Question

My courses / MA102*2-01-2019.3 / General / Quiz 7 (30 July) Calculate the integral f(x) = x^2 +1 from 1 to 1 using the fundamental theorem of calculus. Calculate the same integral using Simpson's rule for n = 2, 4, 6, 8 and 10. What happens when n increases? How much is the error for each na 7 A- B I

I have 2 hours left please solve this question in a short time.Thank you.

Answer 1

f(x) = Sec+1) de Hence solving above equation f(ne) (²+1) de Core + (!) (-1* - (-2)] 3 f(1) +1 + +-) !ابر olm 2.6666 Hence the integral of above equation is 2.6666 Now solving above equation by using Simpson's sule. Which States that s f(1) du 3 (f(x) + f(x) + 41f(,)+2 7 (12) + <f(Mat .... bf (6n-1) + F(en where bra AR

Hence solving selving for be 77.2 1(-U) > M2 Divide the intoval [dot] into Subindowals of Teryth Hoc = od, with the following concepoints just evaluate the function at these endpoints, f (1) 2 if) = 4f() ܙܐ † (?) fo) 2 putting above values in Simpson's formula 271) dr - À (2+4+2) See SQ 1344 2.6666 Error Calculated valo estinrated Vale calculated val. Error 2.666G 2.6666 0 2.6666

2 solving for n = 21 we azt, band has, n=4 have An 1-(-1) 4 - - - - Hence Endpoints cule - NE 2 3 2 Now we evaluate the function at these endpoints. 5 4f(1₂) f (M4) 5 of (ro) = f(1) = 2 4f (n.) hf () 2 F(12) = 2f (0) 4f ( f(1) = 2 purring above values in son m²+1) de = (2+5+2+5+2) S crtan Simpson's formula. 2.6666 Error Cal. Valo est. Veel cul. val. Ensor. 2 6666 2.6666 - o 2.6666 3 for solving n=6 We have az -] , bat, n = 6

AN +- (-) 1) Cull- So endpoints are -1, ,0.31 Now we evaluate the funct" at these points. = f(-1) 2 5.77 2.22 fro) 4f(xv) = 46() 27(82) : 24 4F (M3) af(24) af ( 3 4F (7155) 4f ( 2 f (no) F(1) 4f(0) = 4 2.22 5.77 2 putting above values in simpson's Resimuler. 1 (x+1) da = ġ (2++ 5.77 + 2.22 +4 +2.22 + 5.77 + 2) S. (+*+1)da S 2.6666 2.6666 2.6666 . Hence Eror 2.6666 Eror o

6 for -solving na 8 we have as-d, bad, na8 son - 1-(-1) - 44 So end points are at .-22 9 Ž. Ž.0,5 Now evaluate the function at these points. f(no) F(-) 2 28/3) = 6.25 2.5 4.25 af (14) 4f(",) 2 F (M2) = if ( 2 / 2 ) 4F (13) = 4F (-24) af (og af (no) af (4) af (46) af (1/2) 4F(",) 4F (2) f (18) (۱) 4.25 = 2:50 6.25 - 2 Stri) die putting above values in simpsonis formula. (2+6.25 2+6.25 +2.5 + 24.25 +2 +4.25 +2.5 +2) Si'cu'zl)odre 12 + 6.25 +2 2.6666 So Error is 2.6666 2.6666 = 2.66G6

5 for solving nelo. we have -d, bad a- n10 Дr 1-(-) 1 / 3 endpoints are -1, Kost So the Now we evaluate the funcon at these endpoints. F (no) - 2 4f (1) af (42) f(t) 4F (-45) = 6:56 2F (5/5) - - 2.72 4f( *215) 4,64 2,08 4 4f (13) 2 f (hu) 26 (115) 4F (M5) = 4f (0) 2F (MG) 4F (14) = 4f (2/5) 2F (M8) 2 f 2f ( 315) af (ng) 4F (415) 2 f (45) - 2.0% 4.64 2.72 - 6.56 f (no) f (+) 2 putting above values in simpson's formules: 2+G:56 + 2.72 +4.64 + 2.0% +4 + 2.08 + 4.64 +2.72 +6.56 +2 +2) - 1 (M²+1) da a 2.GGGG

2.6666 2.66GG And Error 2.6666 if we ses in previouse values of n e in m increases then changes occulus Answer for the giren integral. And n (2,4,6,8,10) -ענף בבוזיוס 40 lat zero. the is always