Integrating Basic Functions (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Integrating Powers of x

How do I integrate powers of x?

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

    • space integral x to the power of n 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 c is the constant of integration)

      • This is valid for any value of n except n equals negative 1

      • So you cannot integrate  integral x to the power of negative 1 end exponent space straight d x equals integral 1 over x space straight d x this way

  • If x to the power of n is multiplied by a constant a then

    • space integral a x to the power of n space straight d x equals a integral x to the power of n 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

      • This also is not valid for n equals negative 1

  • These formulae are not on the exam formula sheet, so you need to remember them

  • 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

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

    • e.g. integral cube root of x space straight d x

      • rewrite cube root of x as x to the power of 1 third end exponent

      • then integrate

  • Fractions with x in the denominator will need to be rewritten as negative powers

    • e.g. integral 1 over x squared space straight d x

      • rewrite 1 over x squared as x to the power of negative 2 end exponent

      • then integrate

How do I integrate sums and differences of powers of x?

  • The formulae can be used to integrate sums or differences of powers ofspace x

    • Just integrate term by term

      • e.g. integral open parentheses 8 x cubed minus 2 x plus 4 close parentheses blank straight d x

      • size 16px equals fraction numerator size 16px 8 size 16px x to the power of size 16px 3 size 16px plus size 16px 1 end exponent over denominator size 16px 3 size 16px plus size 16px 1 end fraction size 16px minus fraction numerator size 16px 2 size 16px x to the power of size 16px 1 size 16px plus size 16px 1 end exponent over denominator size 16px 1 size 16px plus size 16px 1 end fraction size 16px plus size 16px 4 size 16px x size 16px plus size 16px c

      • equals 2 x to the power of 4 minus x squared plus 4 x plus c
                 

  • Products and quotients cannot be integrated this way

    • You need to expand and/or simplify first

      • e.g. integral 8 x squared open parentheses 2 x minus 3 close parentheses space straight d x

      • expand 8 x squared open parentheses 2 x minus 3 close parentheses as 16 x cubed minus 24 x squared

      • then integrate term by term

      • you cannot just multiply the integrals of 8 x squared and 2 x minus 3 together

What might I be asked to do once I’ve integrated?

  • You may be given the derivative of a function and asked to find the function

    • Integration and differentiation are inverse operations so

      • integral straight f to the power of apostrophe open parentheses x close parentheses space straight d x equals straight f open parentheses x close parentheses plus c

      • integral fraction numerator straight d y over denominator straight d x end fraction space straight d x equals y plus c

  • With more information the constant of integration,space c, can be found

  • The area under a curve can also be found using integration

Examiner Tips and Tricks

  • Remember the basic pattern of integrating powers of x

    • 'Raise the power by one and divide by the new power'

    • Lots of practice will improve your speed and accuracy

  • It's easy to check your answer when integrating

    • Just differentiate your answer

    • It should turn back into the function you were integrating in the first place

Worked Example

Given that  fraction numerator straight d y over denominator straight d x end fraction equals 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.

Start by rewriting fraction numerator straight d y over denominator straight d x end fractionentirely in powers of x

By laws of indices  fraction numerator 1 over denominator square root of x end fraction equals x to the power of negative 1 half end exponent

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


Remember  integral fraction numerator straight d y over denominator straight d x end fraction space straight d x equals y plus c

We can integrate term by term using  space integral a x to the power of n 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

table row y equals cell integral open parentheses 2 x squared plus 3 minus x to the power of negative 1 half end exponent close parentheses space straight d x end cell row blank equals cell 2 open parentheses fraction numerator x to the power of 2 plus 1 end exponent over denominator 2 plus 1 end fraction close parentheses plus 3 x minus fraction numerator x to the power of negative 1 half plus 1 end exponent over denominator negative 1 half plus 1 end fraction plus c end cell row blank equals cell 2 open parentheses x cubed over 3 close parentheses plus 3 x minus fraction numerator x to the power of 1 half end exponent over denominator 1 half end fraction plus c end cell row blank equals cell 2 over 3 x cubed plus 3 x minus 2 x to the power of 1 half end exponent plus c end cell end table


We can't find the value of c without further info, so that's the answer to the question

It's 'nice' to turn x to the power of 1 half end exponent back intosquare root of x for the final answer, but you would also get the marks without doing that


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

Integrating Trig Functions

How do I integrate sin and cos?

