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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.

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$V=5.8 m^{3}$

Calculus 2 / BC

Chapter 7

APPLICATIONS OF INTEGRATION

Section 2

Volumes

Applications of Integration

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Lectures

0:00

A log 10 m long is cut at …

A log 10$\mathrm { m }$ lo…

01:10

05:25

A log 15 m long is cut at …

02:06

Express the signed area be…

02:26

Approximate the area under…

01:09

Use a calculating utility …

03:21

In the following exercises…

00:30

00:52

As mentioned earlier, we w…

So what we see is we want to approximate the volume of the log using the midpoint rule and we're splitting it into five chunks. So the volume is going to be equal to the sum from an equals 1-5 of the cross sectional area times the change in the X. The depth. And that's how we'll find volume. So we see that the first chunk will be from 0 to 2 m and the second will be 2 to 44 to six and so on. So that gives us that our delta X equals two m. Our cross sectional area were given in the table, so we refer to the table for all eight to end values. So what that's gonna end up giving us for the volume is two times 20.65 plus 0.61 plus 0.59 plus 0.55 Plus .50. We add all this up and that's going to give us 5.8 and remember this is volume. So it's gonna be meters cute. That will be the final answer for the volume. Um, and that's using uh, some method that eventually will lead more into integration, but this is kind of the first step leading into integration.

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