Year 12 Methods (Unit 3 & 4) Integration

Applications of integration

Q: How do you find the total change from a rate of change?

Integrate the rate over the time interval. If a quantity changes at a rate R of t, the total change from t equals a to t equals b is the definite integral of R with respect to t from a to b. If you also know the starting amount, add it to this integral to get the amount at time b.

Q: How do you find the area between a curve and the x-axis?

When the curve lies above the axis on the interval, the area is the definite integral of the function between the two x-values. First find where the curve meets the axis if the limits are not given, then evaluate the integral between those limits.

Q: What is the average value of a function?

The average value of a function f over an interval a to b is one over b minus a, times the definite integral of f from a to b. In words, it is the area under the curve divided by the width of the interval, and it gives the constant height that would enclose the same area.

Q: What is the difference between displacement and distance travelled?

Displacement is the signed integral of velocity: area above the time axis counts positive and area below counts negative, so they can cancel. Distance travelled is the integral of the speed, the size of the velocity, so every part counts positive. Split the interval wherever the velocity changes sign and add the sizes of the areas.

Q: Why is total change an integral and not just the rate?

The rate tells you how fast the quantity changes at one instant, but the quantity keeps changing over the whole interval. Adding up the change over every instant is exactly what the definite integral does, so the total change is the integral of the rate, not the value of the rate at one time.

Q: How do you find the area under a rate-time graph?

The area under a rate-time graph between two times equals the total change in the quantity over that interval. For a straight-line graph you can use the area of the triangle or trapezium; for a curve you evaluate the definite integral of the rate.

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In Year 12 Mathematical Methods (Queensland, QCAA), integration is used to measure accumulation. The definite integral gives the total change from a rate, the area between a curve and the \(x\)-axis, the average value of a function, and the displacement and distance travelled of a moving object — the applications of Unit 4 Topic 1.

The definite integral \(\displaystyle\int_{a}^{b} f(x)\,dx\) measures accumulation, and it has several everyday readings.

Total change from a rate. If a quantity changes at a rate \(R(t)\), the total change from \(t=a\) to \(t=b\) is \(\displaystyle\int_{a}^{b} R(t)\,dt\). With a known starting amount, add it to reach the amount at time \(b\).

Area. If \(f(x)\ge 0\) on \([a,b]\), then \(\displaystyle\int_{a}^{b} f(x)\,dx\) is the area between \(y=f(x)\) and the \(x\)-axis.

Average value. The average value of \(f\) over \([a,b]\) is \(\dfrac{1}{b-a}\displaystyle\int_{a}^{b} f(x)\,dx\) — the area divided by the width.

Motion. For a velocity \(v(t)\), the displacement is \(\displaystyle\int_{a}^{b} v\,dt\) (a signed area) and the distance travelled is \(\displaystyle\int_{a}^{b} |v|\,dt\).

Key idea. Integrate a rate to get the total change; integrate a positive function to get an area; divide that area by \(b-a\) to get the average value; use the sign of \(v\) to tell displacement (signed) from distance (unsigned).

Total change \(=\displaystyle\int_{a}^{b} R\,dt\) — the area under the rate-time graph

Displacement \(=\displaystyle\int v\,dt\) (signed); distance \(=\displaystyle\int |v|\,dt\) (both areas positive)

Total change from a rate \(R(t)\) over \([a,b]\):

\[\text{total change}=\int_{a}^{b} R(t)\,dt,\qquad Q(b)=Q(a)+\int_{a}^{b} R(t)\,dt\]

Q (b) = Q (a) + \int_{a}^{b} R d t

Area between a curve \(f(x)\ge 0\) and the \(x\)-axis:

\[A=\int_{a}^{b} f(x)\,dx\]

A = \int_{a}^{b} f (x) d x

Average value of \(f\) over \([a,b]\):

\[\bar{f}=\dfrac{1}{b-a}\int_{a}^{b} f(x)\,dx\]

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

Motion. Displacement over \([a,b]\) is \(\displaystyle\int_{a}^{b} v\,dt\) (a signed area); distance travelled is \(\displaystyle\int_{a}^{b} |v|\,dt\). Split the interval at each \(v=0\) and add the sizes of the separate areas.

