Since f{x]=1+ 2x2 is an increasing function on [1,4], f(1)=3 is the minimum, and fl4l=33 is the maximum of the function f on [1.4]. The error of an integration based on a Riemann sum may be given as the difference between the upper sum and lower sum as, error(n) = 2 f (M )x, 2 f(m, )Ax, , where flMi}, and film} are respectively, the i=1 i=1 maximum and minimum of the function, f, over the subinterval [xi1, xi], {a} Show that srror(n) = [r-'1) f(1)]Ax, when the sub-interval At, 2 Ax = (4 _1) , is n of equal length. (Hint: Divide the interval [1, 6] equally into n subintervals, x1, 37,, ...., x,_1,x,,..._. x", where x, =1, and x" = 4.] lb] Compute errorllOO) and errorlZG]. Is error{n] decreasing as n increases from 100 to 200