Negative Integrals (DP IB Applications & Interpretation (AI)): Revision Note

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Negative Integrals

What do we mean by 'negative integrals'?

  • The area between a curve and the x-axis 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

      • In neither case will the integral give you the (positive) answer you are looking for for the area

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

Graph of a curve y=f(x) with a shaded area R under the x-axis that is enclosed by the curve, the x-axis, and the vertical lines x=a and x=b. A label says 'Area R entirely under x-axis'.
  • 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 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

Graph showing function y=f(x) and its modulus y=|f(x)|. Negative areas below x-axis are reflected above, making all areas positive.

 

  • 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

Examiner Tips and Tricks

This modulus version of the integral area formula is given in the formula booklet.

Note that it will also work to find an area that is fully under the x-axis.

  • 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 the integral

Examiner Tips and Tricks

If a diagram is not provided, sketching one can really help in this sort of question. Your GDC can help with this.

  • You can also apply ‘open vertical bar y close vertical bar’ manually, by splitting the integral into positive and negative parts

    • Write an integral and evaluate each part separately

    • Change any negative values found to positive

    • Then add all the positive values together to give the total area

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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Paul

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