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- anonymous

List x1, x2, x3, x4 where xi is the left endpoint of the four equal intervals used to estimate the area under the curve of f(x) between x = 4 and x = 6.

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- anonymous

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- zzr0ck3r

can you break this into \(4\) intervals \((4,6)\)?

- anonymous

I'm not understanding what it's looking for here.....
@zzr0ck3r How would I do that? I don't recall anything like that.

- zzr0ck3r

I think I will just show you, and then it will make sense
\(x_1=4, x_2=4.5, x_3=5,\) and \(x_4=6\)

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- anonymous

So how did you choose those four numbers? I understand they are between 4 to 6 but is a random selection?

- zzr0ck3r

it would be the only way to break the interval \([4,6]\) into \(4\) parts.

- anonymous

Don't think so
https://gyazo.com/f25d718a61e95059646b790eba350188

- zzr0ck3r

I meant \(x_4=5.5\) sorry

- zzr0ck3r

want me to explain why?

- anonymous

Please

- zzr0ck3r

When we take the integral of something we are taking the area under the function, imagine you want to take the integral of the function \(f(x) = 2x+3\) from \(4\) to \(6\)
The graph would look something like this|dw:1441154162616:dw|

- zzr0ck3r

we want the are under the graph from \(x=4, \) to \(x=6\). One way to approximate this area would be to break it up into four sections
|dw:1441154252671:dw|
Now we would estimate the area of each part
We treat each part as a rectangle, so the area is 0.5 (the width of the rectangle) times \(f(4.5)\) . So the area of the first rectangle is estimated by \(0.5*f(4.5)\) and we get this area (approximately)
|dw:1441154442521:dw|

- zzr0ck3r

If we break this into infinite pieces, we would get the exact area, not just the approximate.

- anonymous

I understand why you did so...just how? Like why .5 and not .3?

- zzr0ck3r

if I give you two dollars and ask you to break it into 4 even pieces, what would the size of each one be?

- anonymous

Ok, that's why, the pieces must be even. So why wasn't the 6 included? Is it because it's the last and should already be known?

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