Approximating areas with Rectangles

# Approximating areas with Rectangles - ment the...

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More on Approximating area with rectangles. consider the function: f ( x ) = 16 ° x 2 ; suppose we want to approximate the area under the curve with rectangles on the interval [0 ; 4] with a partition of width 0 : 5 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 2 4 6 8 10 12 14 16 x y The area is approximated by using the y ° coordinate of the left endpoint of each subinterval. That is: 1 2 8 X k =1 " 16 ° ° k ° 1 2 ± 2 # = 1 2 h 16 ° (0) 2 i + 1 2 h 16 ° (0 : 5) 2 i + 1 2 h 16 ° (1) 2 i + 1 2 h 16 ° (1 : 5) 2 i + 1 2 h 16 ° (2) 2 i + 1 2 h 16 ° (2 : 5) 2 i + 1 2 h 16 ° (3) 2 i + 1 2 h 16 ° (3 : 5) 2 i = 93 2 = 46 : 5 1

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Now, if we explore the approximation with rectangles whose height is given by the right endpoint of the subinterval: 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 2 4 6 8 10 12 14 16 x y We notice that the formulation should be similar to the one used for the left endpoint approach to the approximation. 1 2 7 X k =1 " 16 ° ° k 2 ± 2 # = 1 2 h 16 ° (0 : 5) 2 i + 1 2 h 16 ° (1) 2 i + 1 2 h 16 ° (1 : 5) 2 i + 1 2 h 16 ° (2) 2 i + 1 2 h 16 ° (2 : 5) 2 i + 1 2 h 16 ° (3) 2 i + 1 2 h 16 ° (3 : 5) 2 i = 77 2 = 38 : 5 As before, we can see, that the actual area A is somewhere between 38 : 5 and 46 : 5 : 38 : 5 ± A ± 46 : 5 2
It is imperative, that we illustrate the fact, that if we choose a better re°ne-

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Unformatted text preview: ment, the approximation whould be much better. Lets choose a partition, such that the base of the rectangle is of length 1 100 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 2 4 6 8 10 12 14 16 x y We can see, that the sum of the area of the approximating rectangles is "close" compared to the previous re&nement. Since the curve is decreasing by choosing the left endpoint of the interval, the sum of the areas of the approximating rectangles, is greater that the actual area. 1 100 400 X k =1 " 16 & & k & 1 100 ± 2 # = 213 733 5000 ± 42 : 746 6 We will explain in the next couple of sections, how to calculate the actual area, which is A = 128 3 ± 42 : 666 666 666 666 666 667 3 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 2 4 6 8 10 12 14 16 x y 4...
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