? ??? Romberg integration for approximating f(x) dx gives R1 = 8 and R22 = -1 then

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Romberg integration for approximating f(x) dx gives R₂1 = 8 and R22 = -1 then f(2)= O 6.5 O 5.5 O -4.75 O-9.5 Consider the quadrature formula Q[f], of the form: 5 8 5 [ƒ(x)dx ≈ Q[f] := {}ƒ(−√0.6) +ƒ(0) +- ƒ(√0.6) F(x)dx Using this quadrature formula Q[f], the approximation of f*xe²* dx is: O 0.13844565 O 4967.10668 O 14.7766778 O -4.2919282 Consider the quadrature formula Q[f], of the form: 5 8 5 į [ƒ(x)dx ≈ Q[ƒ] =¦ƒ(-√0.6) + §ƒ(0) +ƒ(√0.6) F(x)d Using this quadrature formula Q[ƒ], the approximation of fxe ²x dx is: O 0.13844565 O 4967.10668 O 14.7766778 O -4.2919282 Consider the initial value problem: y' = 2ty + 2t 0≤t≤ 1, y(0) = 1 The approximation of y(1) by using the modified Euler's method with h = 0.5 is most nearly: O 3.40625 O 7.69531 O 4 O 2.85156 Romberg integration for approximating f(x) dx gives R₂1 = 8 and R22 = -1 then f(2)= O 6.5 O 5.5 O -4.75 O-9.5 Consider the quadrature formula Q[f], of the form: 5 8 5 [ƒ(x)dx ≈ Q[f] := {}ƒ(−√0.6) +ƒ(0) +- ƒ(√0.6) F(x)dx Using this quadrature formula Q[f], the approximation of f*xe²* dx is: O 0.13844565 O 4967.10668 O 14.7766778 O -4.2919282 Consider the quadrature formula Q[f], of the form: 5 8 5 į [ƒ(x)dx ≈ Q[ƒ] =¦ƒ(-√0.6) + §ƒ(0) +ƒ(√0.6) F(x)d Using this quadrature formula Q[ƒ], the approximation of fxe ²x dx is: O 0.13844565 O 4967.10668 O 14.7766778 O -4.2919282 Consider the initial value problem: y' = 2ty + 2t 0≤t≤ 1, y(0) = 1 The approximation of y(1) by using the modified Euler's method with h = 0.5 is most nearly: O 3.40625 O 7.69531 O 4 O 2.85156

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