. [2 marks]

• Cue. $\frac{1}{5}\int_0^5 4x\,dx = \frac{1}{5}[2x^2]_0^5 = \frac{50}{5} = 10$.

Q2. Water flows in at $r(t) = 2t + 1$ L/min for $0 \le t \le 6$. Find the total volume added. [3 marks]

• Cue. $\int_0^6 (2t + 1)\,dt = [t^2 + t]_0^6 = 42$ litres.

Q3. A particle has velocity $v(t) = 2t - 8$ m/s for $0 \le t \le 6$ s. Find (a) the displacement and (b) the distance travelled. [2+3 marks]

• Cue. (a) $\int_0^6(2t - 8)\,dt = [t^2 - 8t]_0^6 = -12$ m. (b) $v = 0$ at $t = 4$; $|{-16}| + 4 = 20$ m.