Integral in Economics and Business

Application of Integrals in Economics

Integrals play an important role in analyzing various economic and business phenomena. Imagine a company manager who wants to know the total sales of a product in a certain period, or an economist analyzing national income growth. This is where integrals become a very useful tool.

In an economic context, integrals help us calculate the accumulation of a quantity that changes with respect to time. For example, if we have a sales rate function, then the integral of that function will give us the total sales in a certain period.

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

Let's learn how integrals can help analyze the sales of a product. Suppose a technology company launches a new smartphone, and the analysis team finds that sales follow a certain pattern.

After conducting market research, they found that the sales rate per month can be modeled with the function $f(t) = 2000\sqrt{t + 1} + 800$ units per month, where $t$ is the time in months after launch.

Now, here's the question: what is the total sales in the first 3 months? This is where integrals play a role. We need to integrate the sales rate function:

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

Let's solve this by separating the integral first:

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

S = 2000 \int_{0}^{3} (t + 1)^{1/2} \, dt + 800t \Big|_{0}^{3}

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

S = 2000 \left[\frac{2}{3}u^{3/2}\right]_{1}^{4} + 2400

S = \frac{4000}{3}(8 - 1) + 2400 = \frac{28000}{3} + 2400

The final result is $S = 9333 + 2400 = 11733$ smartphone units sold in the first 3 months.

Profit and Revenue Analysis

Now let's look at a different example. Suppose a technology startup has a revenue growth rate that follows an exponential pattern $R'(t) = 5000e^{0.1t}$ thousand dollars per month, where $t$ is the time in months.

To calculate the total revenue increase in the first 6 months, we integrate the growth rate function:

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

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

R = 5000 \left[\frac{e^{0.1t}}{0.1}\right]_{0}^{6}

R = 50000(e^{0.6} - 1)

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

Production Cost Optimization

In the business world, companies always strive to optimize 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 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}

VC = 800 - 2400 + 4000 = 2400

So, the total variable cost to produce 20 units is 2.4 million dollars.

In economic applications, integrals help transform marginal concepts (rates of change) into total concepts (accumulation). This is very useful for making the right business decisions.

Consumer Surplus Analysis

The concept of consumer surplus can also be calculated using integrals. Imagine there's a market with demand function $P = 100 - 0.5Q$ and the equilibrium price is 60 dollars per unit.

Consumer surplus shows the total benefit obtained by consumers above the price they pay. We can calculate it with:

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

After evaluation:

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

CS = 3200 - 1600 = 1600

A consumer surplus of 1600 units shows the total benefit obtained by consumers above the equilibrium price.

Exercises

A company has a sales rate of $f(t) = 1000 + 500t$ units per month. Calculate the total sales in the first 6 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 units?
The investment growth rate function is $I'(t) = 2000e^{0.05t}$ million dollars per year. Calculate the total investment growth in 4 years!

Answer Key

Calculating total sales

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

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

The result is:

$S = \left[1000t + 250t^2\right]_{0}^{6}$
$S = 6000 + 9000 = 15000$

Total sales in 6 months is 15,000 units.
Calculating variable cost

Integrate the marginal cost function:

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

After evaluation:

$VC = \left[x^2 + 50x\right]_{0}^{25}$
$VC = 625 + 1250 = 1875$

Total variable cost is 1,875 thousand dollars or 1.875 million dollars.
Calculating investment growth

For exponential functions, we integrate:

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

The result is:

$I = 2000 \left[\frac{e^{0.05t}}{0.05}\right]_{0}^{4}$
$I = 40000(e^{0.2} - 1) \approx 40000(0.221) = 8856$

Total investment growth in 4 years is approximately 8.86 billion dollars.

Command Palette