you want 3 trapezoids to cover an interval of width 6, so ∆x = 2
That means the x values are -2, 0, 2, 4
So just plug in your formula. The area is approximated by
2 (f(-2) + 2f(0) + 2f(2) + f(4))/2
That's not too hard, eh?
Estimate the value of the integral from negative 2 to 4 of x^3dx by using the Trapezoidal Rule with n = 3.
a) 252
b) 128
c) 63
d) 72
4 answers
y = x^3 would be your function
I don't know if n = 3 means splitting it into 3 traps or 4 traps.
If you want 4 traps it would be easier, so
so your base would be (4-2)/4 = .5
and our x's would be 2, 2.5, 3, 3.5, and 4
with the y's as 8, 15.625, 27, 42.875, and 64
Area = (8+15.625)(.5)/2 + (15.625+27)(.5)/2 + (27+42.875)(.5)/2 + (42.875+64)(.5)/2
= (1/4)(8 + 2(15.625) + 2(27) + 2(42.875) + 64)
= 60.75
If you want it in 3 traps, each base would be 2/3 units
and my answer would be
728/81 + 56/3 + 2728/81 = 184/3 = appr 61.333...
I don't know if n = 3 means splitting it into 3 traps or 4 traps.
If you want 4 traps it would be easier, so
so your base would be (4-2)/4 = .5
and our x's would be 2, 2.5, 3, 3.5, and 4
with the y's as 8, 15.625, 27, 42.875, and 64
Area = (8+15.625)(.5)/2 + (15.625+27)(.5)/2 + (27+42.875)(.5)/2 + (42.875+64)(.5)/2
= (1/4)(8 + 2(15.625) + 2(27) + 2(42.875) + 64)
= 60.75
If you want it in 3 traps, each base would be 2/3 units
and my answer would be
728/81 + 56/3 + 2728/81 = 184/3 = appr 61.333...
The actual answer using the definite integral is
∫ x^3 dx from 2 to 4
= [ (1/4)x^4] from 2 to 4
= 256/4 - 16/4
= 60
so the choices given really don't match either of my two interpretations.
oobleck went from -2 to 6
∫ x^3 dx from 2 to 4
= [ (1/4)x^4] from 2 to 4
= 256/4 - 16/4
= 60
so the choices given really don't match either of my two interpretations.
oobleck went from -2 to 6
72