F(x) = x^3 (a) computer f(x)dx (b) computer the approximation L, and T, for n=4, 8, and 16 for the integral in part (a). For each of these, compute the corresponding absolute error.

Answer and Step-by-step explanation:

F(x) = x^3 (a)

then

compute f(x) dx for each of these, compute the corresponding absolute error.

Solution:  f(x) = x3

            F(x) d(x) = 3x2

The absolute error are ∆x= |x-ẋ|, ∆f=|f(x) – f(ẋ)|

(b) Compute the approximation L, and T, for n=4, 8, and 16 for an integral part.

The linear (L) approximation of function f(x) = x3, n= 4

  f(4) = (4)3 = 64 and f”(4) = 3(4)2 = 48

The linear approximation is  L4(x) = f (4) + f’(4)(x-4)

                           = 64 + 48 (x-4)

For n= 8,

  f(8) = 512 and f,(8)= 3(8)2= 192

The linear approximate is L8(x) = f(8) + f’(8)(x-8)

Similarly for n= 16 the linear approximate is L16(x) = f (16) + f (16) (x-16)

The approximation for T, trapezoidal rule is:

a ʃb f(x)dx = Tn

for n = 4 the T approximation is:

T4  =a ʃb  x3 dx

For n= 8 :

T8 = a ʃb x3 dx

For n= 16 : T16= a ʃb x3 dx.

b is the final answer! : )

chase is right because he already been to the swings and he know how far it will be to get there

step-by-step explanation: