Some Economic Applications Of Integrals

Integrals are used in economic analysis in various ways. We shall illustrate a few simple applications in the present section and then show the application to the Domar growth model in the next.

From a Marginal Function to a Total Function

Given a total function (e.g., a total-cost function), the process of differentiation can yield the marginal function (e.g., the marginal-cost function). Because the process of integration is the opposite of differentiation, it should enable us, conversely, to infer the total function from a given marginal function.

Example 1 If the marginal cost (MC) of a firm is the following function of output, C'(Q) = 2e02Q, and if the fixed cost is CF = 90, find the total-cost function C(Q). By integrating C'(Q) with respect to Q, we find that

This result may be taken as the desired C(Q) function except that, in view of the arbitrary constant c, the answer appears indeterminate. Fortunately, the information that CF = 90 can be used as an initial condition to definitize the constant. When <2 = 0, total cost C will consist solely of CF. Setting Q = 0 in the result of (13.9), therefore, we should get a value of 90; that is, 10<?° + c = 90. But this would imply that c = 90 - 10 = 80. Hence, the total-cost function is

1 Check the definite integrals given in Exercises 13.3-1 and 13.3-2 to determine whether any of them is improper. If improper, indicate which variety of improper integral each one is.

2 Which of the following integrals are improper, and why?

3 Evaluate all the improper integrals in the preceding problem.

4 Evaluate the integral f2 of Example 5, and show that it is also divergent.

Note that, unlike the case of (13.2), where the arbitrary constant c has the same v alue as the initial value of the variable H(0), in the present example we have c = 80 but C(0) = CF = 90, so that the two take different values. In general, it should not be assumed that the arbitrary constant c will always be equal to the initial value of the total function.

Example 2 If the marginal propensity to save (MPS) is the following function of income, S'(Y) = 0.3 - 0.1Y~1/2, and if the aggregate savings S is nil when income Y is 81, find the saving function S(Y). As the MPS is the derivative of the S function, the problem now calls for the integration of S'( Y):

The specific value of the constant c can be found from the fact that S = 0 when Y = 81. Even though, strictly speaking, this is not an initial condition (not relating to 7= 0), substitution of this information into the above integral will nevertheless serve to definitize c. Since

0 = 0.3(81) - 0.2(9) + c => c = —22.5 the desired saving function is s(y) = o.3y - o.2y|/2 - 22.5

The technique illustrated in the above two examples can be extended directly to other problems involving the search for total functions (such as total revenue, total consumption) from given marginal functions. It may also be reiterated that in problems of this type the validity of the answer (an integral) can always be checked by differentiation.

Investment and Capital Formation

Capital formation is the process of adding to a given stock of capital. Regarding this process as continuous over time, we may express capital stock as a function of time, K(t), and use the derivative dK/dt to denote the rate of capital formation.* But the rate of capital formation at time t is identical with the rate of net investment flow at time t, denoted by /(?)• Thus, capital stock K and net investment / are related by the following two equations:

* As a matter of notation, the derivative of a variable with respect to time often is also denoted by a dot placed over the variable, such as K = dK/dt. In dynamic analysis, where derivatives with respect to time occur in abundance, this more concise symbol can contribute substantially to notational simplicity. However, a dot. being such a tiny mark, is easily lost sight of or misplaced; thus, great care is required in using this symbol.

The first equation above is an identity; it shows the synonymity between net investment and the increment of capital. Since I(t) is the derivative of K(t), it stands to reason that K(t ) is the integral or antiderivative of /(f), as shown in the second equation. The transformation of the integrand in the latter equation is also easy to comprehend: The switch from / to dK/dt is by definition, and the next transformation is by cancellation of two identical differentials, i.e., by the substitution rule.

Sometimes the concept of gross investment is used together with that of net investment in a model. Denoting gross investment by / and net investment by /, we can relate them to each other by the equation

where 8 represents the rate of depreciation of capital and 8K, the rate of replacement investment.

Example 3 Suppose that the net investment flow is described by the equation I{t) = 3r1/2 and that the initial capital stock, at time t = 0, is A"(0). What is the time path of capital K? By integrating I(t) with respect to t, we obtain

Next, letting t = 0 in the leftmost and rightmost expressions, we find AT(0) = c. Therefore, the time path of K is

Observe the basic similarity between the results in (13.10) and in (13.2").

The concept of definite integral enters into the picture when one desires to find the amount of capital formation during some interval of time (rather than the time path of K). Since jl(t) dt = K(t), we may write the definite integral to indicate the total capital accumulation during the time interval [a, b]. Of course, this also represents an area under the I(t) curve. It should be noted, however, that in the graph of the K(t) function, this definite integral would appear instead as a vertical distance—more specifically, as the difference between the two vertical distances K(b) and K(a). (cf. Exercise 13.3-4.)

To appreciate this distinction between K(t) and I(t) more fully, let us emphasize that capital K is a stock concept, whereas investment / is a flow concept. Accordingly, while K(t) tells us the amount of K existing at each point of time, /(?) gives us the information about the rate of (net) investment per year (or per period of time) which is prevailing at each point of time. Thus, in order to calculate the amount of net investment undertaken (capital accumulation), we must first specify the length of the interval involved. This fact can also be seen

c when we rewrite the identity dK/dt = /(?) as dK = I(t) dt, which states that dK, the increment in K, is based not only on I(t), the rate of flow, but also on dt, the time that elapsed. It is this need to specify the time interval in the expression I(t) dt that brings the definite integral into the picture, and gives rise to the area representation under the I(t)—as against the K(t)—curve.

Example 4 If net investment is a constant flow at I(t) = 1000 (dollars per year), what will be the total net investment (capital formation) during a year, from t = 0 to t = 1? Obviously, the answer is $1000; this can be obtained formally as follows:

You can verify that the same answer will emerge if, instead, the year involved is from t = 1 to t = 2.

Example 5 If I(t) = 3tl/2 (thousands of dollars per year)—a nonconstant flow —what will be the capital formation during the time interval [1,4], that is, during the second, third, and fourth years? The answer lies in the definite integral

On the basis of the preceding examples, we may express the amount of capital accumulation during the time interval [0, ?], for any investment rate I(t), by the definite integral

Figure 13.5 illustrates the case of the time interval [0, /0], Viewed differently, the

+1 0

Responses

  • Eddie
    How to find initial condition used to definitize the constant of integration?
    1 year ago
  • Vala
    How to find time path of capital formation?
    6 months ago

Post a comment