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5.03 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. We will look more closely at kinematics - the study of motion of objects including displacement, velocity and acceleration - in our next lesson.

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

$R$R is the revenue from producing and selling $x$x items. The marginal revenue from producing and selling the $x$xth item is modelled by $R'=2.1x\left(x-400\right)$R=2.1x(x400), in dollars per item.

  1. Determine the change in revenue from producing and selling the $31$31st item.

  2. Determine the net change in revenue from selling $40$40 items to selling $70$70 items.

  3. Determine the average change in revenue from selling $40$40 items to selling $70$70 items.

Outcomes

3.2.18

calculate total change by integrating instantaneous or marginal rate of change

3.2.19

calculate the area under a curve

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