08 1approxareas

Approximate Areas enclosed by Functions

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The challenge is to find the area enclosed by a curve and the x-axis, between any two given x-values.

08.1area1.gif

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The exact value of the area can often be found using calculus but there are times when a numerical method must be used to get an approximate value for the area.
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There are a number of different numerical methods of getting an approximate area. We will study two:

  • The Left Rectangle Method
  • The Right Rectangle Method

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The Left Rectangle Method

  • We divide the area into vertical rectangles with equal width.
  • Width of each rectangle = h
  • If asked for n rectangles, the width will be:

… … … … $h = \dfrac{x_{max} - x_{min} }{n}$

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  • The top left corner of each rectangle sits on the curve.
08.1area2.gif
  • The left side of the first rectangle is at $x = a$
    • so the height of the first rectangle is $f \big( a \big)$
    • and the Area of the first rectangle is $A = hf \big( a \big)$

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  • In the same way, the heights of the other rectangles are:
    • $f\big( b \big), \; f\big( c \big), \; f\big(d\big)$
      • {Notice we don't use the last point, e}
      • where $b = a + h, \; c = a + 2h, \; d = a + 3h$

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  • Therefore, the areas of the other rectangles are:
    • $hf\big(b\big), \; hf\big(c\big), \; hf\big(d\big)$

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  • Total area of the rectangles is therefore:

… … … … $A = hf\big(a\big) + hf\big(b\big) + hf\big(c\big) + hf\big(d\big)$

… … … … $A = h \Bigg( f\big(a\big) + f\big(b\big) + f\big(c\big) + f\big(d\big) \Bigg)$
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Rule for Left Rectangle Method

08.1ruleleft.gif

… … use as many terms as needed to cover the domain $[a, \; e]$
… … don't include $f\big(e\big)$
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This is an approximation for the area between the curve and the x-axis.

08.1area2.gif

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Notes:

  • The error is the difference between the approximation and the exact area
    • The error can be seen in the diagram as the purple shaded areas not covered by the rectangles.
  • If you use more and thinner rectangles, the approximation will be more accurate
    • but you have to do more calculations
  • In an increasing function,
    • the left rectangle method will always give an approximation less than the exact area.

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Example 1

Use the left rectangle method $(h = 0.5$) to find

  • the area enclosed by $f\big(x\big) = 0.2x^2 + 3$ and the x-axis,
  • between $x = 1 \text{ and } x = 3$.

Solution

… … This creates 4 rectangles with left corners at

… … … $x = 1.0,\; 1.5,\; 2.0,\; 2.5$ … {Don't use $x = 3.0$}
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… … So the height of each rectangle is:

… … … $f\big(1.0\big),\; f\big(1.5\big),\; f\big(2.0\big),\; f\big(2.5\big)$
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… … Now use $f\big(x\big) = 0.2x^2 + 3$

… … … $f\big(1\big) = 3.2,\; f\big(1.5\big) = 3.45,\; f\big(2.0\big) = 3.8,\; f\big(2.5\big) = 4.25$
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… … Hence the approximate area is:

08.1ruleleft.gif

… … … $\text{Area } \approx 0.5 \big( 3.2 + 3.45 + 3.8 + 4.25 \big)$

… … … $\text{Area } \approx 0.5\big(14.7\big)$

… … … $\text{Area } \approx 7.35$ square units

08.1calcleft.gif

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… … Note: this is below the exact area ($7.733$ square units) because $f\big(x\big)$ is an increasing function.
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NOTE: (for this example)

  • This sort of analysis is not part of the course:
  • For the above example, the exact area is $7.733$ square units. (found using calculus)
  • The error is $7.733 – 7.35 = 0.383$ square units.
  • The percentage error is $\dfrac{0.383}{7.733} \times 100 = 4.96\%$

Note: If we increase the number of columns, we can reduce the error.

08.1tableleft.gif

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  • Every approximate area listed in the table is lower than the exact area ($7.733$ square units) because this is an increasing function.

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The Right Rectangle Method

  • We divide the area into vertical rectangles with equal width.
  • Width = h
  • If asked for n rectangles, the width will be:

… … … … $h = \dfrac{x_{max} - x_{min} }{n}$

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  • The top right corner of each rectangle sits on the curve.
08.1area3.gif
  • The right side of the first rectangle is at $x = b$
    • so the height of the first rectangle is $f\big(b\big)$
    • and the Area of the first rectangle is $A = hf\big(b\big)$

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  • In the same way, the heights of the other rectangles are:
    • $f\big(c\big),\; f\big(d\big),\; f\big(e\big)$
      • {don't use the first point, a}
      • where $b = a + h, \; c = a + 2h,\; d = a + 3h,\; e = a + 4h$

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  • Therefore, the areas of the other rectangles are:
    • $hf\big(c\big),\; hf\big(d\big),\; hf\big(e\big)$

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  • Total area of the rectangles is therefore:

… … … …$A = hf\big(b\big) + hf\big(c\big) + hf\big(d\big) + hf\big(e\big)$

… … … … $A = h\Bigg( f\big(b\big) + f\big(c\big) + f\big(d\big) + f\big(e\big) \Bigg)$
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Rule for Right Rectangle Method

08.1ruleright.gif
  • use as many terms as needed to cover the domain $\big[ a,\; e \big]$
    • don't include $f\big(a\big)$

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This is an approximation for the area between the curve and the x-axis.

