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Example 2 Estimate by the six Rectangle Rules using the regular partition P of the interval [ 0, ] into 6 subintervals. Solution Observe that the function with g(0)=1 is continuous at
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Example 2 Estimate by the six Rectangle Rules using the regular partition P of the interval [0, ] into 6 subintervals. Solution Observe that the function with g(0)=1 is continuous at x=0 because Note that P = {0, /6, /3, /2, 2/3, 5/6, }with each subinterval of width /6. The six subintervals are: [0, /6], [/6, /3], [/3, /2], [/2, 2/3], [2/3, 5/6] and [5/6, ]. Then The Lkare the heights of the rectangles used to approximate this definite integral. In the Left Endpoint Rule the Lkare the values of g on the left endpoints of the six subintervals: g(0), g(/6), g(/3), g(/2), g(2/3), g(5/6).
The six subintervals are: [0,/6], [/6,/3], [/3,/2], [/2, 2/3], [2/3,5/6], [5/6,]. In the Right Endpoint Rule the Lkare the values of g on the right endpoints of the six subintervals: g(/6), g(/3), g(/2), g(2/3), g(5/6), g(). In the Midpoint Rule the Lkare the values of g on the midpoints of the six subintervals: g(/12), g(/4), g(5/12), g(7/12), g(3/4), g(11/12). In the Trapezoid Rule the Lkare the averages of the values of g on the endpoints of each of the six subintervals: ½[g(0)+ g(/6)], ½[g(/6)+ g(/3)], ½[g(/3)+ g(/2)], ½[g(/2)+ g(2/3)], ½[g(2/3)+ g(5/6)], ½[g(5/6)+ g()]. Therefore, the Lower Riemann sum coincides with the estimate of the Right Endpoint Rule and the Upper Riemann sum coincides with estimate of the Left Endpoint Rule. Since the function g is decreasing on [0,1], it has its maximum value at the left endpoint of each subinterval and its minimum value at the right endpoint of each subinterval. The values of the Lkare summarized in the table on the next slide
The Left Endpoint Rule and Upper Riemann Sum give the same estimate: The Right Endpoint Rule and Lower Riemann Sum give the same estimate: The Midpoint Rule gives the estimate: The Trapezoid Rule gives the estimate: