Respuesta :
Answer:
a) Trapezoidal: -1392.087124
b) Midpoint: -1744.434609
c) Simpson: -1626.985447
Step-by-step explanation:
The interval have length 12, so the subintervals have length 12/8 = 1.5. Lets call f(x) = sin(x)x³.
Trapezoidal:
[tex]\int\limits^{12}_0 {x^3sin(x)} \, dx \approx\\ 0.75(f(0) + 2f(1.5)+ 2f(3) + 2f(4.5)+ 2f(6) + 2f(7.5)+2f(9)+2f(10.5)+f(12) ) =\\-1392.087124[/tex]
Midpoint:
[tex]\int\limits^{12}_0 {x^3sin(x)} \, dx \approx\\ \\1.5*(f(0.75) + f(2.25) + f(3.75) + f(5.25) + f(6.75) + f(8.25) + f(9.75) + f(11.25) ) \\= -1744.434609[/tex]
Simpson:
[tex]\int\limits^{12}_0 {x^3sin(x)} \, dx \approx\\ \\0.25( f(0) + 4 f(0.75) + 2f(1.5) + 4 f(2.25) + 2f(3) + 4f(3.75) + 2f(4.5) + 4f(5.25)+2f(6) + 4f(6.75) + 2f(7.5) + 4f(8.25) + 2f(9) + 4f(9.75) + 2f(10.5)+4f(11.25)+f(12)) = -1626.985447[/tex]
