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Level 8 - NCEA Level 3

Applications of integration of various functions


Among other things, integration is used to find areas and volumes of physical shapes. It is also used in the solution of differential equations.

Differential equations

It is often possible, in many fields of investigation, to observe the rate at which some quantity is changing and to see how the rate of change varies with time or distance or some other variable. After interpreting the rate of change as the derivative of an unknown function, an investigator looks for the function by reversing an imagined differentiation. That is, by integration.

Example 1

Think of money in an investment account, earning compound interest at a fixed interest rate. Or, think of bacterial cell division. The rate of change of the amount of money in the account or the rate of growth of the number of bacterial cells depends on the respective amounts already present in a directly proportional manner.

Let the amount present be $y$y and look for a function $y(t)$y(t) that gives the amount $y$y at any time $t$t. We know that the rate of change of $y$y is proportional to $y$y itself. So, we write


We might guess at this point that the exponential function will be involved in some way since we know that $\frac{\mathrm{d}}{\mathrm{d}t}e^t=e^t$ddtet=et. But, we can proceed via an integration.

On dividing through by $y$y, we have 

$\frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}t}=k$1ydydt=k and therefore,
$\int\frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}t}\ \mathrm{d}t=\int k\ \mathrm{d}t$1ydydt dt=k dt.

The integral on the left is equivalent to $\int\frac{1}{y}\ \mathrm{d}y$1y dy. This has antiderivative $\ln y$lny

On the right, we have $kt+c$kt+c. So, $\ln y=kt+c$lny=kt+c and it must be that $y=e^{kt+c}=e^{kt}e^c$y=ekt+c=ektec. Thus, 


is the required function. The coefficients $A$A and $k$k are determined by knowing values of $y$y at two different times $t$t. For example, $A$A is the amount present when $t=0$t=0

We can check by differentiation that the function $y(t)=Ae^{kt}$y(t)=Aekt does indeed satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}t}=ky$dydt=ky.


Areas and volumes

If $y$y is a continuous function of $x$x and $x$x varies between values $a$a and $b$b, the fundamental theorem of calculus asserts that  the area between the graph of the function and the $x$x-axis between $a$a and $b$b is given by $\int_a^b\ y(x)\ \mathrm{d}x$ba y(x) dx

Example 2

The function $y(x)=\sqrt{r^2-x^2}$y(x)=r2x2 between $x=0$x=0 and $x=r$x=r describes a quarter of a circle. We check that integration gives the correct formula for the area.

For this, we write the definite integral $\int_0^r\sqrt{r^2-x^2}\ \mathrm{d}x$r0r2x2 dx. It is not obvious what an antiderivative could be in this case, so we make a suitable change of variable by putting $\frac{x}{r}=\sin\theta$xr=sinθ. And, from this we get $\frac{\mathrm{d}x}{\mathrm{d}\theta}=r\cos\theta$dxdθ=rcosθ. This will enable us to make the formal substitution $\mathrm{d}x=r\cos\theta\ \mathrm{d}\theta$dx=rcosθ dθ.

We also note that when $x=0$x=0, $\theta=0$θ=0 and when $x=r$x=r, $\theta=\frac{\pi}{2}$θ=π2.

We have

$\int_0^r\sqrt{r^2-x^2}\ \mathrm{d}x$r0r2x2 dx $=$= $r\int_0^r\sqrt{1-\left(\frac{x}{r}\right)^2}\ \mathrm{d}x$rr01(xr)2 dx
  $=$= $r\int_0^r\sqrt{1-\sin^2\theta}\ \mathrm{d}x$rr01sin2θ dx
  $=$= $r\int_{\theta=0}^{\theta=\frac{\pi}{2}}\cos\theta\ r\cos\theta\ \mathrm{d}\theta$rθ=π2θ=0cosθ rcosθ dθ
  $=$= $r^2\int_0^{\frac{\pi}{2}}\cos^2\theta\ \mathrm{d}\theta$r2π20cos2θ dθ
  $=$= $r^2\int_0^{\frac{\pi}{2}}\frac{1}{2}(\cos2\theta+1)\ \mathrm{d}\theta$r2π2012(cos2θ+1) dθ
  $=$= $\frac{r^2}{2}\left[\frac{1}{2}\sin2\theta+\theta\right]_0^{\frac{\pi}{2}}$r22[12sin2θ+θ]π20
  $=$= $\frac{\pi r^2}{4}$πr24

This result agrees with the well-known formula for the area of a circle.


