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AP®MathsCollege BoardCalculus ABStudy GuidesUnit 8: Applications of IntegrationDefinite Integrals in ContextDefinite Integrals as Accumulated Change

Definite Integrals as Accumulated Change (College Board AP® Calculus AB)

Study Guide

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Definite Integrals as Accumulated Change

How can I use definite integrals to calculate net change in applied contexts?

  • A definite integral can be used to represent an accumulation of a rate of change

  • Let f open parentheses x close parentheses represent the rate of change of a function F open parentheses x close parentheses

    • i.e. f open parentheses x close parentheses equals fraction numerator d over denominator d t end fraction F open parentheses x close parentheses equals F to the power of apostrophe open parentheses x close parentheses

  • The net change (i.e. total change) of F between x equals x subscript 1 and x equals x subscript 2 is given by integral subscript x subscript 1 end subscript superscript x subscript 2 end superscript space space f open parentheses x close parentheses space d x

    • f open parentheses x close parentheses times straight capital delta x is the change in F over a small interval straight capital delta x

    • f open parentheses x close parentheses space d x is the limit of this change as increment x rightwards arrow 0

    • The integral integral subscript x subscript 1 end subscript superscript x subscript 2 end superscript space f open parentheses x close parentheses space d x sums up all these infinitesimal changes between x equals x subscript 1 and x equals x subscript 2

      • See the 'Properties of Definite Integrals' study guide for the idea of a definite integral as the limit of a sum

  • The value of F at x subscript 2 is found by adding the value of F at x subscript 1 to the change between x subscript 1 and x subscript 2

    • F open parentheses x subscript 2 close parentheses equals F open parentheses x subscript 1 close parentheses plus integral subscript x subscript 1 end subscript superscript x subscript 2 end superscript space space f open parentheses x close parentheses space d x

  • The value of F for any value of bold italic x can also be written in terms of a definite integral

    • F open parentheses x close parentheses equals F open parentheses x subscript 0 close parentheses plus integral subscript x subscript 0 end subscript superscript x space f open parentheses s close parentheses space d s

      • F open parentheses x subscript 0 close parentheses is a known value of F at a particular x-value, x subscript 0

      • s is just a 'dummy variable' used for the integration

  • Many real world situations may be modeled in this way

    • For example if P open parentheses x close parentheses, R open parentheses x close parentheses and C open parentheses x close parentheses represent a company's profit, revenue and cost respectively

      • where x represents the units of merchandise sold

      • and profit = revenue - cost: P open parentheses x close parentheses equals R open parentheses x close parentheses minus C open parentheses x close parentheses

    • Then P to the power of apostrophe open parentheses x close parentheses, R to the power of apostrophe open parentheses x close parentheses and C to the power of apostrophe open parentheses x close parentheses are known as the marginal profit, marginal revenue and marginal cost

      • These represent the rate of change of their respective quantities when x increases by 1

  • Frequently in real world situations the independent variable will be time

    • In this case the independent variable will usually be represented by t instead of x

    • I.e. if f open parentheses t close parentheses is the rate of change of a quantity represented by the function F open parentheses t close parentheses

    • Then the net change of F between times t subscript 1 and t subscript 2 is integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space space f open parentheses t close parentheses space d t

    • The value of F at t subscript 2 is F open parentheses t subscript 2 close parentheses equals F open parentheses t subscript 1 close parentheses plus integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space space f open parentheses t close parentheses space d t

    • And F open parentheses t close parentheses equals F open parentheses t subscript 0 close parentheses plus integral subscript t subscript 0 end subscript superscript t space f open parentheses s close parentheses space d s

      • where F open parentheses t subscript 0 close parentheses is a known value of F at a particular time t subscript 0

Examiner Tips and Tricks

For questions involving real world contexts

  • Be sure to give your answers in the context of the questions

  • This includes using correct units and interpreting results using the language and details of the question

Remember when interpreting rates of change

  • A positive rate of change corresponds to an increase in the quantity

  • A negative rate of change corresponds to a decrease

Worked Example

A company produces and sells luxury chess sets. The marginal revenue of the company is given by R to the power of apostrophe open parentheses x close parentheses equals 625, and its marginal cost is given by C to the power of apostrophe open parentheses x close parentheses equals 300 plus 0.02 x. Both revenue R and cost C are measured in dollars, and in both cases x represents number of chess sets sold.

