Marginal profit, marginal revenue, and marginal cost are given by: [ P'(x) = R'(x) - C'(x) ] These represent the rates of change of their respective quantities when $x$ increases by 1.

Time as an Independent Variable

Often, the independent variable in real-world situations is time. If $f(t)$ is the rate of change of a quantity $F(t)$, then: [ \text{Net change of } F \text{ between times } t_1 \text{ and } t_2 \text{ is } \int_{t_1}^{t_2} f(t) , dt ]

The value of $F$ at $t_2$ is: [ F(t_2) = F(t_1) + \int_{t_1}^{t_2} f(t) , dt ]

And for any $t$: [ F(t) = F(t_0) + \int_{t_0}^{t} f(s) , ds ] where $s$ is a dummy variable.

Exam Tips

Worked Examples

Example 1: Marginal Cost and Profit

A company produces and sells luxury chess sets.

(a) Find the value of $C'(200)$ and interpret it in the context of the problem.

Solution:

Given $C'(x) = 300 + 0.02x$: [ C'(200) = 300 + 0.02 \times 200 = 304 ]

(b) Find the change in the cost, $C$, to the company in going from selling 200 to 300 chess sets.

Solution:

This is the integral from $x = 200$ to $x = 300$: [ \int_{200}^{300} (300 + 0.02x) , dx ]

[ \begin{array}{rcl} \text{Change in cost} & = & \left[ 300x + 0.01x^2 \right]_{200}^{300} \ & = & (300 \times 300 + 0.01 \times 300^2) - (300 \times 200 + 0.01 \times 200^2) \ & = & 90000 + 900 - (60000 + 400) \ & = & 30500 \end{array} ]

(c) Using $\text{profit} = \text{revenue} - \text{cost}$, find the change in the profit, $P$, made by the company in going from selling 200 to 300 chess sets.

Solution:

Given $P(x) = R(x) - C(x)$: [ P'(x) = R'(x) - C'(x) ]

[ P'(x) = 625 - (300 + 0.02x) = 325 - 0.02x ]

The change in profit is the integral from $x = 200$ to $x = 300$: [ \int_{200}^{300} (325 - 0.02x) , dx ]

[ \begin{array}{rcl} \text{Change in profit} & = & \left[ 325x - 0.01x^2 \right]_{200}^{300} \ & = & (325 \times 300 - 0.01 \times 300^2) - (325 \times 200 - 0.01 \times 200^2) \ & = & 97500 - 900 - (65000 - 400) \ & = & 32000 \end{array} ]

Example 2: Food Consumption by Rabbits

When the food bowl in a pet rabbit habitat is filled up, it contains 80 grams of food. The rabbits consume food from the bowl at a rate modeled by: [ f(t) = 4 + 4 \cos \left( \frac{\pi}{4} t \right) \text{ for } 0 < t \leq 20 ] where $t$ is the number of hours after the bowl was filled.

(a) How many grams of food do the rabbits eat during the first four hours after the bowl is filled?

Solution:

This is the definite integral of $f(t)$ between $t = 0$ and $t = 4$: [ \int_{0}^{4} \left( 4 + 4 \cos \left( \frac{\pi}{4} t \right) \right) , dt ]

[ \begin{array}{rcl} \int_{0}^{4} \left( 4 + 4 \cos \left( \frac{\pi}{4} t \right) \right) , dt & = & 4 \int_{0}^{4} \left( 1 + \cos \left( \frac{\pi}{4} t \right) \right) , dt \ & = & 4 \left[ t + \frac{4}{\pi} \sin \left( \frac{\pi}{4} t \right) \right]_{0}^{4} \ & = & 4 \left( 4 + \frac{4}{\pi} \sin(\pi) - \left( 0 + \frac{4}{\pi} \sin(0) \right) \right) \ & = & 4 \left( 4 + 0 - 0 \right) \ & = & 16 \end{array} ]

(b) Find $f' \left( \frac{26}{3} \right)$. Using correct units, explain the meaning of $f' \left( \frac{26}{3} \right)$ in the context of the problem.

Solution:

First, differentiate $f(t)$: [ f'(t) = 4 \left( -\sin \left( \frac{\pi}{4} t \right) \right) \left( \frac{\pi}{4} \right) = -\pi \sin \left( \frac{\pi}{4} t \right) ]

Substitute $t = \frac{26}{3}$: [ f' \left( \frac{26}{3} \right) = -\pi \sin \left( \frac{\pi}{4} \left( \frac{26}{3} \right) \right) = -\pi \sin \left( \frac{13 \pi}{6} \right) = -\pi \left( \frac{1}{2} \right) = -\frac{\pi}{2} ]

(c) Assuming that no more food is added to the bowl, find an expression for $F(t)$ in terms of $t$.

Solution:

This is the integral of $f(t)$, subtracted from the amount of food in the bowl at $t = 0$:

[ \begin{array}{rcl} F(t) & = & F(0) - \int_{0}^{t} f(s) , ds \ & = & 80 - \int_{0}^{t} \left( 4 + 4 \cos \left( \frac{\pi}{4} s \right) \right) , ds \ & = & 80 - 4 \int_{0}^{t} \left( 1 + \cos \left( \frac{\pi}{4} s \right) \right) , ds \ & = & 80 - 4 \left[ s + \frac{4}{\pi} \sin \left( \frac{\pi}{4} s \right) \right]_{0}^{t} \ & = & 80 - 4 \left( t + \frac{4}{\pi} \sin \left( \frac{\pi}{4} t \right) - \left( 0 + \frac{4}{\pi} \sin(0) \right) \right) \ & = & 80 - 4 \left( t + \frac{4}{\pi} \sin \left( \frac{\pi}{4} t \right) - 0 \right) \ & = & 80 - 4t - \frac{16}{\pi} \sin \left( \frac{\pi}{4} t \right) \end{array} ]

Glossary

• Definite Integral: A mathematical concept used to calculate the accumulation of a rate of change over an interval.
• Net Change: The total change in a quantity over an interval.
• Rate of Change: The speed at which a variable changes over a specific period of time.
• Marginal Profit/Revenue/Cost: The rate of change of profit, revenue, or cost with respect to the number of units sold.

Practice Questions

Summary and Key Takeaways

• Definite integrals can be used to calculate the net change of a quantity over an interval.
• The net change formula is $\int_{x_1}^{x_2} f(x) , dx$.
• Real-world applications include calculating changes in cost, profit, and other measurable quantities.
• Always interpret results in the context of the problem, using correct units.
• Understanding the rate of change and its implications is crucial for solving applied problems.

Key Takeaways

• Use definite integrals to find accumulated change.
• Interpret the integrals in the context of real-world problems.
• Practice differentiating and integrating functions to build proficiency.

By mastering these concepts, students can effectively tackle problems involving accumulated change and apply these skills in various contexts.