Integration Using Completing the Square (College Board AP® Calculus AB)

Study Guide

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Integration using completing the square

What is completing the square?

  • Completing the square is a way of rewriting quadratic expressions

    • A quadratic expression is of the form a x squared plus b x plus c

    • Rewriting can make them easier to use for various purposes

  • The simple version of completing the square is

    • x squared plus b x plus c equals open parentheses x plus b over 2 close parentheses squared plus c minus open parentheses b over 2 close parentheses squared

      • E.g. space x squared minus 6 x plus 2 equals open parentheses x minus 3 close parentheses squared plus 2 minus open parentheses negative 3 close parentheses squared equals open parentheses x minus 3 close parentheses squared minus 7

  • When there is a coefficient in front of the bold italic x to the power of bold 2 this becomes

    • a x squared plus b x plus c equals a open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses squared plus c minus a open parentheses fraction numerator b over denominator 2 a end fraction close parentheses squared

      • E.g. space 3 x squared plus 12 x minus 7 equals 3 open parentheses x plus 2 close parentheses squared minus 7 minus 3 open parentheses 2 close parentheses squared equals 3 open parentheses x plus 2 close parentheses squared minus 19

How can I integrate using completing the square?

  • Recall the standard integrals

    • integral fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction space d x equals arcsin x plus C comma space space minus 1 less than x less than 1

    • integral fraction numerator 1 over denominator 1 plus x squared end fraction space d x equals arc tan x plus C

    • See the 'Derivatives & Antiderivatives' study guide

  • Completing the square can be used to get integrals into a form where these results can be used

    • The methods from the 'Integrals of Composite Functions' study guide may be needed as well

  • E.g. integral fraction numerator 1 over denominator x squared minus 6 x plus 13 end fraction space d x

    • Complete the square on the denominator

      • x squared minus 6 x plus 13 equals open parentheses x minus 3 close parentheses squared plus 4

    • Rewrite the function to be integrated

      • fraction numerator 1 over denominator x squared minus 6 x plus 13 end fraction equals fraction numerator 1 over denominator 4 plus open parentheses x minus 3 close parentheses squared end fraction equals 1 fourth times fraction numerator 1 over denominator 1 plus open parentheses fraction numerator x minus 3 over denominator 2 end fraction close parentheses squared end fraction

    • That is now a multiple of fraction numerator 1 over denominator 1 plus open parentheses horizontal ellipsis close parentheses squared end fraction

      • This tells us that arctan open parentheses horizontal ellipsis close parentheses is going to be be our likely answer

    • To see if 'adjust and compensate' is needed, use the chain rule to differentiate arctan open parentheses fraction numerator x minus 3 over denominator 2 end fraction close parentheses

      • fraction numerator d over denominator d x end fraction open parentheses arctan open parentheses fraction numerator x minus 3 over denominator 2 end fraction close parentheses close parentheses equals fraction numerator 1 over denominator 1 plus open parentheses fraction numerator x minus 3 over denominator 2 end fraction close parentheses squared end fraction times 1 equals fraction numerator 1 over denominator 1 plus open parentheses fraction numerator x minus 3 over denominator 2 end fraction close parentheses squared end fraction

    • Integrate

      • integral fraction numerator 1 over denominator x squared minus 6 x plus 13 end fraction space d x equals 1 fourth integral fraction numerator 1 over denominator 1 plus open parentheses fraction numerator x minus 3 over denominator 2 end fraction close parentheses squared end fraction space d x equals 1 fourth arctan open parentheses fraction numerator x minus 3 over denominator 2 end fraction close parentheses plus C

Worked Example

Find the indefinite integral integral fraction numerator 1 over denominator square root of 21 minus x squared plus 4 x end root end fraction space d x.

Answer:

Complete the square on the expression in the square root

21 minus x squared plus 4 x equals negative open parentheses x minus 2 close parentheses squared plus 21 minus open parentheses negative open parentheses 2 close parentheses squared close parentheses equals 25 minus open parentheses x minus 2 close parentheses squared

Rewrite the function to be integrated

fraction numerator 1 over denominator square root of 21 minus x squared plus 4 x end root end fraction equals fraction numerator 1 over denominator square root of 25 minus open parentheses x minus 2 close parentheses squared end root end fraction equals 1 fifth times fraction numerator 1 over denominator square root of 1 minus open parentheses fraction numerator x minus 2 over denominator 5 end fraction close parentheses squared end root end fraction

That is now a multiple of fraction numerator 1 over denominator square root of 1 minus open parentheses horizontal ellipsis close parentheses squared end root end fraction

This tells us that arc sin open parentheses horizontal ellipsis close parentheses is going to be be our likely answer

To see if 'adjust and compensate' is needed, use the chain rule to differentiate arcsin open parentheses fraction numerator x minus 2 over denominator 5 end fraction close parentheses

fraction numerator d over denominator d x end fraction open parentheses arc sin open parentheses fraction numerator x minus 2 over denominator 5 end fraction close parentheses close parentheses equals fraction numerator 1 over denominator square root of 1 minus open parentheses fraction numerator x minus 2 over denominator 5 end fraction close parentheses squared end root end fraction times 1 fifth

Integrate

integral fraction numerator 1 over denominator square root of 21 minus x squared plus 4 x end root end fraction space d x equals integral 1 fifth times fraction numerator 1 over denominator square root of 1 minus open parentheses fraction numerator x minus 2 over denominator 5 end fraction close parentheses squared end root end fraction space d x equals arc sin open parentheses fraction numerator x minus 2 over denominator 5 end fraction close parentheses plus C

integral fraction numerator 1 over denominator square root of 21 minus x squared plus 4 x end root end fraction space d x equals arc sin open parentheses fraction numerator x minus 2 over denominator 5 end fraction close parentheses plus C

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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.