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)

To approximate the integral (0,120) f(x) dx using the Midpoint approximation with n = 4, you need to divide the interval (0, 120) into 4 equal subintervals.

The width of each subinterval would be (120 - 0) / 4 = 30. So, each subinterval would be of width 30.

To calculate the approximation using the Midpoint method, you need to calculate the sum of the function values at the midpoint of each subinterval, multiplied by the width of each subinterval.

The midpoints of the subintervals are obtained by adding the half-width of each subinterval to the left end of each subinterval. In this case, since the widths of the subintervals are 30, the midpoints would be at x = 15, 45, 75, and 105.

Now, let's calculate the approximation step by step:

Subinterval 1: x = 0 to 30 with midpoint at x = 15
f(15) = 15
Approximation of the area in the first subinterval = 30 * f(15) = 450

Subinterval 2: x = 30 to 60 with midpoint at x = 45
f(45) = 20
Approximation of the area in the second subinterval = 30 * f(45) = 600

Subinterval 3: x = 60 to 90 with midpoint at x = 75
f(75) = 23
Approximation of the area in the third subinterval = 30 * f(75) = 690

Subinterval 4: x = 90 to 120 with midpoint at x = 105
f(105) = 22
Approximation of the area in the fourth subinterval = 30 * f(105) = 660

Finally, to find the total approximation using the Midpoint method, you add up all the individual approximations:

Total approximation = 450 + 600 + 690 + 660 = 2400

So, the approximation of the integral (0,120) f(x) dx using the Midpoint method with n = 4 is 2400, not 7200 as you calculated.

It's possible that you made an error in your calculations or used an incorrect formula. Double-check your work, following the steps outlined above, to confirm your answer.