O LevelFurther MathsCambridge (CIE)Additional MathsRevision NotesCalculusIntegrationIntegrating Powers of x

Integrating Powers of x (Cambridge (CIE) O Level Additional Maths) : Revision Note

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Integrating Powers of x

How do I integrate powers of x?

  • Powers ofspace x are integrated according to the following formulae:

    • Ifspace f left parenthesis x right parenthesis equals x to the power of n thenspace integral f left parenthesis x right parenthesis space straight d x equals fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus c wherespace n element of straight rational numbers comma space n not equal to negative 1 andspace c is the constant of integration

    •  For each term …

      • … increase the power (of x) by 1

      • … divide by the new power

    • This does not apply when the original power is -1

      • the new power would be 0 and division by 0 is undefined

Formula for integrating x to the power of a constant

  • If the power ofspace x is multiplied by a constant then the integral is also multiplied by that constant

    • Ifspace f left parenthesis x right parenthesis equals a x to the power of n thenspace integral f left parenthesis x right parenthesis space straight d x equals fraction numerator a x to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus c wherespace n element of straight rational numbers comma space n not equal to negative 1 andspace a is a constant andspace c is the constant of integration

  • Remember the special case:

    • space integral a space straight d x equals a x plus c

      • e.g. space integral 4 space straight d x equals 4 x plus c 

    • This allows constant terms to be integrated

How do I integrate expressions containing powers of x?

  • The formulae for integrating powers ofspace x apply to all rational numbers so it is possible to integrate any expression that is a sum or difference of powers ofspace x

    • e.g.  Ifspace f left parenthesis x right parenthesis equals 8 x cubed minus 2 x plus 4 thenspace integral f stretchy left parenthesis x stretchy right parenthesis blank straight d x equals fraction numerator 8 x to the power of 3 plus 1 end exponent over denominator 3 plus 1 end fraction minus fraction numerator 2 x to the power of 1 plus 1 end exponent over denominator 1 plus 1 end fraction plus 4 x plus c equals 2 x to the power of 4 minus x squared plus 4 x plus c

  • Functions involving roots will need to be rewritten as fractional powers ofspace x first

    • eg. Ifspace f left parenthesis x right parenthesis equals 5 cube root of x then rewrite asspace f left parenthesis x right parenthesis equals 5 x to the power of 1 third end exponent and integrate

  • Functions involving fractions with denominators in terms ofbold space bold italic x will need to be rewritten as negative powers ofspace x first

    • e.g.  Ifspace f left parenthesis x right parenthesis equals 4 over x squared plus x squared then rewrite asspace f left parenthesis x right parenthesis equals 4 x to the power of negative 2 end exponent plus x squared and integrate   

  • Products and quotients cannot be integrated this way so would need expanding/simplifying first

    • e.g.  Ifspace f stretchy left parenthesis x stretchy right parenthesis equals 8 x squared left parenthesis 2 x minus 3 right parenthesis thenError converting from MathML to accessible text.

Examiner Tips and Tricks

  • You can speed up the process of integration in the exam by committing the pattern of basic integration to memory

    • In general you can think of it as 'raising the power by one and dividing by the new power'

    • Practice this lots before your exam so that it comes quickly and naturally when doing more complicated integration questions

Worked Example

Given that

space fraction numerator straight d y over denominator straight d x end fraction equals 3 x to the power of 4 minus 2 x squared plus 3 minus fraction numerator 1 over denominator square root of x end fraction

find an expression forspace y in terms ofspace x.

Rewrite all terms as powers of x using the laws of indices for fractional and negative powers on the last term.

fraction numerator straight d y over denominator straight d x end fraction equals space 3 x to the power of 4 space minus space 2 x squared space plus space 3 space minus space x to the power of negative 1 half end exponent

Find y by integrating each term. 

y space equals space integral space open parentheses 3 x to the power of 4 space minus space 2 x squared space plus space 3 space minus space x to the power of negative 1 half end exponent space close parentheses space straight d x

y space equals fraction numerator space 3 x to the power of 5 over denominator 5 end fraction minus fraction numerator 2 x cubed over denominator 3 end fraction space plus space 3 x space minus fraction numerator space x to the power of 1 half end exponent over denominator 1 half end fraction space plus space c

Rewrite using the same format given in the question. 

bold italic y bold space bold equals bold space bold 3 over bold 5 bold italic x to the power of bold 5 bold space bold minus bold space bold 2 over bold 3 bold italic x to the power of bold 3 bold space bold plus bold space bold 3 bold italic x bold space bold minus bold space bold 2 square root of bold x bold space bold plus bold italic c

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Finding the Constant of Integration

How do I find the constant of integration? 

