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VCE 12 Methods 2023

7.11 Applications of integration

Lesson

Integration is an essential tool in the physical sciences, in economics and in statistics. It is needed when continuously varying quantities are involved in a mathematical model of a real-world process. It has wide ranging applications such as finding complex areas, volumes, probabilities, centre of mass, kinematics, average value of a function and net change for a quantity given the rate of change.

 

Net change

Given a function involving the rate of change of a quantity, integration allows us to calculate the net change of the quantity over a given interval or find an explicit formula for the quantity, given initial conditions. For example, if a function described the rate of water flowing in and out of a tank over time, the integral of such a function would allow us to find the net change in volume of water in the tank for a given interval.

Another example would be the integral of a velocity function (the rate of change of displacement with respect to time) would give us the net change in displacement over a given time interval.

Worked examples

Example 1

A population, $P\left(t\right)$P(t), of fish in a pond is known to vary according to the function $P'\left(t\right)=300\sin\left(\frac{\pi t}{6}\right)$P(t)=300sin(πt6), where $t$t is measured in months since counting began.

(a) What is the net change in population in the first $4$4 months since counting began?

Think: We have been given the rate of change of the population, so to find the net change we need to find the value of the definite integral from $t=0$t=0 to $t=4$t=4.

Do:

Net change $=$= $\int_0^4300\sin\left(\frac{\pi t}{6}\right)dt$40300sin(πt6)dt
  $=$= $\left[-\frac{1800}{\pi}\cos\left(\frac{\pi t}{6}\right)\right]_0^4$[1800πcos(πt6)]40
  $=$= $-\frac{1800}{\pi}\cos\frac{2\pi}{3}+\frac{1800}{\pi}\cos0$1800πcos2π3+1800πcos0
  $\approx$ $859$859

 

Thus, the population increased by approximately $859$859 fish in the first $4$4 months.

(b) What is the average rate of change of the population between $t=3$t=3 and $t=10$t=10?

Think: The average rate of change will be the net change in population over the interval divided by the change in time.

Do:

Average rate of change $=$= $\frac{\int_3^{10}300\sin\left(\frac{\pi t}{6}\right)dt}{10-3}$103300sin(πt6)dt103
  $=$= $\frac{\left[-\frac{1800}{\pi}\cos\left(\frac{\pi t}{6}\right)\right]_3^{10}}{7}$[1800πcos(πt6)]1037
  $=$= $\frac{1}{7}\left(-\frac{1800}{\pi}\cos\frac{5\pi}{3}+\frac{1800}{\pi}\cos\frac{\pi}{2}\right)$17(1800πcos5π3+1800πcosπ2)
  $\approx$ $-40.9$40.9

 

Hence, over the given time frame the fish population decreased by an average of $40.9$40.9 fish per month.

(c) If the initial population when counting began was $2400$2400 fish, find the model for $P(t)$P(t).

Think: Find the indefinite integral of $P'(t)$P(t) and then use the population given to solve for the constant of integration.

Do:

$P\left(t\right)$P(t) $=$= $\int P'(t)\ dt$P(t) dt

 

  $=$= $\int300\sin\left(\frac{\pi t}{6}\right)\ dt$300sin(πt6) dt

 

  $=$= $-\frac{1800}{\pi}\cos\left(\frac{\pi t}{6}\right)+C$1800πcos(πt6)+C

 

Find $C$C using the fact that when $t=0$t=0, $P(t)=2400$P(t)=2400.

$2400$2400 $=$= $-\frac{1800}{\pi}\cos\left(0\right)+C$1800πcos(0)+C

 

$\therefore\ C$ C $=$= $2400+\frac{1800}{\pi}$2400+1800π

 

Hence, the model for the population of fish after $t$t months is: $P\left(t\right)=-\frac{1800}{\pi}\cos\left(\frac{\pi t}{6}\right)+2400+\frac{1800}{\pi}$P(t)=1800πcos(πt6)+2400+1800π

Example 2

The marginal profit from the sale of the $x$xth item is given by $P'\left(x\right)=0.75x^2+2x−6$P(x)=0.75x2+2x6, where $P\left(x\right)$P(x) is the profit from selling $x$x items.

(a) Given that the company incurs a loss of $\$400$$400 if no items are sold, find an expression for $P$P in terms of $x$x.

Think: We have been given the rate of change in profit per item. We can find the indefinite integral of $P'(x)$P(x) and then use the initial profit given to solve for the constant of integration.

