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A log 10 $\mathrm{m}$ long is cut at 1 -meter intervals and its cross-

sectional areas $A$ (at a distance $x$ from the end of the log)

are listed in the table. Use the Midpoint Rule with $n=5$

to estimate the volume of the log.

$V=5.8 m^{3}$

Applications of Integration

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Campbell University

Harvey Mudd College

Idaho State University

Boston College

{'transcript': "So in finding the approximate volume of the log with the midpoint rule, we're going to be splitting the log. So if this is our log here, we're gonna split it into five parts. 12345 Um, and each of them are going to be 2 m long because it's 10 m long log, and then we find the volume of each of these individual chunks and then we add them all up. So we have is the volume from and equals 125 That's our number of chunks. And then we have a and Doctor X where a is going to be where Delta X is going to be R 2 m and A as can be dependent on, um, our area. So now what we have is, um, we have our mid points are going to be at 1357 and 9 m. So what we're going to have is a one equals 10.65 A two equals 61 A three equals 0.59 A four equals 55 A five equals 50 We add all these up and then we multiply it by two because that's our Delta X and we end up getting 58 m cubed, which is the approximate volume of the lock."}

California Baptist University

Applications of Integration