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
72
To estimate the value of the integral using the Trapezoidal Rule, we will divide the interval from -2 to 4 into n subintervals of equal width, where n is the number of trapezoids we will use.
In this case, n = 3, so we will divide the interval into 3 subintervals:
Subinterval 1: [-2, -1]
Subinterval 2: [-1, 0]
Subinterval 3: [0, 1]
Subinterval 4: [1, 2]
Subinterval 5: [2, 3]
Subinterval 6: [3, 4]
Now, we need to determine the width of each subinterval. Since the interval width is 6 (from 4 to -2), we divide it by the number of subintervals, which is n+1. So the width of each subinterval is (6 / (3+1)) = 6 / 4 = 1.5.
Next, we evaluate the function x^3 at the endpoints of each subinterval:
Subinterval 1: f(-2) = (-2)^3 = -8
Subinterval 2: f(-1) = (-1)^3 = -1
Subinterval 3: f(0) = 0^3 = 0
Subinterval 4: f(1) = 1^3 = 1
Subinterval 5: f(2) = 2^3 = 8
Subinterval 6: f(3) = 3^3 = 27
Subinterval 7: f(4) = 4^3 = 64
Now, we calculate the sum of the areas of the trapezoids:
Area of Trapezoid 1 = ((-8 + (-1)) / 2) * 1.5 = (-9/2) * 1.5 = -13.5/2 = -6.75
Area of Trapezoid 2 = ((-1 + 0) / 2) * 1.5 = (-1/2) * 1.5 = -0.75/2 = -0.375
Area of Trapezoid 3 = ((0 + 1) / 2) * 1.5 = (1/2) * 1.5 = 0.75/2 = 0.375
Area of Trapezoid 4 = ((1 + 8) / 2) * 1.5 = (9/2) * 1.5 = 13.5/2 = 6.75
Finally, we sum up the areas of all the trapezoids:
Approximate value = -6.75 + (-0.375) + 0.375 + 6.75 = 0
So, the estimate of the value of the integral from -2 to 4 of x^3 dx using the Trapezoidal Rule with n = 3 is approximately 0.
None of the given options (a), b), c), or d)) match this result.
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
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...
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?