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Integrating with Exponential & Logarithmic Functions (DP IB Analysis & Approaches (AA)): Revision Note

Paul

Written by: Paul

Reviewed by: Dan Finlay

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DP Analysis & Approaches (AA): HLDP Analysis & Approaches (AA): HL

Integrating exponential & logarithmic functions

Exponential functions have the general form y equals a to the power of x.  Special case:  y equals straight e to the power of x.

Logarithmic functions have the general form y equals log subscript a x.  Special case:  y equals log subscript straight e x equals ln space x.

What are the antiderivatives of exponential and logarithmic functions?

  • Those involving the special cases have been met before

    • integral subscript blank superscript blank straight e to the power of x space d x equals straight e to the power of x plus c

    • integral subscript blank superscript blank 1 over x space d x equals ln space vertical line x vertical line plus c

    • These are given in the formula booklet

  • Also

    • integral subscript blank superscript blank a to the power of x space d x equals fraction numerator 1 over denominator ln space a end fraction a to the power of x plus c

    • This is also given in the formula booklet

  • By reverse chain rule

    • integral subscript blank superscript blank fraction numerator 1 over denominator x ln space a end fraction space d x equals log subscript a vertical line x vertical line plus c

    • This is not in the formula booklet

      • but the derivative of log subscript a x is given

  • There is also the reverse chain rule to look out for

    • this occurs when the numerator is (a multiple of) the derivative of the denominator

    • integral space fraction numerator f to the power of apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction space d x equals ln open vertical bar f open parentheses x close parentheses close vertical bar plus c

How do I integrate exponentials and logarithms with a linear function of x involved?

  • For the special cases involving e and ln

    • integral subscript blank superscript blank straight e to the power of a x plus b end exponent space d x equals 1 over a straight e to the power of a x plus b end exponent plus straight c

    • integral space fraction numerator 1 over denominator a x plus b end fraction space d x equals 1 over a ln open vertical bar a x plus b close vertical bar plus c

  • For the general cases

    • integral space a to the power of space p x plus q end exponent space straight d x equals fraction numerator 1 over denominator p space ln space a end fraction a to the power of space p x plus q end exponent plus c

    • integral space fraction numerator 1 over denominator open parentheses p x plus q close parentheses ln space a end fraction space straight d x equals 1 over p log subscript a vertical line p x plus q vertical line plus c

  • These four results are not in the formula booklet but all can be derived using ‘adjust and compensate’ from reverse chain rule

Examiner Tips and Tricks

Remember always to use the modulus sign for logarithmic terms in the antiderivative.

However if you can deduce that  g left parenthesis x right parenthesis in  ln vertical line g left parenthesis x right parenthesis vertical line, say, is guaranteed to always be positive, then you can replace the modulus sign with brackets.

Worked Example

a)       Show that integral subscript 1 superscript 2 4 to the power of x space d x equals fraction numerator 6 over denominator ln space 2 end fraction.

5-9-1-ib-hl-aa-only-we3a-soltn

b)       Find integral fraction numerator 1 over denominator open parentheses 2 x minus 1 close parentheses space ln space 3 end fraction space straight d x.

zgf4~aK9_5-9-1-ib-hl-aa-only-we3b-soltn

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    Paul

    Author: Paul

    Expertise: Maths Content Creator

    Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject 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.