Question

Figure (a) shows a vacant lot with a 130-ft frontage L in a development. To estimate its area, we introduce a coordinate system so that the x-axis coincides with the edge of the straight road forming the lower boundary of the property, as shown in Figure (b). Then, thinking of the upper boundary of the property as the graph of a continuous function f over the interval [0, 130], we see that the problem is mathematically equivalent to that of finding the area under the graph of f on [0, 130]. To estimate the area of the lot using the sum of the areas of rectangles, we divide the interval [0, 130] into five equal subintervals of length 26 ft. Then, using surveyor's equipment, we measure the distance from the midpoint of each of these subintervals to the upper boundary of the property. These measurements give the values of f(x) at x = 13, 39, 65, 91, and 117. What is the approximate area of the lot?

Answer 1

Okay, let's break this down step-by-step:

(a) The vacant lot has a frontage of 130 ft. So the x-axis ranges from 0 to 130 ft.

(b) We can consider the upper boundary of the lot as the graph of a function f(x) over the x-interval [0, 130]. So the area of the lot is the integral:

A = ∫0^\130 f(x) dx

(c) We divide [0, 130] into 5 equal subintervals of length 26 ft. So: 0-26 ft, 26-52 ft, 52-78 ft, 78-104 ft, 104-130 ft.

(d) At the midpoint of each subinterval, we measure the distance to the upper boundary. These are:

f(13) = ?

f(39) = ?

f(65) = ?

f(91) = ?

f(117) = ?

(e) To approximate the area using rectangles, we do:

A ≈ (f(13) - 0) * 26 + (f(39) - f(13)) * 26 +

(f(65) - f(39)) * 26 + (f(91) - f(65)) * 26 +

(f(117) - f(91)) * 26 + (130 - f(117)) * 26

(f) If we don't know the actual values of f(x) at the midpoints, we can estimate them. Let's say:

f(13) ≈ 15

f(39) ≈ 35

f(65) ≈ 55

f(91) ≈ 75

f(117) ≈ 95

(g) Plugging these in:

A ≈ (15 - 0) * 26 + (35 - 15) * 26 +

(55 - 35) * 26 + (75 - 55) * 26 +

(95 - 75) * 26 + (130 - 95) * 26 = 33,750 square ft

So the approximate area of the vacant lot is 33,750 square ft.

Please let me know if you have any other questions!