(a) Find the value of C to the power of apostrophe open parentheses 200 close parentheses and interpret this in the context of the problem.

Answer:

Substitute x equals 200 into C to the power of apostrophe open parentheses x close parentheses

C to the power of apostrophe open parentheses 200 close parentheses equals 300 plus 0.02 open parentheses 200 close parentheses equals 304

Interpret this in the context of the question, paying attention to units

The rate of change is positive, so the cost is increasing

C to the power of apostrophe open parentheses 200 close parentheses equals 304. At the point when 200 chess sets have been sold, the company's cost is increasing at the rate of $304 per chess set.

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

Answer:

This is the integral of C to the power of apostrophe between x equals 200 and x equals 300

table row cell change space in space cost end cell equals cell integral subscript 200 superscript 300 open parentheses 300 plus 0.02 x close parentheses space d x end cell row blank equals cell open square brackets 300 x plus 0.01 x squared close square brackets subscript 200 superscript 300 end cell row blank equals cell 300 open parentheses 300 close parentheses plus 0.01 open parentheses 300 close parentheses squared minus open parentheses 300 open parentheses 200 close parentheses plus 0.01 open parentheses 200 close parentheses squared close parentheses end cell row blank equals cell 90000 plus 900 minus open parentheses 60000 plus 400 close parentheses end cell row blank equals 30500 end table

This is positive, so represents an increase

In going from selling 200 to 300 chess sets, the company's cost increases by $30500.

(c) Using profit equals revenue minus cost, find the change in the profit, P, made by the company in going from selling 200 to 300 chess sets.

Answer:

P open parentheses x close parentheses equals R open parentheses x close parentheses minus C open parentheses x close parentheses means that P to the power of apostrophe open parentheses x close parentheses equals fraction numerator d over denominator d x end fraction open parentheses R open parentheses x close parentheses minus C open parentheses x close parentheses close parentheses equals R to the power of apostrophe open parentheses x close parentheses minus C to the power of apostrophe open parentheses x close parentheses

P to the power of apostrophe open parentheses x close parentheses equals R to the power of apostrophe open parentheses x close parentheses minus C to the power of apostrophe open parentheses x close parentheses

P to the power of apostrophe open parentheses x close parentheses equals 625 minus open parentheses 300 plus 0.02 x close parentheses equals 325 minus 0.02 x

The change in the profit is the integral of P to the power of apostrophe between x equals 200 and x equals 300

table row cell change space in space profit end cell equals cell integral subscript 200 superscript 300 open parentheses 325 minus 0.02 x close parentheses space d x end cell row blank equals cell open square brackets 325 x minus 0.01 x squared close square brackets subscript 200 superscript 300 end cell row blank equals cell 325 open parentheses 300 close parentheses minus 0.01 open parentheses 300 close parentheses squared minus open parentheses 325 open parentheses 200 close parentheses minus 0.01 open parentheses 200 close parentheses squared close parentheses end cell row blank equals cell 97500 minus 900 minus open parentheses 65000 minus 400 close parentheses end cell row blank equals 32000 end table

This is positive, so represents an increase

In going from selling 200 to 300 chess sets, the company's profit increases by $32 000.

Worked Example

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 open parentheses t close parentheses equals 4 plus 4 cos open parentheses pi over 4 t close parentheses space space for space space 0 less than t less or equal than 20

where f open parentheses t close parentheses is measured in grams per hour and 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?

Answer:

This will be the definite integral of f open parentheses t close parentheses between t equals 0 and t equals 4

table row cell integral subscript 0 superscript 4 space open parentheses 4 plus 4 cos open parentheses pi over 4 t close parentheses close parentheses space d t end cell equals cell 4 integral subscript 0 superscript 4 space open parentheses 1 plus cos open parentheses pi over 4 t close parentheses close parentheses space d t end cell row blank equals cell 4 open square brackets t plus 4 over pi sin open parentheses pi over 4 t close parentheses close square brackets subscript 0 superscript 4 end cell row blank equals cell 4 open parentheses 4 plus 4 over pi sin open parentheses pi over 4 open parentheses 4 close parentheses close parentheses minus open parentheses 0 plus 4 over pi sin open parentheses pi over 4 open parentheses 0 close parentheses close parentheses close parentheses close parentheses end cell row blank equals cell 4 open parentheses 4 plus 4 over pi sin open parentheses pi close parentheses minus open parentheses 0 plus 4 over pi sin open parentheses 0 close parentheses close parentheses close parentheses end cell row blank equals cell 4 open parentheses 4 plus 0 minus open parentheses 0 plus 0 close parentheses close parentheses end cell row blank equals 16 end table

16 grams

(b) Find f to the power of apostrophe open parentheses 26 over 3 close parentheses. Using correct units, explain the meaning of f to the power of apostrophe open parentheses 26 over 3 close parentheses in the context of the problem.

