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.