Interpreting a result in context | Year 12

01 / LEARN

Understand the idea.

Interpret the integrand and the differential

Read the units of both the integrand and the differential. Integrating a flow in litres per minute over minutes gives litres; integrating cross-sectional area in square metres along metres gives volume. The differential represents the small interval over which each contribution accumulates.

An integral of a rate gives change, not automatically a final amount. Add the initial amount when it is needed. For a physical model, state the interval and use the interpretation of the rate to decide what a negative contribution means.

Check the result in context

An integral and an average have different units. The average value on $[a, b]$ is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ , for $a < b$ , so it has the same units as $f$ . In motion, average speed uses distance; average velocity uses signed displacement.

Report a result with units and a sentence describing what it measures. Use a rough bound or average to check its scale. Keep exact values during calculation, and avoid extending a model beyond the interval where its assumptions are plausible.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

A pipe has cross-sectional area A(x)=4x+1 square metres for 0<=x<=3 metres. Find its volume.

Worked solution

1
Accumulate cross-sectional area along the pipe.
$V = \int_{0}^{3} A (x) d x = \int_{0}^{3} (4 x + 1) d x$
Area in square metres multiplied by slice length in metres gives volume in cubic metres.
2
Find and evaluate the antiderivative.
$V = {[2 x^{2} + x]}_{0}^{3} = (18 + 3) - 0 = 21 m^{3}$
Endpoint subtraction adds the volumes of all slices over the stated 3 metre length.
3
Check using the average area.
$\bar{A} = \frac{1 + 13}{2} = 7 m^{2}, V = 7 \cdot 3 = 21 m^{3}$
For a linearly varying area, the average is the mean of its endpoint values.

Answer

$21 m^{3}$ .

EXAMPLE 02

Worked example 2

A force F(x)=3x squared+2 N acts along a straight displacement from x=0 to 2 m. Find the work.

Worked solution

1
Use the work integral.
$W = \int_{0}^{2} F (x) d x = \int_{0}^{2} (3 x^{2} + 2) d x$
The force acts along the displacement, so force times each displacement increment contributes work.
2
Evaluate the signed accumulation.
$W = {[x^{3} + 2 x]}_{0}^{2} = (8 + 4) - 0 = 12 N m$
The antiderivative differentiates to the stated force throughout the interval.
3
Express the physical result.
$W = 12 J$
One newton metre is one joule; positive force along positive displacement gives positive work.

Answer

$12 J$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A rate in metres per second is integrated over seconds. What are the resulting units?

Show answer & reasoning

Metres.

Multiplying $m s^{- 1}$ by $s$ leaves metres. For signed velocity, the integral measures displacement; integrating speed instead measures total distance.

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6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Water enters at $2 t + 3$ litres/minute for $0 \leq t \leq 4$ . Initially there are 10 litres. Find the final amount.
2 marks
Find the average temperature $20 + 2 t$ degrees C over $0 \leq t \leq 4$ hours.
2 marks
Velocity is $v = t - 2$ m/s for $0 \leq t \leq 4$ . Find displacement and total distance.
2 marks
A population increases at $100 e^{0.1 t}$ people/year for $0 \leq t \leq 10$ . Find the increase.
2 marks
Marginal cost is $6 q + 10$ dollars/unit. Find the cost increase when output rises from $q = 0$ to 5.
2 marks
A solid has cross-section area $π {(1 + x)}^{2}$ for $0 \leq x \leq 2$ metres. Find its volume.
2 marks

Read the worked solutions

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

Water enters at $2 t + 3$ litres/minute for $0 \leq t \leq 4$ . Initially there are 10 litres. Find the final amount.
2 marks
The added amount is ${[t^{2} + 3 t]}_{0}^{4} = 28$ .
Add the initial 10 litres.
$38$ litres
Find the average temperature $20 + 2 t$ degrees C over $0 \leq t \leq 4$ hours.
2 marks
Integrate to obtain 96 degree-hours.
Divide by the 4-hour interval.
$24^{\circ} C$ .
Velocity is $v = t - 2$ m/s for $0 \leq t \leq 4$ . Find displacement and total distance.
2 marks
The signed integral cancels.
Split at $t = 2$ and integrate speed on each side.
Displacement 0 m; distance 4 m.
A population increases at $100 e^{0.1 t}$ people/year for $0 \leq t \leq 10$ . Find the increase.
2 marks
An antiderivative is $1000 e^{0.1 t}$ .
Subtract its $t = 0$ value from its $t = 10$ value.
$1000 (e - 1)$ people, approximately 1718.
Marginal cost is $6 q + 10$ dollars/unit. Find the cost increase when output rises from $q = 0$ to 5.
2 marks
Integrate marginal cost over output.
${[3 q^{2} + 10 q]}_{0}^{5} = 125$ .
$125$ .
A solid has cross-section area $π {(1 + x)}^{2}$ for $0 \leq x \leq 2$ metres. Find its volume.
2 marks
Integrate $π$ times $(1 + 2 x + x^{2})$ .
Evaluate to $π$ times $(2 + 4 + \frac{8}{3})$ .
$\frac{26 π}{3} m^{3}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics integration techniques and applications, including inverse-trigonometric patterns, substitution, parts, simple partial fractions, areas and solids of revolution.

Official curriculum source ↗

In this lesson

Understand the idea.

Interpret the integrand and the differential

Check the result in context

See the method in action.

Worked example 1

Accumulate cross-sectional area along the pipe.

Find and evaluate the antiderivative.

Check using the average area.

Worked example 2

Use the work integral.

Evaluate the signed accumulation.

Express the physical result.

Try it yourself.

Check your understanding

Keep practising