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.

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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.
• 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

… … 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.

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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:

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

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

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

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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.

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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.
• 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

• 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.

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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:

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

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

… … … $\text{Area } \approx 8.15$ 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 $f\big(x + h\big)$

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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

• 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

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

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