Question

Question 2. Consider the approximation of the definite integral () (a) Begin by using 2 points/nodes (i.e., n + 1 = 2, with the two points being x = a and r = b). Replace f(x) by the constant /(a+b)/2] on the entire interval a <<b. Show that this leads to the numerical integration formula M,()) = (b − a) ) Graphically illustrate this approximation. (b) In analogy with the derivation of the Trapezoidal rule and Simpson's rule, generalize part (a) to write down a numerical integration formula M.() for arbitrary n 1 (ie, when using more than 2 points). (Hint: Consider using the step size h = (b-a)/n and the nodes x, = a + (-1/2)h for j = 1,...,n). (Note: This is called the midpoint rule; it is a popular alternative to the Trapezoidal rule). (c) For the integral I = 1 da, calculate M.(/) and M3(/) and the corresponding approximation errors. Compare your results to those for the Trapezoidal rule (sce worksheet from class). Due: 10/08/2019 by 4:30pm

Answer 1

@ I(4) = 2 x dx Replacing fx by the constant fcath), we gel M, CA) = b fcato) dx - fcat sa da - flatb] [216 = (b-a). f(atb) #x27 f(x) Mi it is, IC the astea under f(x) from X= a to x=b is approxim.de by the area of the rectangle of side (bra) and of catb). 6 Dividing the interval [a,b] in m ) equal subinterval with step. size h= bra. So, the intervals aste, [a, ki], [x, , Hz], [x2, x3] ..., [a, ath], [atch], [a +3h]. arre [a, ath] , [ath, ath], ..., [at(n.bh, 67 We will apply poort (as in every subintervals. Now, the mid points of every sub-intervals are XIF atatha ath, x₂ = athyarhaath, x = atzhthth = at Sh ... xa at (n-Dhtb at(n-1h 2 2 . + B +nh 9+2n-! that is, xj = at(3-4); j = 1, 2,..., et n f(x1) th f(x2) et hf(x) t o thf (an) a h 4(xj) n. 2, f@ +(5^2)n) Min = h. & f [at (5-2) n] , na bra

© ISS de la(2) H() = (-0.4 (140) = 1.46) te Ź . The corresponding error = | ln (2) – 2 1 0.02648 Here, n=3, aso, b=), :. h= bra = 12 i M3 (f) - h. f(x1) at f(x) + f(x3) 7 = 33. [fCotto) + f (0+ 2) + f (0+5) 3. [ f(&) + f ( 37 ) &&(s)] = 3 [itro at 143/C + 15 16] - / (+ 8 + ) 1 1 & 478 693 :. The corresponding error = 1 in (2) - 02 0.60339249