Question

(1 point) Let f(x) = x2 (a) Compute S.' f(x) dx. 0.25 (b) Compute the approximations L, R , Tn, and M, for n = 4, 8, and 16 for the integral in part (a). For each of these, compute the corresponding absolute error. Note: Make sure all answers are correct to six decimal places. | L4 = 0.140625 \ELI = 0.109375 R4 = TERI = 0.265625
$L16 = \ELİ = R16 = Er] = T6 = Erl = M16 = |EM= Note: Observe what happens to the errors when n is doubled.$

Answer 1

Given that flad 2x3. I (0)12 Sfade - Sada (4), 4-0 120.25 - (b) for nay on Co.nl, we have AX2 10 - Yy. So, the partitions (0iu].[ 3](3, 9 442 AX (ro)* +794fc4a34 free)) (400) + f(u) + +(63) ++*)) = 10+ eu + g +24) L4 - 0. [HD65 TEL1=1 I - Lul:10:25 -0.14062513 0.109375 Ru - Ax ( + (%) *)++(3/4)44409) -ül But $ + 2y +1) boky = 0.390625 TERI = TRY-0.2812 10.390625-0.25/30140625 Tya 41 160) +2444)+2+(92) 42-4(3/4)441) | : ( 0 1 2x + 2 (%)12 (24) +1) os ly = 0.265625 TER | Ty - 0.28) = 10.265625 -0.21) = 0.015625 Mys. Ax ( +(4) +-1%) 44({}+13)). , My 0.2421875 TEMA | MY-0.20) = 10-242870 -0.28130-00481,

For na8 on Coill., we have Ara 10 1/8. So, the partitions are [0,XJCX,11][28, %][318,4] [45][54,64 [66: W][78,13 * lg2 ax (40) ++l%&)*+P%) + f(%) + f(487+465k) ++143) ottyy Holg = 0.19140625 IELI=148-1) = 10.19140628-0.28)= 0.05859375 Rg2An (flye) «t(218 )+f316) +fL48)+(5/6)+F(616) 44(X)f(9) = $( 2 944 +1) R8 = 0.31640625. (ER)=1R8 -1) =10.3164068-0.21): 0.06640625. b) +2.401k )42.4(918) +2.4(3/8) 2+ (468) +26 (54) +24183)+28(718) +410) uts = 0.25390625 IET) = 1T8-I) = 10.26390625 -0.28) = 0.00 39005. Mg - Aa (Unk)+(316) +f(546) +1V16)++C/16) +f("16) +41 318) ++/":/4)) 2 MB = 0.2480 4687 IEP= 1 Mg-1) = 10.24804687-0251 = 0.00195313. For na16 on [0,1], we have Ax= 100 = 1/16, So ithe Amp intervals are Co./16] [%16, 116][ho, ho ][ 316, ) [11c, Sho][576, 414 ] [ 618,16][176, 8 he][8/16, 96] [M6, ]'

(10/6,"1181C413,416C%201131] [itic, he ] ['10"] [5/12 Libe Ax[fo) ++lY16) ++49rb)+----++(576) 416 = 0.21972656 TEL/= 146-1) = 10-219726-0-21) = 0.030273 R16 = An (filib) ++(216) +---+ 40)] : Rib = 0.282226. TERI=1 R16"I] = 10.382226-0.25)= 0.032226 T16 = A (+10)+2 (f(416) ++(40) +---++(15/3)] ++4) i. Ti = 0.250976. JETI = 116-IV = 10.250976-0-231= 0.000976 M16 - Ad (418a) + 43132) +---++(31132)) ..M16= 0-2495117 JEM) = 1 M6-I1 = 10.249849-0-28/0.0004888