Answer:
1,107 cc
Step-by-step explanation:
The 10 intervals of the scanning are
[0,1.5), [1.5,3), [3,4.5), [4.5,6), [6,7.5), [7.5,9), [9,10.5), [10.5,12), [12,13.5), [13.5,15)
The ideal thing would be to estimate the volume with the Midpoint Rule should be take n=10.
As we must use n=5, we divide the interval [0,15] in five intervals of length 15/5 = 3
[0,3], [3,6], [6,9], [9,12], [12,15] and take their midpoints
1.5, 4.5, 7.5, 10.5, and 13.5.
Now we estimate the volume V of 5 cylinders of height h=3 and area of the base A equals to the area given for the scanning where that midpoints fall
Cylinder 1
Midpoint=1.5, interval of scanning 2nd
A = 18, V= height*area of the base = 18*3 =54 cc
Cylinder 2
Midpoint=4.5, interval of scanning 4th
A = 78, V= height*area of the base = 78*3 =234 cc
Cylinder 3
Midpoint=7.5, interval of scanning 6th
A = 106, V= height*area of the base = 106*3 =318 cc
Cylinder 4
Midpoint=10.5, interval of scanning 8th
A = 129, V= height*area of the base = 129*3 =387 cc
Cylinder 5
Midpoint=13.5, interval of scanning 10th
A = 38, V= height*area of the base = 38*3 =114 cc
And the estimate volume is
54+234+318+387+114=1,107 [tex]cm^3[/tex]