Do:

$P\left(x\right)$P(x) $=$= $\int P'(x)\ dx$P(x) dx

 

  $=$= $\int0.75x^2+2x−6\ dx$0.75x2+2x6 dx

 

  $=$= $0.25x^3+x^2-6x+C$0.25x3+x26x+C

 

Find $C$C using the fact that when $x=0$x=0, $P(x)=-400$P(x)=400.

$-400$400 $=$= $0.25(0)^3+(0)^2-6(0)+C$0.25(0)3+(0)26(0)+C

 

$\therefore\ C$ C $=$= $-400$400

 

Hence, the profit from selling $x$x items is given by $P(x)=0.25x^3+x^2-6x-400$P(x)=0.25x3+x26x400

(b) Hence, determine the profit from selling $20$20 items.

Think: We want to find the value of $P\left(20\right)$P(20).

Do:

$P\left(20\right)$P(20) $=$= $0.25(20)^3+(20)^2+6(20)-400$0.25(20)3+(20)2+6(20)400

 

  $=$= $1880$1880

 

Thus, the profit from selling $20$20 items is $\$1880$$1880.

(c) Find the net change in profit if the number of items sold changes from $10$10 to $50$50 items.

Think: We can use the profit function we now have to find the profit for $10$10 and $50$50 items and then find the difference or equivalently, we can find the integral of $P'\left(x\right)$P(x) over the given interval.

Do:

Net change $=$= $\int_{10}^{50}\left(0.75x^2+2x-6\right)dx$5010(0.75x2+2x6)dx
  $=$= $\left[0.25x^3+x^2-6x\right]_{10}^{50}$[0.25x3+x26x]5010
  $=$= $33450-290$33450290
  $=$= $33160$33160

 

Thus, the net change in profit from selling $10$10 to $50$50 items is $\$33160$$33160.

Practice questions

Question 1

An object is cooling and its rate of change of temperature, after $t$t minutes, is given by $T'=-10e^{-\frac{t}{5}}$T=10et5.

  1. Determine the instantaneous rate of change of the temperature after $8$8 minutes. Give your answer correct to two decimal places where appropriate.

  2. Determine the total change of temperature after $9$9 minutes. Give your answer correct to two decimal places.

  3. Hence, determine the average change in temperature over the first $9$9 minutes.

    Give your answer correct to two decimal places where appropriate.

Question 2

The rate of flow of water into a supply tank is given by $V'=2000-30t^2+5t^3$V=200030t2+5t3, for $0\le t\le7$0t7, where $V$V is the amount of water (in litres) in the tank $t$t hours after midnight.

  1. Determine the initial flow rate.

  2. Complete the table of values.

    $t$t $-1$1 $0$0 $1$1 $4$4 $6$6
    $V''$V $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
  3. Hence state the time $t$t when the flow rate is a maximum, and the maximum flow rate at this time.

    Maximum flow rate occurs at $t=\editable{}$t=

    Maximum flow rate = $\editable{}$ litres/hour

  4. Which of the following is the graph of $V'$V?

    Loading Graph...

    A

    Loading Graph...

    B

    Loading Graph...

    C

    Loading Graph...

    D
  5. Find the area bound by the graph of $V'$V and the $t$t-axis between $t=0$t=0 and $t=3$t=3.

  6. What does the area found in part (f) represent?

    The average flow rate in the first $3$3 hours.

    A

    The amount of water in the tank at $3$3am.

    B

    The average amount of water in the tank in the first $3$3 hours.

    C

    The total change in the amount of water in the tank in the first $3$3 hours.

    D

Question 3

The marginal cost for the production of the $x$xth item is modelled by $C'=8x+909$C=8x+909, in dollars per item.

  1. Determine the net change in cost for producing between $13$13 and $19$19 items.

  2. Determine the average change in cost for producing between $13$13 and $19$19 items.

Question 4

The total revenue, $R$R (in thousands of dollars), from producing and selling a new product, $t$t weeks after its launch, is such that $\frac{dR}{dt}=401+\frac{500}{\left(t+1\right)^3}$dRdt=401+500(t+1)3.

  1. Determine the revenue function $R$R in terms of $t$t, expressing your answer in positive index form.

    Use $C$C as the constant of integration.

  2. Given that the initial revenue at the time of launch was $0$0, solve for the constant $C$C and hence state the revenue function.

  3. Determine the average revenue earned over the first $5$5 weeks. Round your answer to 2 decimal places.

  4. Determine the revenue earned in the $6$6th week.

Outcomes

U34.AoS3.10

application of integration to problems involving finding a function from a known rate of change given a boundary condition, calculation of the area of a region under a curve and simple cases of areas between curves, average value of a function and other situations.

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