Integration_Trapezoid(3).ppt

# T ln t f 8 9 2100 140000 140000 2000 ln f 8 8 9 8

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t ln ) t ( f 8 9 2100 140000 140000 2000 ) ( . ) ( ln ) ( f 8 8 9 8 2100 140000 140000 2000 8 ) ( . ) ( ln ) ( f 30 8 9 30 2100 140000 140000 2000 30 s / m . 27 177 s / m . 67 901

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11 Solution (cont) 2 67 901 27 177 8 30 . . ) ( I m 11868 a) b) The exact value of the above integral is 30 8 8 9 2100 140000 140000 2000 dt t . t ln x m 11061
12 Solution (cont) b) Value e Approximat Value True E t 11868 11061 m 807 c) The absolute relative true error , , would be t 100 11061 11868 11061 t % . 2959 7

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13 Multiple Segment Trapezoidal Rule In Example 1, the true error using single segment trapezoidal rule was large. We can divide the interval [8,30] into [8,19] and [19,30] intervals and apply Trapezoidal rule over each segment. t . t ln ) t ( f 8 9 2100 140000 140000 2000 30 19 19 8 30 8 dt ) t ( f dt ) t ( f dt ) t ( f 2 30 19 19 30 2 19 8 8 19 ) ( f ) ( f ) ( ) ( f ) ( f ) (
14 Multiple Segment Trapezoidal Rule With s / m . ) ( f 27 177 8 s / m . ) ( f 75 484 19 s / m . ) ( f 67 901 30 2 67 . 901 75 . 484 ) 19 30 ( 2 75 . 484 27 . 177 ) 8 19 ( ) ( 30 8 dt t f m 11266 Hence:

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15 Multiple Segment Trapezoidal Rule 11266 11061 t E m 205 The true error is: The true error now is reduced from -807 m to -205 m. Extending this procedure to divide the interval into equal segments to apply the Trapezoidal rule; the sum of the results obtained for each segment is the approximate value of the integral.
16 Multiple Segment Trapezoidal Rule Figure 4: Multiple (n=4) Segment Trapezoidal Rule Divide into equal segments as shown in Figure 4. Then the width of each segment is: n a b h The integral I is: b a dx ) x ( f I

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b h ) n ( a h ) n ( a h ) n ( a h a h a h a a dx ) x ( f dx ) x ( f ... dx ) x ( f dx ) x ( f 1 1 2 2 17 Multiple Segment Trapezoidal Rule The integral I can be broken into h integrals as: b a dx ) x ( f
• Fall '19
• 205 M, 1.8534%, 7.2959%, 79.506%

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