Question

Use the upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width)

y=sqrt(1-x^2)

https://www.webassign.net/larson/4_02-30.gif

I was able to find the upper sum, which was 0.895, but I cannot find the lower sum. I have gotten 0.824 and 0.659 as answers, but neither are considered correct.

Answer 1

From the figure shown, the interval is divided into 5 equal parts making each subinterval to be 0.2.

Part A:


`y= √(1-x^2)`

The approximate the area of the region shown in the figure using the lower sums is given by:


`Area= [y(0.2)*0.2]+[y(0.4)*0.2]+[y(0.6)*0.2]+[y(0.8)*0.2] \\ +[y(1)*0.2] \\ \\ =[√(1-(0.2)^2)*0.2]+[√(1-(0.4)^2)*0.2]+[√(1-(0.6)^2)*0.2] \\ +[√(1-(0.8)^2)*0.2]+[√(1-(1)^2)*0.2] \\ \\ =(0.9798*0.2)+(0.9165*0.2)+(0.8*0.2)+(0.6*0.2)+(0*0.2) \\ \\ =0.196+0.183+0.16+0.12=0.659`



Part B:

The approximate the area of the region shown in the figure using the lower sums is given by:


`Area= [y(0)*0.2]+[y(0.2)*0.2]+[y(0.4)*0.2]+[y(0.6)*0.2] \\ +[y(0.8)*0.2] \\ \\ =[√(1-(0)^2)*0.2]+[√(1-(0.2)^2)*0.2]+[√(1-(0.4)^2)*0.2] \\ +[√(1-(0.6)^2)*0.2] +[√(1-(0.8)^2)*0.2] \\ \\ =(1*0.2)+(0.9798*0.2)+(0.9165*0.2)+(0.8*0.2)+(0.6*0.2) \\ \\ =0.2+0.196+0.183+0.16+0.12=0.859`



Part C:

The approximate area of the given region is given by


`Area= (0.659+0.859)/(2) = (1.518)/(2) =0.759`