08.1area3.gif

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Notes:

  • The error can be seen in the diagram as the parts of the rectangles above the curve.
  • If you use more and thinner rectangles, the approximation will be more accurate
    • but you have to do more calculations
  • In an increasing function,
    • the right rectangle method will always give an approximation greater than the exact area.

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Example 2

Use the right rectangle method ($h = 0.5$) to find

  • the area enclosed by $f\big(x\big) = 0.2x^2 + 3$ and the x-axis,
  • between $x = 1 \text{ and } x = 3$.

Solution

… … This creates 4 rectangles with right corners at

… … … $x = 1.5,\; 2.0,\; 2.5,\; 3.0$ … {Don't use $x = 1.0$}
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… … So the height of each rectangle is:

… … … $f\big(1.5\big),\; f\big(2.0\big),\; f\big(2.5\big),\; f\big(3.0\big)$
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… … Now use $f\big(x\big) = 0.2x^2 + 3$

… … … $f\big(1.5\big) = 3.45,\; f\big(2.0\big) = 3.8,\; f\big(2.5\big) = 4.25,\; f\big(3.0\big) = 4.8$
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… … Hence the approximate area is:

08.1ruleright.gif

… … … $\text{Area } \approx 0.5 \big(3.45 + 3.8 + 4.25 + 4.8\big)$

… … … $\text{Area } \approx 0.5\big(16.3\big)$

… … … $\text{Area } \approx 8.15$ square units
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… … Note: this value is above the exact area ($7.733$ square units) because this is an increasing function
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NOTE: (for this example)

  • This sort of analysis is not part of the course:
  • For the above example, the exact area is $7.733$ square units.
  • The error is $8.15 – 7.733 = 0.417$ square units.
  • The percentage error is $\dfrac{0.417}{7.733} \times 100 = 5.39\%$
  • Compare this to an error of $4.96\%$ obtained using the left rectangle method on the same problem.
  • Both methods are similar in accuracy.

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Average of Left and Right Rectangle Results

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  • The accuracy of the final result can be improved by averaging the left rectangle and the right rectangle results.
  • This is called the Trapezoidal Method
    • because it is the equivalent of drawing a trapezium instead of each rectangle

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In the example above:

  • Left Rectangle result = $7.35$ square units
  • Right Rectangle result = $8.15$ square units

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  • Average = $7.75$ square units
  • Exact Area = $7.733$ square units

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  • Error = $0.017$ square units
  • %Error = $0.22\%$

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  • Compare this to errors of approx $5\%$ using either rectangle method for this example

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  • The Trapezoidal Method is significantly more accurate than the Left and Right Rectangle Methods

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  • Other methods exist which also give more accurate approximations than the left and right rectangle method, but they are not a part of this course.

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Trapezoidal Method

  • A more formal investigation of The Trapezoidal Method is covered in the Year 11 Maths Methods Course
    • but not in the Year 12 Methods Course

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  • Rather than calculating and then averaging the results from the Left Rectangle and the Right Rectangle Methods,
    • We can draw a trapezium in place of each rectangle
    • each trapezium having a width of h
    • and having top left corner height defined by $f\big(x\big)$
    • and top right corner height defined by $f\big(x + h\big)$
08.1area4.gif
08.1area5.gif

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  • The area of one trapezium is therefore given by

… … $\text{Area(trapezium) } = \dfrac{h}{2} \Big( f(x) + f(x + h) \Big)$
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  • If we have 4 trapeziums with corners at a, b, c, d, e, then the areas are:

… … $\text{Area}(ab) = \dfrac{h}{2} \Big( f(a) + f(b) \Big)$

… … $\text{Area}(bc) = \dfrac{h}{2} \Big( f(b) + f(c) \Big)$

… … $\text{Area}(cd) = \dfrac{h}{2} \Big( f(c) + f(d) \Big)$

… … $\text{Area}(de) = \dfrac{h}{2} \Big( f(d) + f(e) \Big)$
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  • Adding these together, gives us the rule:

… … $\text{Area } = \dfrac{h}{2} \Bigg( f\big(a\big) + 2f\big(b\big) +2f\big(c\big) +2f\big(d\big) +f\big(e\big)\Bigg)$

Rule for Trapezoidal Method

08.1ruletrapezoidal.gif
  • use as many middle terms as needed to cover the domain $\big[a,\; e\big]$

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Example 3

Use the Trapezoidal Method ($h = 0.5$) to find

  • the area enclosed by $f\big(x\big) = 0.2x^2 + 3$ and the x-axis,
  • between $x = 1 \text{ and } x = 3$.

Solution

… … This creates 4 trapeziums with corners at

… … … $x = 1.0, \; 1.5, \; 2.0, \; 2.5, \; 3.0$
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… … So the height of each point is:

… … … $f\big(1.0\big),\; f\big(1.5\big),\; f\big(2.0\big),\; f\big(2.5\big),\; f\big(3.0\big)$
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… … Now use $f\big(x\big) = 0.2x^2 + 3$

… … … $f\big(1.0\big) = 3.2,\; f\big(1.5\big) = 3.45,\; f\big(2.0\big) = 3.8,\; f\big(2.5\big) = 4.25,\; f\big(3.0\big) = 4.8$
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… … Hence the approximate area is:

… … … $\text{Area } \approx 0.25 \big( 3.2 + 2(3.45) + 2(3.8) + 2(4.25) + 4.8\big)$

… … … $\text{Area } \approx 0.25\big(31\big)$

… … … $\text{Area } \approx 7.75$ square units

08.1calctrap.gif

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… … As shown above, this value has an error of $0.22\%$

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