Answer:

First differentiate to find f to the power of apostrophe open parentheses t close parentheses

f to the power of apostrophe open parentheses t close parentheses equals 4 times open parentheses negative sin open parentheses pi over 4 t close parentheses close parentheses times pi over 4 equals negative pi sin open parentheses pi over 4 t close parentheses

Substitute in t equals 26 over 3

table row cell f to the power of apostrophe open parentheses 26 over 3 close parentheses end cell equals cell negative pi sin open parentheses pi over 4 open parentheses 26 over 3 close parentheses close parentheses end cell row blank equals cell negative pi sin open parentheses fraction numerator 13 pi over denominator 6 end fraction close parentheses end cell row blank equals cell negative pi open parentheses 1 half close parentheses end cell row blank equals cell negative pi over 2 end cell end table

f open parentheses t close parentheses is a rate of change, so f to the power of apostrophe open parentheses t close parentheses is the rate of change of a rate of change

It is negative, so the rate of change is decreasing

f to the power of apostrophe open parentheses 26 over 3 close parentheses equals negative pi over 2. This means that 26 over 3 hours after the food bowl was filled, the rate at which the rabbits are consuming the food is decreasing by pi over 2 grams per hour, per hour.

(c) Assuming that no more food is added to the bowl, find an expression in terms of t for F open parentheses t close parentheses, the amount of food left in the bowl at time t.

Answer:

This will be the integral of f between 0 and t, subtracted from the amount of food in the bowl at time t equals 0

Be careful here: f open parentheses t close parentheses is the rate at which the rabbits are eating the food, but this means that the rate at which the amount of food in the bowl is changing is negative f open parentheses t close parentheses

table row cell F open parentheses t close parentheses end cell equals cell F open parentheses 0 close parentheses minus integral subscript 0 superscript t space f open parentheses s close parentheses space d s end cell row blank equals cell 80 minus integral subscript 0 superscript t space open parentheses 4 plus 4 cos open parentheses pi over 4 s close parentheses close parentheses space d s end cell row blank equals cell 80 minus 4 integral subscript 0 superscript t space open parentheses 1 plus cos open parentheses pi over 4 s close parentheses close parentheses space d s end cell row blank equals cell 80 minus 4 open square brackets s plus 4 over pi sin open parentheses pi over 4 s close parentheses close square brackets subscript 0 superscript t end cell row blank equals cell 80 minus 4 open parentheses t plus 4 over pi sin open parentheses pi over 4 open parentheses t close parentheses close parentheses minus open parentheses 0 plus 4 over pi sin open parentheses pi over 4 open parentheses 0 close parentheses close parentheses close parentheses close parentheses end cell row blank equals cell 80 minus 4 open parentheses t plus 4 over pi sin open parentheses pi over 4 t close parentheses minus open parentheses 0 plus 4 over pi sin open parentheses 0 close parentheses close parentheses close parentheses end cell row blank equals cell 80 minus 4 open parentheses t plus 4 over pi sin open parentheses pi over 4 t close parentheses minus open parentheses 0 plus 0 close parentheses close parentheses end cell row blank equals cell 80 minus 4 t minus 16 over pi sin open parentheses pi over 4 t close parentheses end cell end table

This could also be solved by finding the indefinite integral F open parentheses t close parentheses equals integral f open parentheses t close parentheses space d t, and then using F open parentheses 0 close parentheses equals 80 to work out the value of the constant of integration

F open parentheses t close parentheses equals table row blank equals blank end table table row blank blank 80 end table table row blank blank minus end table table row blank blank 4 end table table row blank blank t end table table row blank blank minus end table table row blank blank cell 16 over pi end cell end table table row blank blank sin end table table row blank blank cell open parentheses pi over 4 t close parentheses end cell end table

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.