## 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.

$$$\frac{dN}{dt} = kN \tag{1.1}$$$

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

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

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

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

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 $$df/dt$$.