Integral in Economics and Business

Accumulating Sales and Costs from Their Rates

Economic data often describe a rate, such as units sold per month or additional cost per unit produced. An integral adds these changes over a period of time or a range of production.

For example, integrating a sales rate over three months gives the total number of units sold during those three months.

This concept is similar to calculating the area under a curve, where the horizontal axis represents time and the vertical axis represents the rate of change of an economic quantity.

Sales Analysis Using Integrals

Suppose a technology company launches a new smartphone. Its monthly sales rate is modeled by $f(t) = 2000\sqrt{t + 1} + 800 \text{ units}$ per month, where $t$ is the time in months after launch.

To find the total sales during the first $3 \text{ months}$ , integrate the sales rate function:

S = \int_{0}^{3} (2000\sqrt{t + 1} + 800) \, dt

Solve this by separating the integral first:

S = \int_{0}^{3} 2000\sqrt{t + 1} \, dt + \int_{0}^{3} 800 \, dt

For integrals containing square roots, we can use substitution. Let $u = t + 1$ , then $du = dt$ . Don't forget to change the integration limits too!

S = 2000 \int_{1}^{4} u^{1/2} \, du + 2400

The continuous model predicts $S = \frac{35200}{3} \approx 11733.33$ smartphone units in the first $3 \text{ months}$ . Interpreted as whole devices, that is approximately $11733$ smartphones.

Revenue Growth Analysis

A second model gives a technology startup's revenue growth rate as $R'(t) = 5000e^{0.1t}$ thousand dollars per month, where $t$ is measured in months.

To calculate the total revenue increase in the first $6 \text{ months}$ , we integrate the growth rate function:

R = \int_{0}^{6} 5000e^{0.1t} \, dt

With $e^{0.6} \approx 1.822$ , the total revenue increase is approximately $50000 \times 0.822 = 41100$ thousand dollars or $41.1 \text{ million}$ dollars.

In business, companies often need to estimate production costs. Suppose the marginal cost to produce a good is $MC(x) = 0.3x^2 - 12x + 200$ thousand dollars per unit, where $x$ is the number of units produced.

Now, if the company wants to know the total variable cost to produce the first $20 \text{ units}$ , they simply integrate the marginal cost function:

VC = \int_{0}^{20} (0.3x^2 - 12x + 200) \, dx

After we integrate and evaluate, we obtain:

VC = \left[0.1x^3 - 6x^2 + 200x\right]_{0}^{20}

Assuming $VC(0)=0$ , the total variable cost to produce $20 \text{ units}$ is $2.4 \text{ million}$ dollars. Any fixed cost is separate and is not included in this integral.

In economic applications, an integral converts a marginal quantity or rate of change into an accumulated total. Integrating marginal cost, for example, gives the change in total cost.

Consumer Surplus Analysis

Consumer surplus can also be calculated with an integral. Consider a market with demand function $P = 100 - 0.5Q$ and equilibrium price $60 \text{ dollars per unit}$ . The equilibrium quantity on this demand curve follows from:

\begin{aligned} &60=100-0.5Q \\ &\Longrightarrow \\ &Q=80 \end{aligned}

Consumer surplus adds, across the $80$ units traded, the difference between buyers' willingness to pay given by the demand curve and the market price they actually pay:

\begin{aligned} CS &= \int_{0}^{80} (100 - 0.5Q - 60) \, dQ \\ &= \int_{0}^{80} (40 - 0.5Q) \, dQ \end{aligned}

After evaluation:

CS = \left[40Q - 0.25Q^2\right]_{0}^{80}

A consumer surplus of $1600 \text{ dollars}$ measures the total monetary benefit consumers receive above the equilibrium price.

Exercises

Each problem gives a rate such as marginal cost or marginal revenue and asks for the accumulated total, so integrate the rate over the stated interval and then read the result in the original units.

A company has a sales rate of $f(t) = 1000 + 500t \text{ units}$ per month. Calculate the total sales in the first $6 \text{ months}$ !
If the marginal cost of a product is $MC(x) = 2x + 50$ thousand dollars per unit, what is the total variable cost to produce $25 \text{ units}$ ?
The investment growth rate function is $I'(t) = 2000e^{0.05t}\text{ million}$ dollars per year. Calculate the total investment growth in $4 \text{ years}$ !

Worked Solutions

Calculating total sales

Since we have the sales rate function, just integrate it:

$S = \int_{0}^{6} (1000 + 500t) \, dt$

The result is:
$S = \left[1000t + 250t^2\right]_{0}^{6}$
Total sales in $6 \text{ months}$ is $15{,}000 \text{ units}$ .
Calculating variable cost

Integrate the marginal cost function:

$VC = \int_{0}^{25} (2x + 50) \, dx$

After evaluation:
$VC = \left[x^2 + 50x\right]_{0}^{25}$
Total variable cost is $1{,}875$ thousand dollars or $1.875 \text{ million}$ dollars.
Calculating investment growth

For exponential functions, we integrate:

$I = \int_{0}^{4} 2000e^{0.05t} \, dt$

The result is:
$I = 2000 \left[\frac{e^{0.05t}}{0.05}\right]_{0}^{4}$
Total investment growth in $4 \text{ years}$ is approximately $8.86 \text{ billion dollars}$ .