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Year 12 Maths - Methods (Unit 3 & Unit 4) Integration

Applications of integration

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Theory

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 as the area under a rate-time graphA rising rate-time graph R of t. The region under the graph between t equals a and t equals b is shaded; its area equals the total change in the quantity over that interval. t R R(t) a b area
Total change \(=\displaystyle\int_{a}^{b} R\,dt\) — the area under the rate-time graph
Displacement as the signed area under a velocity-time graphA velocity-time line that starts below the time axis and rises above it, crossing at one point. The area below the axis counts negative and the area above counts positive; their sum is the displacement, while their combined size is the distance travelled. t v v(t) − +
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)+∫abR dt

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

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

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

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

  1. 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.
  2. 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).
  3. Antidifferentiate and evaluate. Integrate to the standard forms, then apply \(\displaystyle\int_{a}^{b} f = F(b)-F(a)\).
  4. 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
Set up — total volume is the integral of the rate over \([0,4]\):
\(V\)\(=\)\(\displaystyle\int_{0}^{4}(2t+3)\,dt\)
\(=\)\(\big[\,t^{2}+3t\,\big]_{0}^{4}\)
Evaluate — \(F(4)-F(0)\):
\(=\)\((4^{2}+3\times 4)-0\)
\(=\)\(16+12=28\)
\(\therefore\ V=28\) litres
V=28
Example 2 — Area under a curve
Find the area between \(y=6x-x^{2}\) and the \(x\)-axis.
Solution
Limits — find the \(x\)-intercepts:
\(6x-x^{2}\)\(=\)\(0\)
\(x(6-x)\)\(=\)\(0\Rightarrow x=0,\ 6\)
Set up — the curve is above the axis on \([0,6]\):
\(A\)\(=\)\(\displaystyle\int_{0}^{6}(6x-x^{2})\,dx\)
\(=\)\(\big[\,3x^{2}-\dfrac{x^{3}}{3}\,\big]_{0}^{6}\)
Evaluate — \(F(6)-F(0)\):
\(=\)\(\big(3(36)-\dfrac{216}{3}\big)-0\)
\(=\)\(108-72=36\)
\(\therefore\ A=36\ \text{m}^{2}\)
A=36
Example 3 — Average value
Find the average value of \(f(x)=x^{2}\) over \([0,3]\).
Solution
Set up — average value is \(\dfrac{1}{b-a}\displaystyle\int_a^b f\):
\(\bar{f}\)\(=\)\(\dfrac{1}{3-0}\displaystyle\int_{0}^{3} x^{2}\,dx\)
Integrate — power rule, then evaluate:
\(\displaystyle\int_{0}^{3} x^{2}\,dx\)\(=\)\(\left[\dfrac{x^{3}}{3}\right]_{0}^{3}=\dfrac{27}{3}=9\)
Divide by the width \(b-a=3\):
\(\bar{f}\)\(=\)\(\dfrac{1}{3}\times 9\)
\(=\)\(3\)
\(\therefore\ \bar{f}=3\)
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
Sign of \(v\) — find where the velocity is zero:
\(2t-4\)\(=\)\(0\Rightarrow t=2\)

\(v<0\) on \([0,2]\) and \(v>0\) on \([2,3]\).

Displacement — signed integral
\(s\)\(=\)\(\displaystyle\int_{0}^{3}(2t-4)\,dt\)
\(=\)\(\big[\,t^{2}-4t\,\big]_{0}^{3}\)
\(=\)\((9-12)-0=-3\)
\(\therefore\) displacement \(=-3\) m
Distance — add the sizes of the two areas
\(\displaystyle\int_{0}^{2}(2t-4)\,dt\)\(=\)\(\big[\,t^{2}-4t\,\big]_{0}^{2}=-4\)
\(\displaystyle\int_{2}^{3}(2t-4)\,dt\)\(=\)\(\big[\,t^{2}-4t\,\big]_{2}^{3}=1\)
\(d\)\(=\)\(|-4|+|1|=4+1=5\)
\(\therefore\) distance \(=5\) m
Velocity-time graph v=2t-4The line v=2t-4 crosses the t-axis at t=2; the area below the axis from 0 to 2 is 4, the area above from 2 to 3 is 1. t v t=2 4 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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