Question

Problem #4: (16 points) Estimate the following integral using a three point Gauss Legendre rule I = | xe* +3dz Gauss zeroes and weights - 3/5 5/9 10 8/9 3 + 3/5 5/9 22. what is the required mapping (3) (a) x(E)= $ + 3.5 x()= 1.75 $ +3.75 x(E)= 3.5 $ +1.5 (d) x(E)= 1.5 & + 3.5 (e) none of the above 23. provide the values of (E1), 2(62) and (61) = 2.394 ,(62) = 3.750 (b) (81) = 2.013 (62) = 4.567 (c) (i) = 2.568 , 3(E) = 3.102 (d) (EI) = 1.346 ,*(62) = 2.105 (e) none of the above. (E3)you'll need these to evaluate the integral. , and (63) = 5.106 , and 3(83) = 5.478 , and (3) = 5.496 , and (3) = 4.786 24. determine the Jacobian of the transformation (a) J = 5.5 (b) J = 3.75 (c) J = 3.5 (d) J = 1.5 none of the above 25. determine the value of the integral. (a I = 1.104 x 103 I= 1.103 x 103 (c) I = 630.14 (d) I = - 1.104 x 103 (e) none of the above

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Answer 1

Theree point frauss Legendse Rule I= [ (x&%+3) doc Garss zesses and weights It is sig Change of variables x= atb + balis a=2 6:55 1x: 2495 + (552) B) = 3.45 +1.453 Mus, 197 1 limit change from. (2,5.5) tof-1,1) foreste si a face sige 31.75 (f(1.452+3.75 )aby +1.75) ff (3.45+1.45 5) ds 15,= -85 f(3.45 -1.75 F5) = f(2.33445) = 2 (3.) )$=0. 1(31) = 3.75 - 1,45(0) - 3.75 1763.87+(%)145) = f(5.10c) = 2(3)

I above 24 Jacobian transformation (2,5,5) = (-1,0 Ans. b-a: 1:75 tris. d)/ none of 22 tc): xe*+ 3 f(2.3947 - 29.2316 F(3.75) - 162.454 f(5.106)- 845.5359 Is 175 Tw, f(2.394) + W4 F(3.45)+Wf(5.100) +1.75 (E1) (29.2316)+ (75) (162.454)+61)(645.Sesi 2175(6470521) - 1182.74 = 1.103 x103 Ans. (6)