Calculating Areas (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Definite Integrals

What is a definite integral?

  • A definite integral is defined by the following formula

    • integral subscript a superscript b straight f to the power of apostrophe open parentheses x close parentheses space straight d x equals open square brackets straight f open parentheses x close parentheses close square brackets subscript a superscript b equals straight f open parentheses b close parentheses minus straight f open parentheses a close parentheses

      • i.e. integrate as usual to find straight f open parentheses x close parentheses

      • then substitute to find straight f open parentheses b close parentheses and straight f open parentheses a close parentheses

      • and subtract straight f open parentheses a close parentheses from straight f open parentheses b close parentheses to find the value of the definite integral

    • a and b are numbers and are called the integration limits

      • a is the lower limit

      • b is the upper limit

      • the integral is 'from a to b'

    • A constant of integration (“plus c”) is not needed with definite integrals

  • Note that the answer to a definite integral is a number

    • The answer to an indefinite integral is another function

Examiner Tips and Tricks

  • Be careful when substituting in to find straight f open parentheses b close parentheses minus straight f open parentheses a close parentheses

    • It's quite easy to make mistakes here

    • Especially when fractions and negative numbers are involved

  • Your calculator may be able to find the value of definite integrals

    • You can use this to check your work

  • Look out for phrases in exam questions like "Use algebraic integration" or "Using calculus"

    • These mean that full working out of the integral 'by hand' is required

    • A calculator answer without working would not score marks

Worked Example

Show that

integral subscript 2 superscript 4 3 x left parenthesis x squared minus 2 right parenthesis space straight d x equals 144


Start by expanding the brackets inside the integral

integral subscript 2 superscript 4 open parentheses 3 x cubed minus 6 x close parentheses space straight d x


Integrate as usual (here it's a 'powers of x' integration)

Write the answer in square brackets with the integration limits outside

table row cell integral subscript 2 superscript 4 open parentheses 3 x cubed minus 6 x close parentheses space straight d x end cell equals cell open square brackets 3 open parentheses fraction numerator x to the power of 3 plus 1 end exponent over denominator 3 plus 1 end fraction close parentheses minus 6 open parentheses fraction numerator x to the power of 1 plus 1 end exponent over denominator 1 plus 1 end fraction close parentheses close square brackets subscript 2 superscript 4 end cell row blank equals cell open square brackets 3 over 4 x to the power of 4 minus 3 x squared close square brackets subscript 2 superscript 4 end cell end table

Now substitute 4 into that function
And subtract from it the function with 2 substituted in

table row cell open square brackets 3 over 4 x to the power of 4 minus 3 x squared close square brackets subscript 2 superscript 4 end cell equals cell open parentheses 3 over 4 open parentheses 4 close parentheses to the power of 4 minus 3 open parentheses 4 close parentheses squared close parentheses minus open parentheses 3 over 4 open parentheses 2 close parentheses to the power of 4 minus 3 open parentheses 2 squared close parentheses close parentheses end cell row blank equals cell open parentheses 192 minus 48 close parentheses minus open parentheses 12 minus 12 close parentheses end cell row blank equals cell 144 minus 0 end cell row blank equals 144 end table


And that's the answer we were asked to show!

bold integral subscript bold 2 superscript bold 4 bold 3 bold italic x stretchy left parenthesis x squared minus 2 stretchy right parenthesis bold space bold d bold italic x bold equals bold 144

Calculating Areas with Definite Integrals

How can I calculate areas using definite integrals?

  • Region R in the following diagram is the 'area under a curve' between x equals a and x equals b

    • It is the region bounded by

      • the curve y equals straight f open parentheses x close parentheses

      • the bold italic x-axis

      • and the lines x equals a and and x equals b

Area under a curve found by integration
  • The exact area of a region like region R in the diagram can be found by evaluating the definite integral

    • Area equals integral subscript a superscript b straight f open parentheses x close parentheses space straight d x

      • i.e. definite integrals can be used as area calculators!

    • Note that for this to work, a must be 'on the left' and b must be 'on the right'

      • i.e. bold italic a bold less or equal than bold italic b

How do I form a definite integral to find the area under a curve?