  • You can integrate sin x and cos x by using the formulae

    • bold integral bold sin bold space bold italic x bold space bold d bold italic x bold equals bold minus bold cos bold space bold italic x bold plus bold italic c

    • bold integral bold cos bold space bold italic x bold space bold d bold italic x bold equals bold sin bold space bold italic x bold plus bold italic c

      • c is the constant of integration

  • If x is multiplied by a constant a then

    • bold integral bold sin bold space bold italic a bold italic x bold space bold d bold italic x bold equals bold minus bold 1 over bold italic a bold cos bold space bold italic a bold italic x bold plus bold italic c

    • bold integral bold cos bold space bold italic a bold italic x bold space bold d bold italic x bold equals bold 1 over bold italic a bold sin bold space bold italic a bold italic x bold plus bold italic c

  • None of these formulae are on the exam formula sheet, so you need to remember them

  • For calculus with trigonometric functions angles must be measured in radians

    • Make sure you know how to change the angle mode on your calculator

Examiner Tips and Tricks

  • Remember to include 'c', the constant of integration, for any indefinite integrals

  • As soon as you see a question involving integration and trigonometry

    • put your calculator into radians mode 

Worked Example

Given that  straight f to the power of apostrophe open parentheses x close parentheses equals 4 cos 3 x minus 1 half sin 2 x, find an expression for straight f open parentheses x close parentheses in terms of x.

Remember that  integral straight f to the power of apostrophe open parentheses x close parentheses space straight d x italic equals straight f stretchy left parenthesis x stretchy right parenthesis italic plus c

We can integrate term by term using the integration formulae for sin a x and cos a x

table attributes columnalign right center left columnspacing 0px end attributes row cell straight f open parentheses x close parentheses end cell equals cell integral open parentheses 4 cos 3 x minus 1 half sin 2 x close parentheses space straight d x end cell row blank equals cell 4 open parentheses 1 third sin 3 x close parentheses minus 1 half open parentheses negative 1 half cos 2 x close parentheses plus c end cell row blank equals cell 4 over 3 sin 3 x plus 1 fourth cos 2 x plus c end cell end table


We can't find the value of c without further info, so that's the answer to the question


bold f stretchy left parenthesis x stretchy right parenthesis bold equals bold 4 over bold 3 bold sin bold 3 bold italic x bold plus bold 1 over bold 4 bold cos bold 2 bold italic x bold plus bold italic c

Integrating e^x

How do I integrate exponentials?

  • straight e to the power of x can be integrated using the formula

    • bold integral bold space bold e to the power of bold italic x bold space bold d bold italic x bold equals bold space bold e to the power of bold italic x bold plus bold italic c

      • c is the constant of integration

  • If x is multiplied by a constant a then

    •  bold integral bold e to the power of bold italic a bold italic x end exponent bold space bold d bold italic x bold equals bold 1 over bold italic a bold e to the power of bold italic a bold italic x end exponent bold plus bold italic c

  • These formulae are not on the exam formula sheet, so you need to remember them

Examiner Tips and Tricks

  • Because 'straight e to the power of x is its own integral' it is quite easy to integrate exponentials

    • Just be careful dealing with the constant in straight e to the power of a x end exponent 

Worked Example

Given that  straight f to the power of apostrophe open parentheses x close parentheses equals fraction numerator straight e to the power of 2 x end exponent minus straight e to the power of negative 3 x end exponent over denominator 2 end fraction, find an expression for straight f open parentheses x close parentheses in terms of x.

Remember that  integral straight f to the power of apostrophe open parentheses x close parentheses space straight d x italic equals straight f stretchy left parenthesis x stretchy right parenthesis italic plus c

We can integrate term by term using the integration formula for straight e to the power of a x end exponent

You might find it easier to split the fraction into two separate fractions first

table row cell straight f open parentheses x close parentheses end cell equals cell integral open parentheses 1 half straight e to the power of 2 x end exponent minus 1 half straight e to the power of negative 3 x end exponent close parentheses space straight d x end cell row blank equals cell 1 half open parentheses 1 half straight e to the power of 2 x end exponent close parentheses minus 1 half open parentheses fraction numerator 1 over denominator negative 3 end fraction straight e to the power of negative 3 x end exponent close parentheses plus c end cell row blank equals cell 1 fourth straight e to the power of 2 x end exponent plus 1 over 6 straight e to the power of negative 3 x end exponent plus c end cell end table


We can't find the value of c without further info, so that's the answer to the question


bold f stretchy left parenthesis x stretchy right parenthesis bold equals bold 1 over bold 4 bold e to the power of bold 2 bold x end exponent bold plus bold 1 over bold 6 bold e to the power of bold minus bold 3 bold x end exponent bold plus bold italic 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.