Chap5_Sec2

# Chap5_Sec2 - x =(2 1/5 = 1/5 s So the Midpoint Rule gives 2...

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EVALUATING INTEGRALS Evaluate the following integrals by interpreting each in terms of areas. a. b. 1 2 0 1 x dx - 3 0 ( 1) x dx - Example 4 (5.2)

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EVALUATING INTEGRALS Since y 2 = 1 - x 2 , we get: x 2 + y 2 = 1 s This shows that the graph of f is the quarter-circle with radius 1. Example 4 a 1 2 2 1 4 0 1 (1) 4 x dx π - = = Since we can interpret this integral as the area under the curve from 0 to 1. 2 ( ) 1 0 f x x = - 2 1 y x = -
EVALUATING INTEGRALS The graph of y = x – 1 is the line with slope 1 shown here. s We compute the integral as the difference of the areas of the two triangles: 3 1 1 1 2 2 2 0 ( 1) (2 2) (1 1) 1.5 x dx A A - = - = - = Example 4 b

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MIDPOINT RULE Use the Midpoint Rule with n = 5 to approximate s The endpoints of the five subintervals are: 1, 1.2, 1.4, 1.6, 1.8, 2.0 s So, the midpoints are: 1.1, 1.3, 1.5, 1.7, 1.9 s The width of the subintervals is:

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Unformatted text preview: x = (2 - 1)/5 = 1/5 s So, the Midpoint Rule gives: 2 1 1 dx x âˆ« Example 5 (5.2) [ ] 2 1 1 (1.1) (1.3) (1.5) (1.7) (1.9) 1 1 1 1 1 1 5 1.1 1.3 1.5 1.7 1.9 0.691908 dx x f f f f f x â‰ˆ Î” + + + + = + + + + â‰ˆ âˆ« MIDPOINT RULE As f ( x ) = 1/ x for 1 â‰¤ x â‰¤ 2, the integral represents an area, and the approximation given by the rule is the sum of the areas of the rectangles shown. Example 5 PROPERTIES OF INTEGRALS If it is known that find: 10 8 ( ) 17 and ( ) 12 f x dx f x dx = = âˆ« âˆ« Example 7 (5.2) 10 8 ( ) f x dx âˆ« 8 10 10 8 ( ) ( ) ( ) f x dx f x dx f x dx + = âˆ« âˆ« âˆ« 10 10 8 8 ( ) ( ) ( ) 17 12 5 f x dx f x dx f x dx =-=-= âˆ« âˆ« âˆ«...
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## This note was uploaded on 05/23/2011 for the course MAC 2311 taught by Professor Noohi during the Fall '08 term at FSU.

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Chap5_Sec2 - x =(2 1/5 = 1/5 s So the Midpoint Rule gives 2...

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