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)dt10−3 |
| $=$= | $\frac{\left[-\frac{1800}{\pi}\cos\left(\frac{\pi t}{6}\right)\right]_3^{10}}{7}$[−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+2x−6, 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$$400 if no items are sold, find an expression for $P$P in terms of $x$x.
Think: We have been given the rate of change in profit per item. We can find the indefinite integral of $P'(x)$P′(x) and then use the initial profit given to solve for the constant of integration.
Do:
| $P\left(x\right)$P(x) | $=$= | $\int P'(x)\ dx$∫P′(x) dx |
|
| $=$= | $\int0.75x^2+2x−6\ dx$∫0.75x2+2x−6 dx |
|
|
| $=$= | $0.25x^3+x^2-6x+c$0.25x3+x2−6x+c |
|
Find $c$c using the fact that when $x=0$x=0, $P(x)=-400$P(x)=−400.
| $-400$−400 | $=$= | $0.25(0)^3+(0)^2-6(0)+c$0.25(0)3+(0)2−6(0)+c |
|
| $\therefore\ c$∴ c | $=$= | $-400$−400 |
|
Hence, the profit from selling $x$x items is given by $P(x)=0.25x^3+x^2-6x-400$P(x)=0.25x3+x2−6x−400
(b) Hence, determine the profit from selling $20$20 items.
Think: We want to find the value of $P\left(20\right)$P(20).
Do:
| $P\left(20\right)$P(20) | $=$= | $0.25(20)^3+(20)^2+6(20)-400$0.25(20)3+(20)2+6(20)−400 |
|
| $=$= | $1880$1880 |
|
Thus, the profit from selling $20$20 items is $\$1880$$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+2x−6)dx |
| $=$= | $\left[0.25x^3+x^2-6x\right]_{10}^{50}$[0.25x3+x2−6x]5010 | |
| $=$= | $33450-290$33450−290 | |
| $=$= | $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′=−10e−t5.
Determine the instantaneous rate of change of the temperature after $8$8 minutes. Give your answer correct to two decimal places where appropriate.
Determine the total change of temperature after $9$9 minutes. Give your answer correct to two decimal places.
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′=2000−30t2+5t3, for $0\le t\le7$0≤t≤7, where $V$V is the amount of water (in litres) in the tank $t$t hours after midnight.
Determine the initial flow rate.
Complete the table of values.
$t$t $-1$−1 $0$0 $1$1 $4$4 $6$6 $V''$V′′ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ 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
Which of the following is the graph of $V'$V′?
Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DFind the area bound by the graph of $V'$V′ and the $t$t-axis between $t=0$t=0 and $t=3$t=3.
What does the area found in part (f) represent?
The average flow rate in the first $3$3 hours.
AThe amount of water in the tank at $3$3am.
BThe average amount of water in the tank in the first $3$3 hours.
CThe 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.
Determine the net change in cost for producing between $13$13 and $19$19 items.
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(x−400), in dollars per item.
Determine the change in revenue from producing and selling the $31$31st item.
Determine the net change in revenue from selling $40$40 items to selling $70$70 items.
Determine the average change in revenue from selling $40$40 items to selling $70$70 items.