1.3 Fundamentals

The inverse operation to differentiation is called antidifferentiation or, more commonly, integration. In models of exponential growth we assume that the growth rate is proportional to the population size.

$$\begin{equation} \frac{dN}{dt} = kN \tag{1.1} \end{equation}$$

where $N$ is the population size, $t$ is time and $k$ is a proportionality constant. We could solve for $N$ as a function of time by integrating the derivative. However, we know that the exponential function is the only function which equals its derivative.

$$\frac{d(e^t)}{dt} = e^t$$

Thus, we can modify the function to give the solution

$$N = N_0 e^{kt}$$

since differentiating $N_0 e^{kt}$ gives eqn. (1.1).

$$\frac{dN}{dt} = \frac{d(N_0 e^{kt})}{dt} = kN_0 e^{kt} = kN$$

If we write eqn. (1.1) as

$$\begin{equation} \frac{dN}{dt} = kN_0 e^{kt} \tag{1.2} \end{equation}$$

then the “antidifferentiation” becomes more obvious: find $N$ so that its derivative equals $kN_0 e^{kt}$. If we define

$$g(t) = kN_0 e^{kt}, \;\;\;\;\;f(t) = N_0 e^{kt}$$

then eqn. (1.2) becomes

$$g(t) = df/dt$$

Thus $g(t)$ is the derivative of $f(t)$ and, conversely, $f(t)$ is an antiderivative of $g(t)$. But there is a slight problem. The antiderivative is not unique. If we define

$$x(t) = N_0 e^{kt} + 10, \;\;\;\;y(t) = N_ e^{kt} +354$$

then $x(t)$ and $y(t)$ are also antiderivatives of $g(t)$, as can be checked by differentiating: $dx/dt = g(t), dy/dt = g(t)$. One interpretation of this is that the graphs of $x(t)$ and $y(t)$ have the same slope (derivative) for any given value of $t$. Since the functions differ by only a constant, we can write the general form of the antiderivative as $f(t) + C$ where $C$ can be any constant. We usually write the antiderivative as the indefinite integral

$$\begin{equation} \int g(t)dt = f(t) + C \tag{1.3} \end{equation}$$

and call $C$ the integration constant. Recall that the term “$dt$” identifies the variable of integration just as it does the variable of differentiation in the derivative $df/dt$.

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