  • The curve y equals f left parenthesis x right parenthesis and the x-axis should be obvious boundaries

    • but the trick can be identifying a and b

      • i.e. the lower and upper limits of the definite integral 

  • If a diagram is given in the question, this can help locate the limits

    • If a diagram is not given, then sketch one

  • The 'left' and 'right' boundaries may be vertical lines

    • In that case their equations give the integral limits

      • x equals a and x equals b

  • The bold italic y-axis may be one of the (vertical) boundaries

    • in that case one of the limits will be bold italic x bold equals bold 0

  • The 'left' and 'right' boundaries don't have to be vertical lines

    • One or both of them could be where y equals straight f open parentheses x close parentheses intercepts the bold italic x-axis

      • i.e., one of the roots of the equation straight f left parenthesis x right parenthesis equals 0

    • In this case solve the equationspace straight f left parenthesis x right parenthesis equals 0 to find the limit(s)

Examiner Tips and Tricks

  • Look out for questions that ask you to find an indefinite integral in one part

    • where 'plus c'  is needed in the answer

    • then in a later part use the same integral as a definite integral

      • where 'plus c' is not needed

  • Add information to any diagram provided in the question

    • axes intercepts

    • values of limits

    • mark and shade the area you’re trying to find

  • If no diagram is provided, sketch one!

Worked Example

The following diagram shows a part of the graph of the curve with equation  y equals 3 plus 2 x minus x squared.  The region labelled R is bounded by the curve, the positive y-axis, and the positive x-axis.

Graph of 3+2x-x^2


Find the exact area of region R.

Start by finding the integration limits.

The y-axis is the left boundary, so x equals 0 will be the lower integration limit

The right boundary is where the curve intercepts the x-axis
Solve the equation  y equals 0  to find its x-coordinate

table row cell 3 plus 2 x minus x squared end cell equals 0 row cell x squared minus 2 x minus 3 end cell equals 0 row cell open parentheses x plus 1 close parentheses open parentheses x minus 3 close parentheses end cell equals 0 end table


x equals negative 1 space space or space space x equals 3


We want the positive intercept, so the upper integral limit will be x equals 3

Put all that info into the definite integral

table attributes columnalign right center left columnspacing 0px end attributes row Area equals cell integral subscript 0 superscript 3 open parentheses 3 plus 2 x minus x squared close parentheses space straight d x end cell row blank equals cell open square brackets 3 x plus 2 open parentheses fraction numerator x to the power of 1 plus 1 end exponent over denominator 1 plus 1 end fraction close parentheses minus open parentheses fraction numerator x to the power of 2 plus 1 end exponent over denominator 2 plus 1 end fraction close parentheses close square brackets subscript 0 superscript 3 end cell row blank equals cell open square brackets 3 x plus x squared minus 1 third x cubed close square brackets subscript 0 superscript 3 end cell row blank equals cell open parentheses 3 open parentheses 3 close parentheses plus open parentheses 3 close parentheses squared minus 1 third open parentheses 3 close parentheses cubed close parentheses minus open parentheses 3 open parentheses 0 close parentheses plus open parentheses 0 close parentheses squared minus 1 third open parentheses 0 close parentheses cubed close parentheses end cell row blank equals cell open parentheses 9 plus 9 minus 9 close parentheses minus open parentheses 0 plus 0 minus 0 close parentheses end cell row blank equals cell 9 minus 0 end cell row blank equals 9 end table


bold 9 bold space bold units to the power of bold 2

Negative Integrals

What do we mean by a 'negative integral'

  • The answer to a definite integral is a number

    • This number can be positive or  negative (or zero!)

  • The area between a curve and the bold italic x-axis may lie fully or partially below the x-axis

    • This occurs when the functionspace straight f left parenthesis x right parenthesis takes negative values within the boundaries of the area

  • A definite integral used to find such an area

    • will be negative if the area is fully under thespace x-axis

    • possibly negative if the area is partially under thespace x-axis

      • even if positive, the integral will not calculate the correct area in this case

      • the 'negative areas' will subtract from the 'positive areas'

How do I find the area under a curve when the curve is fully under the x-axis?

 

Area found with integration, entirely under the x-axis
  • STEP 1
    Write the definite integral to find the area as usual

    • This may involve finding the lower and upper integration limits

  • STEP 2
    The answer to the definite integral will be negative

    • But the answer will have the same 'size' as the area

    • So just remove the minus sign to get the area

      • e.g.  if the value of the integral is negative 36 

      • then the area will be 36 (square units)

How do I find the area under a curve when the curve is partially under the x-axis?

 

Area found with integration, partially under the x-axis
  • STEP 1
    Split the area into parts

    • the area(s) that are above the x-axis

    • and the area(s) that are below the x-axis

  • STEP 2
    Write the definite integral for each part

    • This may involve finding the lower and upper integration limits for each part

      • e.g. solving straight f open parentheses x close parentheses equals 0 to find where y equals straight f open parentheses x close parentheses crosses the x-axis 

  • STEP 3
    Find the value of each definite integral separately

     

  • STEP 4
    Change the negative values to positive

    • Then find the total area by summing the 'positive versions' of each integral

Examiner Tips and Tricks

  • If no diagram is provided, sketch one

    • This lets you see where the curve is above and below the x-axis

    • Then you can split up your integrals accordingly

Worked Example

The diagram below shows the graph of