Question

QUESTION 5 The integral I= = 1 ata de is to be approximated numerically. (a) Find the least integer M and the appropriate step size h so that the global error for the composite (10) Trapezoidal rule, given by ban2f"(0), a<<< is less than 10-s for the approximation of I. (b) Use the two-term Gaussian quadrature formula and 6 decimal place arithmetic to approximate 1. (Hint: Parameters are g = 1, i = 1, 2;rı = 0.5773502692, r2 = -0.5773502692) (9) (c) Evaluate the integral analytically and find the error in the approximation in (a) above. (6) [25] TOTAL MARKS: (100)

Answer 1

(6) First approammating the integral f dre (x+4) Teapezoidal Trule. 1o Sub-intervals each 80 numerically by using Divide the integral [02] into of width h= 0.2 No=0, x= 0.2, ₂ = 0.4, z = 0.6, ny = 0.8, s = 1.0 n6= 1.2, na = 1.4, p = 1.6, ng=1.8, 10 = 2.0. The values of the function at points of division are as under : No=0 n = 0.2 g = 0.4 = 0.6 y = 0.8 5 = 1.0. ya nty 0.25 0.238 0.0227 0.217 0208 0.2 26 n6 = 1.2 = 114 g = 116 na = 1.8 1.- 210 - y= x+y 0.192 0.1850.178 0.172 0.167 Notnh For Trapezoidal rule, I fou du = h (Sum of first and last redinater) +2 (Sum of all the intermediate sto credinater)

Sa dore = (2+4) 2 0.2 [10.25 +0.167) +2(0.236 +0.227 + 0.217 +0.208 + 0.2+0.192 +0.185+0.178 +0:172) ->>] 417 +2 [0.47 [047 +33 63 = 0 = 01 (4.051) = 0.4051 Now, f(x) = (+4). fin) = -1 and "W= 2 (n+4)3 2 max a sneb max o en c2 0.0 3125 - En (8)= -K (b-a) f"(2) and 50) + hº (2-0) (0.03125) 12 +62 (0.03125) < J.-5 Acc. to Question, th² 6 x 10-5 0.03125

h²c 192 x 10-5 3 h 0.0438 and 2 > 45. 662 h (b) evaluate Staty da gusing two-term Gaussian - Quadercature formula Here f(x) = 1 144 2 n=2 First we transform the interval [2] to [-] by using the substitution x = b ta tb-a, u 2 2 = 2tot 2-0 .. 2 Itu da-du Now, I = 1 S du _dre - (x+4) (1+u+4) is cum du (4+5) .. I Scots de = a + (-) + 0() where (si)= rit 5

Now, using the given parameters. I- - dn = (+4) 0 0.5773502 692 +5 + (1.000) -0.5773 502692 +5 or Ia Pontu -dre (2+4) 5.5773502692 4.422 649731 0 0.17929661 +0.226108794 = 0.405405 404 2 0.405 405 Analytical evaluation of integral 2 dx = f (n+4) log (nty) log 6 - log 4 = 0.6778/5125 -0.602059991 = 0.17 6091258 7 0.1761

.. Frash True value - approximate value in (a) 0.1761 0.4051 0.229.
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