Figure (a) shows a vacant lot with a 100-ft frontage 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, 100], we see that the problem is mathematically equivalent to that of finding the area under the graph of f on [0, 100]. To estimate the area of the lot using a Riemann sum, we divide the interval [0, 100] into five equal subintervals of length 20 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 = 10, 30, 50, 70, and 90. What is the approximate area of the lot?
A) 8,100 sq ft
B) 8,400 sq ft
C) 8,600 sq ft
D) 7,900 sq ft

Question

Figure (a) shows a vacant lot with a 100-ft frontage 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, 100], we see that the problem is mathematically equivalent to that of finding the area under the graph of f on [0, 100]. To estimate the area of the lot using a Riemann sum, we divide the interval [0, 100] into five equal subintervals of length 20 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 = 10, 30, 50, 70, and 90. What is the approximate area of the lot?  
A) 8,100 sq ft 
B) 8,400 sq ft 
C) 8,600 sq ft 
D) 7,900 sq ft

Quizplus · Accepted Answer

The area can be approximated by the Riemann sum:
$A\approx \frac{20}{2}(f(10)+f(30)+f(50)+f(70)+f(90))$
Plugging in the given values for $f(x)$, we get:
$A\approx \frac{20}{2}(30+60+50+40+45)=8,400$ sq ft. Thus, the answer is B.

Figure (A) Shows a Vacant Lot with a 100-Ft Frontage