Contextual integral models | Year 12

01 / LEARN

Understand the idea.

Define the quantity before integrating

Begin by naming the quantity and the rate being integrated. A marginal cost is dollars per additional item; a flow rate is volume per unit time. The independent variable determines what the bounds mean and which units cancel.

Distinguish change, final amount and average before calculating. Include any known starting amount when a final total is requested. State relevant assumptions, such as treating production quantity as continuous, and interpret the result within the supplied model.

Average value is total divided by interval length

The average value on $[a, b]$ , with $a < b$ , is the integral divided by interval length. It is the constant height producing the same signed area across that interval. A time-average rate has rate units, not accumulated-quantity units.

For a curved function, averaging only its endpoint values generally gives a different result. An exact integral is exact within the mathematical model; measured data, rounded coefficients and changing conditions may still limit its physical prediction.

f_{average} = \frac{1}{b - a} \int_{a}^{b} f (x) d x

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

A product has marginal cost $C^{'} (q) = 0.03 q^{2} + 2$ dollars/item and fixed cost $C (0) = 100$ dollars. Find $C (10)$ .

Worked solution

1
Express total cost as fixed cost plus increase.
$C (10) = 100 + \int_{0}^{10} (0.03 q^{2} + 2) d q$
The integral of marginal cost gives the additional cost of producing from zero to ten items.
2
Find the cost increase.
$Δ C = {[0.01 q^{3} + 2 q]}_{0}^{10}$
The power rule divides the quadratic coefficient by 3.
3
Evaluate the variable cost.
$Δ C = 0.01 (10^{3}) + 2 (10) = 10 + 20 = 30 dollars$
Marginal dollars per item integrated over item quantity gives dollars.
4
Add the fixed cost and interpret.
$C (10) = 100 + 30 = 130 dollars$
This model treats output as continuous and includes the cost incurred even at zero output.

Answer

130 dollars

EXAMPLE 02

Worked example

Find the average value of $f (t) = t^{2}$ on $[0, 3]$ .

Worked solution

1
Write the average-value expression.
$f_{average} = \frac{1}{3 - 0} \int_{0}^{3} t^{2} d t$
Average value is accumulated signed area divided by the interval length.
2
Evaluate the integral.
$\int_{0}^{3} t^{2} d t = {[\frac{t^{3}}{3}]}_{0}^{3} = 9$
The reverse power rule gives the exact accumulated value.
3
Divide by the interval length.
$f_{average} = \frac{9}{3} = 3$
A rectangle of height 3 and width 3 has the same area 9.
4
Compare with the endpoint mean.
$\frac{f (0) + f (3)}{2} = \frac{0 + 9}{2} = \frac{9}{2}$
The endpoint mean is different because it does not account for all values of the curved function.

Answer

$3$

03 / YOUR TURN

Try it yourself.

Check your understanding

A rate is $2 + t$ L/min on $[0, 4]$ . Find the average rate.

Show answer & reasoning

$4$ L/min

The volume supplied is $\int_{0}^{4} (2 + t) d t = {[2 t + \frac{t^{2}}{2}]}_{0}^{4} = 16$ L. Dividing by 4 min gives the average rate $4$ L/min.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Find the average value of $f (x) = 3 x^{2}$ on $[0, 2]$ .
3 marks
$C^{'} (q) = 2 q + 5$ dollars/item and $C (0) = 40$ . Find $C (6)$ .
3 marks
Water flows at $12 e^{- t}$ L/min for $0 \leq t \leq ln 2$ . Find the supplied volume.
3 marks
A force $F (x) = 4 x$ newtons moves an object from $x = 1$ to $x = 3$ metres. Find $\int_{1}^{3} F (x) d x$ and its unit.
3 marks
A rate model $r (t) = 10 - 2 t$ items/hour is used for production until $t = 8$ hours. Explain a limitation.
3 marks
The average value of $f (x) = k x$ on $[0, 4]$ is 6. Find $k$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the average value of $f (x) = 3 x^{2}$ on $[0, 2]$ .
3 marks
The integral is 8; divide by interval length 2.
$4$
$C^{'} (q) = 2 q + 5$ dollars/item and $C (0) = 40$ . Find $C (6)$ .
3 marks
Change $= {[q^{2} + 5 q]}_{0}^{6} = 36 + 30 = 66$ .
Add 40.
106 dollars
Water flows at $12 e^{- t}$ L/min for $0 \leq t \leq ln 2$ . Find the supplied volume.
3 marks
${[- 12 e^{- t}]}_{0}^{ln 2} = - 6 + 12 = 6$ .
$6$ L
A force $F (x) = 4 x$ newtons moves an object from $x = 1$ to $x = 3$ metres. Find $\int_{1}^{3} F (x) d x$ and its unit.
3 marks
${[2 x^{2}]}_{1}^{3} = 18 - 2 = 16$ N·m.
One newton-metre is one joule.
16 joules
A rate model $r (t) = 10 - 2 t$ items/hour is used for production until $t = 8$ hours. Explain a limitation.
3 marks
The zero is $t = 5$ ; a revised model or valid interval is needed beyond that time.
The model predicts negative production after 5 hours.
The average value of $f (x) = k x$ on $[0, 4]$ is 6. Find $k$ .
3 marks
$\frac{1}{4} \int_{0}^{4} k x d x = 2 k$ .
Solve $2 k = 6$ .
$k = 3$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods integration and logarithmic antiderivatives; original learning and assessment material, with by-hand methods and explicitly described numerical models.

Official curriculum source ↗

In this lesson

Understand the idea.

Define the quantity before integrating

Average value is total divided by interval length

See the method in action.

Worked example

Express total cost as fixed cost plus increase.

Find the cost increase.

Evaluate the variable cost.

Add the fixed cost and interpret.

Worked example

Write the average-value expression.

Evaluate the integral.

Divide by the interval length.

Compare with the endpoint mean.

Try it yourself.

Check your understanding

Keep practising