Further Applications of Integration (DP IB Maths: AI HL)

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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

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 all, or some, of the curve is below the x-axis?

  • Use the modulus function
    • The modulus is also called the absolute value (Abs)
    • Essentially the modulus function makes all function values positive
    • Graphically, this means any negative areas are reflected in the x-axis

5-4-3-ib-hl-ai-aa-extraaa-ai-run-fig1-neg-area

 

  • A GDC will recognise the modulus function
    • look for a key or on-screen icon that says 'Abs' (absolute 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
STEP 1
If a diagram is not given, use a GDC to draw the graph ofspace y equals f left parenthesis x right parenthesis
If not identifiable from the question, use the graph to find the limits a and b
 
STEP 2
Write down the definite integral needed to find the required area
Remember to include the modulus ( | ... | ) symbols around the function
Use the GDC to evaluate it 

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
    • You should use your GDC to do this 

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.

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

QYldKmqS_5-4-3-ib-hl-ai-aa-extraaa-ai-we2-soltn

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Area Between Curve & y-axis

What is meant by the area between a curve and the y-axis?

5-9-4-ib-hl-ai-aa-extraaa-fig1-area-yaxis

  • The area referred to is the region bounded by
    • the graph ofspace y equals f left parenthesis x right parenthesis
    • thespace y-axis
    • the horizontal linespace y equals a
    • the horizontal linespace y equals b
  • The exact area can be found by evaluating a definite integral

How do I find the area between a curve and the y-axis?

  • Use the formula

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

    • This is given in the formula booklet
    • The function is normally given in the formspace y equals f left parenthesis x right parenthesis
      • so will need rearranging into the formspace x equals g left parenthesis y right parenthesis
    • a and b may not be given directly as could involve the the x-axis (y equals 0) and/or a root ofspace x equals g left parenthesis y right parenthesis
      • use a GDC to plot the curve and find roots as necessary
STEP 1

If a diagram is not given, use a GDC to draw the graph ofspace y equals f left parenthesis x right parenthesis
(orspace x equals g left parenthesis y right parenthesis if already in that form)
If not identifiable from the question, use the graph to find the limits a and b

STEP 2

If needed, rearrangespace y equals f left parenthesis x right parenthesis into the formspace x equals g left parenthesis y right parenthesis

STEP 3
Write down the definite integral needed to find the required area
Use a GDC to evaluate it
A GDC is likely to require the function written with ‘x’ as the variable (not ‘y’)
Remember to include the modulus ( | … | ) symbols around the function
Modulus may be called ‘Absolute value (Abs)’ on some GDCs

  • In trickier problems some (or all) of the area may be 'negative'
    • this would be any area that is to the left of the y-axis (negative x values)
    • open vertical bar x close vertical bar makes such areas 'positive' by reflecting them in the y-axis
      • a GDC will apply open vertical bar x close vertical bar automatically as long as the modulus ( | ... | )symbols are included



Examiner Tip

  • If no diagram is provided, quickly sketch one so that you can see where the curve is to the left and right of the y - axis and split up your integrals accordingly
    • You should use your GDC to do this

Worked example

Find the area enclosed by the curve with equationspace y equals 2 plus square root of x plus 4 end root, the y-axis and the horizontal lines with equations y equals 3 and y equals 6.

5-4-3-ib-hl-ai-aa-extraaa-ai-we1-soltn

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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
      • using a GDC, one method is not particularly trickier than the other
  • The total area required could be the sum or difference of the area under the curve and the area under the 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 a diagram is not given, use a GDC to draw the graphs of the curve and line and identify the area to be found
 
STEP 2
Use a GDC to find the root(s) of the curve, the root of the line, and the x-coordinates of any intersections between the curve and the line.
 
STEP 3
Use the graph to determine whether areas will need adding or subtracting
Deduce the limits and thus the definite integral(s) to find the area(s) under the curve and the line
Use a GDC to calculate the area under the curve
integral subscript a superscript b open vertical bar y close vertical bar space straight d x
Remember to include the modulus (|...|) symbols around the function
Use a GDC to calculate the area under the line - this could be another definite integral orspace 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
 
STEP 4
Add or subtract areas accordingly to obtain a final answer

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, use your GDC to graph one and if you have time copy the sketch into your working

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.

Find the area of the region R.

5-4-3-ib-hl-ai-adapted-we3-soltn

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Definite Integrals

What is a definite integral?

  • This is known as the Fundamental Theorem of Calculus
  • a and b are called limits
    • a is the lower limit
    • b is the upper limit
  • space straight f left parenthesis x right parenthesisis the integrand
  • space straight F left parenthesis x right parenthesisis an antiderivative ofspace straight f left parenthesis x right parenthesis
  • The constant of integration (“+c”) is not needed in definite integration
    • “+c” would appear alongside both F(a) and F(b)
    • subtracting means the “+c”’s cancel

How do I find definite integrals analytically (manually)?

STEP 1
Give the integral a name to save having to rewrite the whole integral every time
If need be, rewrite the integral into an integrable form

space I equals integral subscript a superscript b straight f left parenthesis x right parenthesis space straight d x

STEP 2
Integrate without applying the limits; you will not need “+c
Notation: use square brackets [ ] with limits placed at the end bracket
 
STEP 3
Substitute the limits into the function and evaluate

Examiner Tip

  • If a question does not state that you can use your GDC then you must show all of your working clearly, however it is always good practice to check you answer by using your GDC if you have it in the exam

Worked example

a)
Show that

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

 5-4-3-ib-sl-aa-only-we1-soltn-a

b)
Use your GDC to evaluate

space integral subscript 0 superscript 1 3 e to the power of x squared sin space x end exponent space straight d x

giving your answer to three significant figures.

5-4-3-ib-sl-aa-only-we1-soltn-b

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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.