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anonymous
 one year ago
can anyone please explain the concept of reiman sums to me?
anonymous
 one year ago
can anyone please explain the concept of reiman sums to me?

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theEric
 one year ago
Best ResponseYou've already chosen the best response.3Riemann sums are not to bad, by themselves. If you've got into "integration" in calculus, then it's probably easy for things to start getting mixed up. There are a few kinds of Riemann sums. I can first tell you what they all are, vaguely. Then there's one little detail that we decide on to have a particular Riemann sum. It's like, I can explain what a dog is, with fur and all, but there are many kinds of dogs, and you need some detail to explain any one kind of dog. So! The Riemann sum looks at a function. This is the standard, "I have this set of values, and it maps to this other set of values!" sort of thing. So, that's something we can draw. Don't worry about the numbers involved  it's just any function.dw:1434006654496:dw The Riemann sum draws rectangles, in a way. You pick a place for the Riemann sum.

theEric
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434006771453:dw So there is where I want to find a Riemann sum, for example. Now I break it up into smaller sections!

theEric
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434006905547:dw However many sections you want, really. And you can make them as wide or skinny as you want. Now, you consider a rectangle in each section, that is as wide as it's section. How tall is it? \(\it {That}\) is the specific part. And it always depends on the function. So, now you know the vague description. You have a function. You look at some piece of the function. Then you split it up into sections, and think that rectangles must sit in those sections (on the axis). But, how tall are they? So, to find the height of the rectangle, for the simple cases, you really just ask, what is the value of the function at \(\it {this}\) part of each section? Example: what is the value at the \(\it{left\ side}\) of each section?

theEric
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434007283820:dwdw:1434007401993:dwAnd that will set the height for each section.

theEric
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434007501204:dw The Riemann sum can find the total area. It looks at its rectangles. Are you familiar with the summation expression, \(\sum\)? Well, a rectangle has area of \(width\times height\). The width is the distance from the left side of a section to the right side. If we say that the right side of the section on an \(x\) axis is \(x_1\), and the lefts side is \(x_0\), then we easily find a width of \(x_1x_0\). This is likedw:1434007891314:dw

theEric
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434007965064:dw It's as if we had these \(x\)'s all along it.dw:1434008045960:dw More generally, the width for some section \(i\) is \(\left(x_ix_{i1}\right)\). So, for section\(1\) specifically, \(i=1\), so \(\left(x_ix_{i1}\right)=\left(x_1x_{11}\right)=\left(x_1x_0\right)\), so it's just like we said before. The height comes from the function. Let's keep using the left side of the function. Say our function is named \(f\). And we'll use it on our \(x\). So, we're looking at \(f(x)\). If we keep with how we were labeling the xaxis, the first section has \(x_0\) on the left. So the height is \(f(x_0)\). This is the first section, so \(i=1\). The left side is one less than the number of the section, how we wrote it. Generally, we're looking at \(f(x_{i1})\). See, for the first section it works out to be \(f(x_{i1})=f(x_{11})=f(x_0)\) So, we have a width of \(x_1x_0\) and height \(f(x_0)\) for the first section. For any section \(i\), we have the width \(\left(x_ix_{i1}\right)\) and the height \(f(x_{i1})\) if we get the height with the left side of each section. Any rectangle then has the area \(\left(x_1x_0\right)\times f(x_{i1})\) To sum the areas of some number \(n\) of the rectangles up, we just do this! \[\sum_i^n\left[\left(x_1x_0\right)\times f(x_{i1})\right]\] If we based the height differently, then the height part would change. Like, if we used the right side, you can check to see that we're using \(f(x_i)\). We could also find the middle of the section, halfway between left and right, by using \(f\left(\dfrac{x_i+x_{i1}}2\right)\). The middle value is the average of the side values, there. Any questions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@theEric thank you so much! How do you know when do use the summation rules or if you just find the area of each individual rectangle and add them together? Are you only allowed to use the summation rules if you don't have the value of n??

theEric
 one year ago
Best ResponseYou've already chosen the best response.3You're welcome! When you apply the Riemann sum, it's like an estimation of the area from the axis to the function for some part of the domain.dw:1434012514231:dw You usually want more rectangles when you want to be accurate, and you'll always add them all up. As for the part about \(n\), that's really there for the "summation notation" to say that we'll look at all the sections (all of the \(i\) values) including the \(n\)th (\(\it{last}\)) section. I don't know what kind of problem we might apply this to, but maybe we can choose how many sections we want. Then we're choosing \(n\). So, I would guess that\(n\) would be easy to know.

theEric
 one year ago
Best ResponseYou've already chosen the best response.3Does that make sense, at the moment?
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