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Indefinite Integral Rules (College Board AP® Calculus AB)

Study Guide

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Indefinite integrals of sums, differences and constant multiples

How do I integrate sums, differences and constant multiples of terms?

  • When integrating sums or differences of terms

    • the integral is simply the sum (or difference) of the integrals of the terms

    • This may be expressed as integral open parentheses f open parentheses x close parentheses plus-or-minus g open parentheses x close parentheses close parentheses space d x space equals space integral f open parentheses x close parentheses space d x plus-or-minus integral g open parentheses x close parentheses space d x

      • E.g. integral open parentheses x squared plus cos x close parentheses space d x equals integral x squared space d x plus integral cos x space d x equals 1 third x cubed plus sin x plus C

      • We still only need a single constant of integration

    • Note that products and quotients of terms cannot be integrated in this way

      • They may need to be expanded or simplified first

      • Or a more advanced integration technique may need to be used

        • See the 'Methods of Integration' study guide

  • When integrating a constant multiple of a term

    • the term may be brought out in front of the integral as a multiplier

    • This may be expressed as integral k f open parentheses x close parentheses space d x space equals space k integral f open parentheses x close parentheses space d x

      • E.g. integral 5 e to the power of x space d x equals 5 integral e to the power of x space d x equals 5 e to the power of x plus C

  • The rules for sums, differences and constant multiples can be combined

    • If p and q are constants, then

      • integral open parentheses p f open parentheses x close parentheses plus-or-minus q g open parentheses x close parentheses close parentheses space d x equals p integral f open parentheses x close parentheses space d x plus-or-minus q integral g open parentheses x close parentheses space d x

      • This idea can be extended for any number of terms inside the integral

Worked Example

Find the indefinite integral integral open parentheses 3 x cubed minus 2 sin 3 x plus 5 e to the power of 4 x end exponent close parentheses space d x.

Answer:

Write as a sum or difference of integrals, with constants pulled in front as multipliers

3 integral x cubed space d x minus 2 integral sin 3 x space d x plus 5 integral e to the power of 4 x end exponent space d x

Once you've become confident with integration, you can skip this step

Now integrate the terms one by one

3 times 1 fourth x to the power of 4 minus 2 times open parentheses negative 1 third cos 3 x close parentheses plus 5 times 1 fourth e to the power of 4 x end exponent plus C

Don't forget the constant of integration

Multiply out and simplify

3 over 4 x to the power of 4 plus 2 over 3 cos 3 x plus 5 over 4 e to the power of 4 x end exponent plus C

Simplifying expressions to find indefinite integrals

How can I simplify expressions to help me find indefinite integrals?

  • Sometimes expressions need to be simplified before you can integrate them

  • This may involve expanding brackets

    • E.g. integral open parentheses x squared plus 2 close parentheses squared space d x

    • Expand the brackets

      • open parentheses x squared plus 2 close parentheses squared equals x to the power of 4 plus 4 x squared plus 4

    • Now integrate in the usual way

      • integral open parentheses x squared plus 2 close parentheses squared space d x equals integral open parentheses x to the power of 4 plus 4 x squared plus 4 close parentheses space d x equals 1 fifth x to the power of 5 plus 4 over 3 x cubed plus 4 x plus C

  • Or rearranging fractions (including using laws of exponents)

    • E.g. integral fraction numerator 5 x cubed minus 3 over denominator x squared end fraction space d x

    • Rearrange the fraction

      • fraction numerator 5 x cubed minus 3 over denominator x squared end fraction equals fraction numerator 5 x cubed over denominator x squared end fraction minus 3 over x squared equals 5 x minus 3 over x squared

    • Use laws of exponents

      • 5 x minus 3 over x squared equals 5 x minus 3 x to the power of negative 2 end exponent

    • Now integrate in the usual way

      • integral fraction numerator 5 x cubed minus 3 over denominator x squared end fraction space d x equals integral open parentheses 5 x minus 3 x to the power of negative 2 end exponent close parentheses space d x equals 5 over 2 x squared plus 3 x to the power of negative 1 end exponent plus C

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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.