Asked by Pax
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)
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)
Answers
Answered by
Reiny
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
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
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