Example 3

In statistics, the definite integral is used in calculating certain probabilities from a known probability density function. 

Suppose, in a certain experiment, any non-negative real number can occur but small numbers occur with greater likelihood than large numbers and, to be more precise, the relative density of occurrence of numbers over the range goes according to the rule $g(x)=e^{-x}$g(x)=ex.

The probability that any particular number occurs in the experiment is zero. However, the probability that a number in some interval $(a,b)$(a,b) in the domain occurs is non-zero.

The probability is small for a small interval $\delta x$δx and it depends on the current value of $g(x)$g(x). In fact, we understand the probability density at $x$x to be the limiting value of the ratio $\frac{\delta p}{\delta x}$δpδx at $x$x. We write $\delta p=g(x)\delta x$δp=g(x)δx.

We sum the small probabilities over the increments $\delta x$δx by evaluating the definite integral $p=\int_a^bg(x)\ \mathrm{d}x$p=bag(x) dx. That is, $p=\int_a^be^{-x}\ \mathrm{d}x$p=baex dx.


Example 4

Verify that $g(x)=e^{-x}$g(x)=ex can be a probability density function over the domain $(0,\infty)$(0,).

If a random variable $X$X takes values only in the domain $(0,\infty)$(0,), the total probability over this interval must be $1$1. So, we check whether $\int_0^{\infty}g(x)\ \mathrm{d}x=1$0g(x) dx=1.

$\int_0^{\infty}g(x)\ \mathrm{d}x$0g(x) dx $=$= $\int_0^{\infty}e^{-x}\ \mathrm{d}x$0ex dx
  $=$= $\left[-e^{-x}\right]_0^{\infty}$[ex]0
  $=$= $1$1

We used the fact that $x\rightarrow\infty$x implies $e^{-x}\rightarrow0$ex0.

Example 5

In an experiment, a certain event occurs repeatedly. The waiting times between recurrences of the event are measured and are found to be increasing in a way that can be related to the time elapsed since the beginning of the experiment. 

Let $t$t be the time (in seconds) elapsed since the start of the experiment.  The experimenter finds that the waiting time between events is related to the elapsed time by a function $w(t)$w(t), where $w(t)$w(t) is the wait time between the occurrences of the event at or immediately before time $t$t and the next one.

As an approximation, we can say that at time $t$t, the event is recurring at the rate of once every $w(t)$w(t) seconds. That is, the number $x$x of events since the beginning of the experiment is growing at a rate that can be written as $\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{w(t)}$dxdt=1w(t).

The experimenter would like to find from this differential equation the function $x(t)$x(t) that relates the total number of events observed to the elapsed time $t$t

Integrating both sides with respect to $t$t, we have

$\int\frac{\mathrm{d}x}{\mathrm{d}t}\ \mathrm{d}t=\int\frac{1}{w(t)}\ \mathrm{d}t$dxdt dt=1w(t) dt

or, more simply,

$\int\mathrm{d}x=\int\frac{1}{w(t)}\ \mathrm{d}t$dx=1w(t) dt

Therefore, $x=\int\frac{1}{w(t)}\ \mathrm{d}t+C$x=1w(t) dt+C for some constant $C$C.

To solve the problem completely, the function $w(t)$w(t) that the experimenter found from the experimental data is needed and also some extra information that will allow the determination of the constant $C$C.

Suppose $w(t)=0.15t$w(t)=0.15t and the event had occurred $21$21 times when the elapsed time $t$t was $120$120 seconds. Then,

$x$x $=$= $\int\frac{1}{w(t)}\ \mathrm{d}t+C$1w(t) dt+C
  $=$= $\int\frac{1}{0.15t}\ \mathrm{d}t+C$10.15t dt+C
  $=$= $\frac{\log_e0.15t}{0.15}+C$loge0.15t0.15+C

When $t=120$t=120, $x=21$x=21. So, $21=\frac{\log_e0.15\times120}{0.15}+C=19.3+C$21=loge0.15×1200.15+C=19.3+C. From this, it is deduced that $C=1.7$C=1.7. Thus the required function is



In the calculations for this experiment, a discrete process of counting was treated as though it were continuous. The antidifferentiation only works for continuous functions and in this case, the experimenter would need to consider how well the continuous model reflected the discrete reality after various amounts of elapsed time.




Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods


Apply integration methods in solving problems

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