Measurements of a lake’s width were taken at 15-foot intervals, as shown:
x= 0 15 30 45 60 75 90 105 120
f(x)= 0 15 18 20 19 23 24 22 12
Estimate integral (0,120) f(x) dx with n = 4, using Midpoint approximation.
For this question, I ended up with 7200 but compared to the other approx's I had used (left, right, trapezoidal) this number seems way too high. (my other numbers were 1830, 2190, and 2010 respectively). Could someone check me on this and either explain why I'm getting such a high number or let me know if i'm just overthinking my answer.
my work is the integral from 0 to 120 of 30(15+45+75+105)
I will assume your n = 4 implies you want midpoints as follows
0 -->18 = 9
18 --> 19 = 18.5
19 --> 24 = 21.5
24 --> 12 = 18
so I would have
30(9+18.5+21.5+18) = 2010
To estimate the integral using the Midpoint approximation with n = 4, you need to divide the interval [0, 120] into 4 equal subintervals.
First, find the width of each subinterval:
Subinterval width = (120 - 0) / 4 = 30
Next, find the midpoint of each subinterval:
Midpoint of subinterval 1: (0 + 30) / 2 = 15
Midpoint of subinterval 2: (30 + 60) / 2 = 45
Midpoint of subinterval 3: (60 + 90) / 2 = 75
Midpoint of subinterval 4: (90 + 120) / 2 = 105
Now, evaluate the function at each midpoint:
f(15) = 15
f(45) = 18
f(75) = 23
f(105) = 22
Finally, calculate the approximation of the integral using the Midpoint rule:
Approximation = Subinterval width * (f(15) + f(45) + f(75) + f(105))
= 30 * (15 + 18 + 23 + 22)
= 30 * 78
= 2340
Therefore, the estimated integral (0,120) f(x) dx using the Midpoint approximation with n = 4 is 2340.
Based on your calculations, it seems that you have mistakenly multiplied the subinterval width by the sum of the midpoints instead of the sum of the function values at the midpoints. Make sure to evaluate the function at the midpoints and multiply the correct values to obtain the correct approximation.
To estimate the integral using the Midpoint approximation with n = 4, you need to divide the interval from 0 to 120 into 4 equal subintervals.
The width of each subinterval is given by (120 - 0) / 4 = 30.
Now, for each subinterval, you need to find the midpoint and evaluate the function at that point.
The midpoints of the subintervals are: 7.5, 22.5, 37.5, 52.5, 67.5, 82.5, 97.5, 112.5.
Evaluate the function at each midpoint:
f(7.5) = 0
f(22.5) = 15
f(37.5) = 18
f(52.5) = 20
f(67.5) = 19
f(82.5) = 23
f(97.5) = 24
f(112.5) = 22
Now, multiply each function value by the width of the subinterval (30) and sum them up:
30 * (f(7.5) + f(22.5) + f(37.5) + f(52.5) + f(67.5) + f(82.5) + f(97.5) + f(112.5)) = 30 * (0 + 15 + 18 + 20 + 19 + 23 + 24 + 22) = 30 * 141 = 4230.
Therefore, the estimate of the integral (0, 120) f(x) dx using Midpoint approximation with n = 4 is 4230.
It seems like there might be an error in your calculation. Double-check your work and make sure you are evaluating the function correctly at the midpoints and multiplying by the width of each subinterval.