1 Example 1 Estimate by the six Rectangle Rules using the regular partition P of the interval [0,1] into 4 subintervals. Solution This definite integral gives the area of a quarter circle of radius 1 and therefore has value  /4. Hence estimating this integral means estimating the value of  /4. Note that P = {0, ¼, ½, ¾, 1} with each subinterval of width ¼. The four subintervals are: [0, ¼ ], [¼, ½], [½,¾] and [¾, 1] while the function is Then The L k are the heights of the 4 rectangles used to approximate this definite integral. In the Left Endpoint Rule the L k are the values of f on the left endpoints of the four subintervals: f(0), f(¼), f(½), f(¾). In the Right Endpoint Rule the L k are the values of f on the right endpoints of the four subintervals: f(¼), f(½), f(¾), f(1).

2 In the Midpoint Rule the L k are the values of f on the midpoints of the four subintervals: f(1/8), f(3/8), f(5/8), f(7/8). In the Trapezoid Rule the L k are the averages of the values of f on the endpoints of each of the four subintervals: ½[f(0)+f(¼)], ½[f(¼)+f(½)], ½[f(½)+f(¾)], ½[f(¾)+f(1)]. 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 f 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 L k are summarized in the table on the next slide The 4 subintervals are: [0, ¼ ], [¼, ½], [½,¾], [¾, 1].

3 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: Left Endpoint Rule Right Endpoint Rule Midpoint Rule Trapezoid Rule Lower Riemann Sum Upper Riemann Sum L1L2L3L4L1L2L3L4 f(0)=1 f(¼) .968 f(½) .866 f(¾) .661 f(¼) .968 f(½) .866 f(¾) .661 f(1)=0 f(1/8) .992 f(3/8) .927 f(5/8) .781 f(7/8) .484 ½(1+.968) ½( ) ½( ) ½(.661+0) f(¼) .968 f(½) .866 f(¾) .661 f(1)=0 f(0)=1 f(¼) .968 f(½) .866 f(¾) .661