The pictures below show the left end point approximations to the area, A, with Ar = 1/8, 1/16 and 1/128 respectively. L8 = .7265625000, L16 = .6972656250, L128 = .6705627441 We see that the room for error decreases as the number of subintervals increases or as Ar - 0. Thus we see that A = lim Rn = lim En = lim f(x;)Ax = lim f(x;)Ax 1- +00 i=1 where r; is any point in the interval [ri-1, I;]. In fact this is our definition of the area under the curve on the given interval. (If A denotes the area beneath f(x) 2 0 where f is continuous on the interval [a, b], then each interval [I;-1, I;] has length Ar = " for any given n.) . Example Estimate the area under the graph of f(x) = 1/x from x = 1 to x = 4 using six pproximating rectangles and Ar = bo = 1-1 = }, where [a, b] = [1, 4] and n = 6. Lark the points 20, $1, 12, -.., T's which divide the interval [1, 4] into six subintervals of equal length on the following axis: Fill in the following tables: 20 = 1 21 = 3/2 12 = 2 13 = 5/2 14 = 3 15 = 7/2426-4 f(It) = 1/24 (a) Find the corresponding right endpoint approximation to the area under the curve y = 1/x on the interval [1, 4]. Po = f(21) Ax + f(22)Ax + f(23)Ax + f(x4)Ax + f(25)Ax + f(x6)4x (b) Find the corresponding left endpoint approximation to the area under the curve y = 1/ on the interval [1, 4]. Lo = f(:o) Ax + f (:1) A.r + f(::2) Ax + f(:3) A.F + f(:4) Aa: + f(::5) Ax