  • STEP 1 

    • Rewrite the function into a more easily integrable form

      • Each term needs to be a power of x (or a constant)

  • STEP 2 

    • Integrate each term and remember “+c”

      • Increase power by 1 and divide by new power

  • STEP 3

    • Substitute the coordinates of a given point in to form an equation in c

      • Solve the equation to find c

Notes fig3, A Level & AS Level Pure Maths Revision Notes

 

Example of finding the constant of integration

 

Examiner Tips and Tricks

  • If a constant of integration can be found then the question will need to give you some extra information

    • If this is given then make sure you use it to find the value of c

Worked Example

Given straight f apostrophe open parentheses x close parentheses space equals space fraction numerator open parentheses x plus 3 close parentheses squared over denominator square root of x end fraction and straight f open parentheses 1 close parentheses space equals space 25, find straight f open parentheses x close parentheses.   

Rewrite straight f to the power of apostrophe open parentheses x close parentheses in a form the can be integrated more easily. 

table row cell straight f to the power of apostrophe open parentheses x close parentheses space end cell equals cell space fraction numerator open parentheses x plus 3 close parentheses open parentheses x plus 3 close parentheses over denominator square root of x end fraction end cell row blank equals cell space fraction numerator x to the power of 2 space end exponent plus space 6 x space plus space 9 over denominator x to the power of 1 half end exponent end fraction end cell row space equals cell fraction numerator space x squared over denominator x to the power of 1 half end exponent end fraction space plus fraction numerator space 6 x over denominator x to the power of 1 half end exponent end fraction space plus fraction numerator space 9 over denominator x to the power of 1 half end exponent end fraction end cell row blank equals cell space x to the power of 3 over 2 space end exponent plus space 6 x to the power of 1 half end exponent space plus space 9 x to the power of negative 1 half end exponent end cell end table   

Integrate each term, remember to include a constant of integration.

straight f open parentheses x close parentheses space equals fraction numerator space x to the power of 5 over 2 end exponent over denominator 5 over 2 end fraction plus fraction numerator 6 x to the power of 3 over 2 end exponent over denominator 3 over 2 end fraction plus fraction numerator 9 x to the power of 1 half end exponent over denominator 1 half end fraction plus c   

Simplify.

straight f open parentheses x close parentheses space equals 2 over 5 x to the power of 5 over 2 end exponent plus 4 x to the power of 3 over 2 end exponent plus 18 x to the power of 1 half end exponent plus c

Use f(1) =25 to find the value of c.

straight f open parentheses 1 close parentheses space equals 2 over 5 open parentheses 1 close parentheses to the power of 5 over 2 end exponent plus 4 open parentheses 1 close parentheses to the power of 3 over 2 end exponent plus 18 open parentheses 1 close parentheses to the power of 1 half end exponent plus c

table row cell 2 over 5 plus 4 plus 18 space plus c space end cell equals cell space 25 end cell row cell 22 2 over 5 space plus space c space end cell equals cell space 25 end cell row cell c space end cell equals cell fraction numerator space 13 space over denominator 5 end fraction end cell end table

bold f stretchy left parenthesis x stretchy right parenthesis bold space bold equals bold 2 over bold 5 bold italic x to the power of bold 5 over bold 2 end exponent bold space bold plus bold space bold 4 bold italic x to the power of bold 3 over bold 2 end exponent bold space bold plus bold space bold 18 bold italic x to the power of bold 1 over bold 2 end exponent bold space bold plus fraction numerator bold space bold 13 over denominator bold 5 end fraction

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Paul

Author: Paul

Expertise: Maths Content Creator (Previous)

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.

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