How to apply the definite integral

Identify what the integral means. A rate \(\Rightarrow\) total change; a positive function \(\Rightarrow\) area; area \(\div (b-a)\) \(\Rightarrow\) average value; a velocity \(\Rightarrow\) displacement or distance.
Find the limits. Read them from the interval, or find where the curve meets the \(x\)-axis (for an area) or where \(v=0\) (for distance).
Antidifferentiate and evaluate. Integrate to the standard forms, then apply \(\displaystyle\int_{a}^{b} f = F(b)-F(a)\).
Interpret with units. Add any initial value, divide by the width for an average, or add the area sizes for a distance — and state the units.

Distance tip. If \(v\) changes sign in the interval, the displacement can be small (or zero) while the distance is large. Always test whether \(v=0\) inside the interval before computing a distance.

Example 1 — Total change

Water flows into a tank at \(R=2t+3\) litres/min. Find the total volume in the first \(4\) minutes.

Solution

Integrate the rate over \([0,4]\).

\(V\)	\(=\)	\(\displaystyle\int_{0}^{4}(2t+3)\,dt\)
\(\)	\(=\)	\(\big[\,t^{2}+3t\,\big]_{0}^{4}\)
\(\)	\(=\)	\(16+12=28\) litres

V = 28

Example 2 — Area under a curve

Find the area between \(y=6x-x^{2}\) and the \(x\)-axis.

Solution

Roots \(x=0\) and \(x=6\); integrate between them.

\(A\)	\(=\)	\(\displaystyle\int_{0}^{6}(6x-x^{2})\,dx\)
\(\)	\(=\)	\(\big[\,3x^{2}-\dfrac{x^{3}}{3}\,\big]_{0}^{6}\)
\(\)	\(=\)	\(108-72=36\ \text{m}^{2}\)

A = 36

Example 3 — Average value

Find the average value of \(f(x)=x^{2}\) over \([0,3]\).

Solution

Divide the integral by the width \(3\).

\(\bar{f}\)	\(=\)	\(\dfrac{1}{3-0}\displaystyle\int_{0}^{3} x^{2}\,dx\)
\(\)	\(=\)	\(\dfrac{1}{3}\times\dfrac{27}{3}\)
\(\)	\(=\)	\(\dfrac{1}{3}\times 9=3\)

\bar{f} = 3

Example 4 — Displacement vs distance

A particle has velocity \(v=2t-4\) m/s on \([0,3]\). Find its displacement and the distance travelled.

Solution

\(v=0\) at \(t=2\), so split the interval there.

displacement	\(=\)	\(\displaystyle\int_{0}^{3}(2t-4)\,dt=-3\) m
distance	\(=\)	\(4+1=5\) m

Area below the axis on \([0,2]\) is \(4\); area above on \([2,3]\) is \(1\).

s = - 3, d = 5

Common pitfalls

Total change is the integral, not the rate. Do not substitute \(t=b\) into \(R(t)\); the total change accumulates over the whole interval, so it is \(\displaystyle\int_{a}^{b} R\,dt\).

Do not forget the \(\dfrac{1}{b-a}\) for an average. The bare integral is the area; the average value divides that area by the width of the interval.

Displacement is not distance. When \(v\) changes sign, split at each \(v=0\) and add the sizes of the areas for the distance; the signed integral alone gives only the displacement.

Frequently asked questions

How do you find the total change from a rate of change?

Integrate the rate over the interval: the total change from \(t=a\) to \(t=b\) is \(\displaystyle\int_{a}^{b} R(t)\,dt\). Add the starting amount to reach the amount at time \(b\).

How do you find the area between a curve and the x-axis?

When the curve is above the axis, the area is \(\displaystyle\int_{a}^{b} f(x)\,dx\). Find the intercepts first if the limits are not given.

What is the average value of a function?

It is \(\dfrac{1}{b-a}\displaystyle\int_{a}^{b} f(x)\,dx\) — the area under the curve divided by the width of the interval.

What is the difference between displacement and distance travelled?

Displacement is the signed integral \(\displaystyle\int v\,dt\); distance is \(\displaystyle\int |v|\,dt\). They differ whenever \(v\) changes sign.

Why is total change an integral and not just the rate?

The rate is the change at one instant; adding the change over every instant of the interval is exactly the definite integral.

How do you find the area under a rate-time graph?

The area between two times equals the total change over that interval — a triangle or trapezium for a straight line, or \(\displaystyle\int_{a}^{b} R\,dt\) for a curve.

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