Further Applications of Integration (DP IB Maths: AA SL)

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Paul

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Negative Integrals

  • The area under a curve may appear fully or partially under the x-axis
    • This occurs when the functionspace f left parenthesis x right parenthesis takes negative values within the boundaries of the area
  • The definite integrals used to find such areas
    • will be negative if the area is fully under thespace x-axis
    • possibly negative if the area is partially under thespace x-axis
      • this occurs if the negative area(s) is/are greater than the positive area(s), their sum will be negative
  • When using a GDC use the modulus (absolute value) function so that all definite integrals have a positive value

space A equals integral subscript a superscript b open vertical bar y close vertical bar space straight d x 

    • This is given in the formula booklet

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

 ib-aa-sl-5-4-4-negative-areas-diagram

STEP 1
Write the expression for the definite integral to find the area as usual
This may involve finding the lower and upper limits from a graph sketch or GDC and f(x) may need to be rewritten in an integrable form
 
STEP 2
The answer to the definite integral will be negative
Area must always be positive so take the modulus (absolute value) of it
e.g.  Ifspace I equals negative 36 then the area would be 36 (square units)

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

 ib-aa-sl-5-4-4-negative-areas-diagram2

  • For questions that allow the use of a GDC you can still use

space A equals integral subscript a superscript c open vertical bar f left parenthesis x right parenthesis close vertical bar space straight d x

  • To find the area analytically (manually) use the following method
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 expression for the definite integral for each part (give each part a name, I1, I2, etc)
This may involve finding the lower and upper limits of each part from a graph sketch or a GDC, finding the roots of the function (i.e. wherespace f left parenthesis x right parenthesis equals 0) and rewritingspace f left parenthesis x right parenthesis in an integrable form
 
STEP 3
Find the value of each definite integral separately
 
STEP 4
Find the area by summing the modulus (absolute values) of each integral
(Mathematically this would be writtenspace A equals open vertical bar I subscript 1 close vertical bar plus open vertical bar I subscript 2 close vertical bar plus open vertical bar I subscript 3 close vertical bar plus...)         

Examiner Tip

  • If no diagram is provided, quickly sketch one so that you can see where the curve is above and below the x - axis and split up your integrals accordingly

Worked example

The diagram below shows the graph ofspace y equals f left parenthesis x right parenthesis wherespace f left parenthesis x right parenthesis equals left parenthesis x plus 4 right parenthesis left parenthesis x minus 1 right parenthesis left parenthesis x minus 5 right parenthesis.

 5-4-4-ib-sl-aa-only-we1-qu-img

The regionspace R subscript 1 is bounded by the curvespace y equals f left parenthesis x right parenthesis, the x-axis and the y-axis.

The regionspace R subscript 2 is bounded by the curvespace y equals f left parenthesis x right parenthesis, the x-axis and the linespace x equals 3.

a)
Determine the coordinates of the point labelledspace P.

5-4-4-ib-sl-aa-only-we1-soltn-a

b)
i)
Find a definite integral that would help find the area of the shaded regionspace R subscript 2 and briefly explain why this would not give the area of the regionspace R subscript 2.
ii)
Find the exact area of the shaded regionspace R subscript 2.

5-4-4-ib-sl-aa-only-we1-soltn-b

c)
Find the exact total area of the shaded regions,space R subscript 1 and R subscript 2.

5-4-4-ib-sl-aa-only-we1-soltn-c

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Area Between a Curve and a Line

  • Areas whose boundaries include a curve and a (non-vertical) straight line can be found using integration
    • For an area under a curve a definite integral will be needed
    • For an area under a line the shape formed will be a trapezium or triangle
      • basic area formulae can be used rather than a definite integral
      • (although a definite integral would still work)
  • The area required could be the sum or difference of areas under the curve and line

5-4-4-ib-sl-aa-only-curv-line-part-1

5-4-4-ib-sl-aa-only-curv-line-part-2

How do I find the area between a curve and a line?

STEP 1
If not given, sketch the graphs of the curve and line on the same diagram
Use a GDC to help with this step
 
STEP 2
Find the intersections of the curve and the line
If no diagram is given this will help identify the area(s) to be found
 
STEP 3
Determine whether the area required is the sum or difference of the area under the curve and the area under the line
Calculate the area under a curve using a integral of the form
integral subscript a superscript b y space straight d x
Calculate the area under a line using eitherspace A equals 1 half b h for a triangle orspace A equals 1 half h left parenthesis a plus b right parenthesis for a trapezium (y-coordinates will be needed)
 
STEP 4
Evaluate the definite integrals and find their sum or difference as necessary to obtain the area required

Examiner Tip

  • Add information to any diagram provided
  • Add axes intercepts, as well as intercepts between lines and curves
  • Mark and shade the area you’re trying to find
  • If no diagram is provided, sketch one!

Worked example

The regionspace R is bounded by the curve with equation y equals 10 x minus x squared minus 16 and the line with equationspace y equals 8 minus x.

space R lies entirely in the first quadrant.

a)
Using your GDC, or otherwise, sketch the graphs of the curve and the line on the same diagram.
Identify and label the regionspace R on your sketch and use your GDC to find thespace x-coordinates of the points of intersection between the curve and the line.

5-4-4-ib-sl-aa-only-we2-soltn-a

b)
i)
Write down an integral that would find the area of the regionspace R.
ii)
Find the area of the regionspace R.

5-4-4-ib-sl-aa-only-we2-soltn-b

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Area Between 2 Curves

  • Areas whose boundaries include two curves can be found by integration
    • The area between two curves will be the difference of the areas under the two curves
      • both areas will require a definite integral
    • Finding points of intersection may involve a more awkward equation than solving for a curve and a line

Notes curv_hard

How do I find the area between two curves?

STEP 1
If not given, sketch the graphs of both curves on the same diagram
Use a GDC to help with this step
 
STEP 2
Find the intersections of the two curves
If no diagram is given this will help identify the area(s) to be found
 
STEP 3
For each area (there may only be one) determine which curve is the ‘upper’ boundary
For each area, write a definite integral of the form
space integral subscript a superscript b left parenthesis y subscript 1 minus y subscript 2 right parenthesis space straight d x
wherespace y subscript 1 is the function for the ‘upper’ boundary andspace y subscript 2 is the function for the ‘lower’ boundary
Be careful when there is more than one region – the ‘upper’ and ‘lower’ boundaries will swap
 
STEP 4
Evaluate the definite integrals and sum them up to find the total area
(Step 3 means no definite integral will have a negative value)

Examiner Tip

  • If no diagram is provided sketch one, even if the curves are not accurate
  • Add information to any given diagram as you work through a question
  • Maximise use of your GDC to save time and maintain accuracy:
    • Use it to sketch the graphs and help you visualise the problem
    • Use it to find definite integrals

Worked example

The diagram below shows the curves with equationsspace y equals straight f left parenthesis x right parenthesis and space y equals g left parenthesis x right parenthesis where

space straight f left parenthesis x right parenthesis equals left parenthesis x minus 2 right parenthesis left parenthesis x minus 3 right parenthesis squared

space straight g left parenthesis x right parenthesis equals x squared minus 5 x plus 6  

Find the area of the shaded region.

5-4-4-ib-sl-aa-only-we3-qu-img

5-4-4-ib-sl-aa-only-we3-soltn

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Paul

Author: Paul

Expertise: Maths

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 – one of the many reasons he is excited to be a member of the SME team.