09Dapproxareas

= Approximate Areas enclosed by Functions =



The challenge is to find the __**area**__ between a curve and the x-axis and between any two given x-values.

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.

There are a number of different methods of getting an approximate area. We will study two.


 * 1) The left rectangle method
 * 2) The right rectangle method


 * The Left Rectangle Method **




 * We divide the area into vertical rectangles with equal width.


 * Width = h


 * If asked for n rectangles, the width will be:

math . \qquad h = \dfrac{x_{max} - x_{min} }{n} \qquad. math


 * 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(a)
 * and the ** Area ** of the first rectangle is A = hf(a)


 * In the same way, the ** heights ** of the other rectangles are:
 * f(b), f(c), f(d) ** {Notice we don't use the last point, e} **


 * where b = a + h, c = a + 2h, d = a + 3h


 * Therefore, the ** areas ** of the other rectangles are:
 * hf(b), hf(c), hf(d)


 * Total area of the rectangles is therefore:
 * A = hf(a) + hf(b) + hf(c) + hf(d)
 * use as many terms as needed to cover the domain [a, e]
 * don't include f(e)


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


 * Notes: **
 * The error (the difference between the approximation and the exact area) 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 __ **lower** __ than the exact area.


 * Example 1 **


 * Use the left rectangle method (h = 0.5) to find the area between y = 0.2x 2 + 3 and the x-axis, between x = 1 and x = 3.

... ... ... x = 1.0, 1.5, 2.0, 2.5 ... ** {Don't use x = 3.0} **
 * ** This creates 4 rectangles with left corners at **


 * ** So the __height__ of each rectangle is: **

... ... ... f(1.0), f(1.5), f(2.0), f(2.5)


 * ** Now use y = 0.2x 2 + 3 **

... ... ... f(1) = 3.2, f(1.5) = 3.45, f(2.0) = 3.8, f(2.5) = 4.25


 * ** Hence the __approximate area__ is: **

... ... ... Area = 0.5 [ 3.2 + 3.45 + 3.8 + 4.25 ] ... ... ... Area = 0.5(14.7) ... ... ... Area = 7.35 square units

... ... ... Note that this is __below__ the exact area (7.733) because this is an increasing function.

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 0.383/7.733*100 = 4.96%

... ...


 * Notice that every approximate area listed is lower than the exact area (7.733) because this is an increasing function.


 * 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:

math . \qquad h = \dfrac{x_{max} - x_{min} }{n} \qquad. math


 * 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(b)
 * and the ** Area ** of the first rectangle is A = hf(b)


 * In the same way, the ** heights ** of the other rectangles are:
 * f(c), f(d), f(e) ... ** {Notice we don't use the first point, a} **


 * where b = a + h, c = a + 2h, d = a + 3h, e = a + 4h


 * Therefore, the ** areas ** of the other rectangles are:
 * hf(c), hf(d), hf(e)


 * Total area of the rectangles is therefore:
 * A = hf(b) + hf(c) + hf(d) + hf(e)
 * use as many terms as needed to cover the domain [a, e]
 * don't include f(a)


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


 * Notes: **
 * The error (the difference between the approximation and the exact area) 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 __ **larger** __ than the exact area.


 * Example 2 **


 * Use the right rectangle method (h = 0.5) to find the area between y = 0.2x 2 + 3 and the x-axis, between x = 1 and x = 3.


 * ** This creates 4 rectangles with right corners at **

... ... ... x = 1.5, 2.0, 2.5, 3.0 ... **{Don't use x = 1.0}**


 * ** So the __height__ of each rectangle is: **

... ... ... f(1.5), f(2.0), f(2.5), f(3.0)


 * ** Now use y = 0.2x 2 + 3 **

... ... ... f(1.5) = 3.45, f(2.0) = 3.8, f(2.5) = 4.25, f(3.0) = 4.8


 * ** Hence the __approximate area__ is: **

... ... ... Area = 0.5 [ 3.45 + 3.8 + 4.25 + 4.8 ] ... ... ... Area = 0.5(16.3) ... ... ... Area = 8.15 square units


 * Notice that this value is __ above __ the exact area (7.733) because this is an increasing function

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 0.417/7.733*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.


 * Average of Left and Right Rectangle Results **


 * 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


 * In the example above:
 * Left Rectangle result = 7.35 square units
 * Right Rectangle result = 8.15 square units


 * Average = 7.75 square units
 * Exact Area = 7.733 square units


 * Error = 0.017 square units
 * %Error = 0.22%
 * Compare this to errors of approx 5% using either rectangle method for this example


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


 * 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.


 * Trapezoidal Method **


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


 * 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 corner heights defined by f(x) and f(x + h)

math . \qquad \qquad \text{Area(trapezium) } = \dfrac{h}{2} \Big( f(x) + f(x + h) \Big) \qquad. math
 * The area of one trapezium is therefore given by


 * If we have 4 trapeziums with corners at a, b, c, d, e, then the areas are:

math . \qquad \qquad \text{Area}(ab) = \dfrac{h}{2} \Big( f(a) + f(b) \Big) \qquad. \\ .\\ . \qquad \qquad \text{Area}(bc) = \dfrac{h}{2} \Big( f(b) + f(c) \Big) \\.\\ . \qquad \qquad \text{Area}(cd) = \dfrac{h}{2} \Big( f(c) + f(d) \Big) \\.\\ . \qquad \qquad \text{Area}(de) = \dfrac{h}{2} \Big( f(d) + f(e) \Big) \\.\\ math


 * Adding these together, gives us the rule:
 * use as many middle terms as needed to cover the domain [a, e]


 * Example 3 **


 * Use the Trapezoidal Method (h = 0.5) to find the area between y = 0.2x 2 + 3 and the x-axis, between x = 1 and x = 3.


 * ** This creates 4 trapeziums with corners at **

... ... ... x = 1.0, 1.5, 2.0, 2.5, 3.0


 * ** So the __height__ of each point is: **

... ... ... f(1.0), f(1.5), f(2.0), f(2.5), f(3.0)
 * ** Now use y = 0.2x 2 + 3 **

... ... ... f(1.0) = 3.2, f(1.5) = 3.45, f(2.0) = 3.8, f(2.5) = 4.25, f(3.0) = 4.8


 * ** Hence the __approximate area__ is: **

... ... ... Area = 0.25 [ 3.2 + 2(3.45) + 2(3.8) + 2(4.25) + 4.8 ] ... ... ... Area = 0.25(31) ... ... ... Area = 7.75 square units

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