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 June 15th, 2010, 02:11 PM #1 Member   Joined: Feb 2010 Posts: 53 Thanks: 0 Integral as continous summation vs. area computation?? Okay, so I understand that in the follwoing integral: $\int[f(x)]dx$ that $dx$ stands for those infinitetismally small widths of rectangles, and this inturn gives us a continuous summation of the rectangles under f(x). But what if I wanted to use the integral to add up the values of a function at everypoint? Say I had a function $R(t)$ which takes as input time and outputs the amount of snow I had removed from my driveway at that time. If I wanted to add up all the values of $R(x)$ with a continuous summation, from when I started shoveling to 5 hours lates I would use this integral right: $\int_0^5R(t)dt$. Heres what I dont understand, and yet understand, and then dont understand again. Since I just want to add up the values, and not find the area, whats the need for the $dt$, or more appropriately the value of $\Delta t$ in the $\sum$ counterpart of the integral, but whats the need for it? Dont I just want to add up all the values from 0 to 5? So then I try this and realize that I simply get more and more values as I divide the function into smaller and smaller pieces. At first I may be adding $1+2+3= 6$ , but dividing it up more I may be adding $1+1.5+2+2.25+2.6+3= 12.35$, and even more $1+1.23+1.56+1.67+2.3+2.5+2.7+2.8+2.88+2.89+2.9+3= 27.43$ !! This approach will simply go to infinity right!? So I understand that the $dt$ stops this from happening, and I also understand its geomtric interpretation as the width of infinitismally small rectangles. My issues comes with understanding what this ellusive $dt$ means when we are dealing with a continuous summation of a function like my snow shoveling one $R(t)$? Is it an averager? Does it act to divide the function into its average? I know this cant be true because their is a seperate definition for the average of a function right!? So what the hell is $dt$ anyway!? And please refrain from saying "the width of the rectangles", "a really small number", or "the back side of the integration function" as an answer. Thanks so much in advance.
 June 15th, 2010, 03:55 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Re: Integral as continous summation vs. area computation?? You say R(t) is the amount of snow shoveled at time t. If it is instantaneous it would be 0, so I presume you mean R(t) to be the amount shoveled in some interval. That interval is ?t. I you mean R(t) to be a rate, then you need to put in the time unit.
June 15th, 2010, 04:36 PM   #3
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Re: Integral as continous summation vs. area computation??

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 Originally Posted by mathman You say R(t) is the amount of snow shoveled at time t. If it is instantaneous it would be 0, so I presume you mean R(t) to be the amount shoveled in some interval. That interval is ?t. I you mean R(t) to be a rate, then you need to put in the time unit.
You are correct my apologies, yes I mean that R(t) stand for the amount of snow that has been shoveled at that period of time. So maybe at the end of first hour (i.e t= 1) I'd only sovled 3 units of snow, but by the 5th our (i.e. t=5) I may have shoveled 12 untis of snow. That kind of function. What does the amount of snow shoveled multiplied by the infinitismal time increment $\Delta t$, what does this value represent? In mathematical terms, and in terms of the snow shoveled.

June 15th, 2010, 08:53 PM   #4
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Re: Integral as continous summation vs. area computation??

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 Originally Posted by mfetch22 You are correct my apologies, yes I mean that R(t) stand for the amount of snow that has been shoveled at that period of time. So maybe at the end of first hour (i.e t= 1) I'd only sovled 3 units of snow, but by the 5th our (i.e. t=5) I may have shoveled 12 untis of snow.
So in that case you have $R(t)=\int_0^tr(x)\,dx$, where r(t) is the rate of snow shoveling.

Quote:
 Originally Posted by mfetch22 What does the amount of snow shoveled multiplied by the infinitismal time increment $\Delta t$, what does this value represent? In mathematical terms, and in terms of the snow shoveled.
Let's suppose for a moment that you shoveled at a constant rate of 3 units of snow per hour for the first two hours, then slowed to a rate of 2 units of snow per hour afterward. It's easy to see that this gives 3 units of snow at time 1 and 12 units of snow at time 5.

A first approximation of the amount of snow you shoveled (if we didn't know that it was 12 units) would be to take a sample point in the range, say at time 2.5, and multiply that rate by the amount of time, 5 hours. That would give 2 units per hour * 5 hours = 10 units. A better approximation would split it in two parts centered around, say, time 1.25 and time 3.75, with rates 3 units per hour and 2 units per hour; this would give 3 units per hour * 2.5 hours + 2 units per hour * 2.5 hours = 12.5 units. Splitting into finer and finer intervals will eventually get close to the true answer, 12 units.

In these examples, Delta-t was 5 and 2.5, respectively. If you split the range from x = a to b into k parts, Delta-x will be (b - a)/k.

June 16th, 2010, 09:47 AM   #5
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Re: Integral as continous summation vs. area computation??

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Originally Posted by CRGreathouse
Quote:
 Originally Posted by mfetch22 You are correct my apologies, yes I mean that R(t) stand for the amount of snow that has been shoveled at that period of time. So maybe at the end of first hour (i.e t= 1) I'd only sovled 3 units of snow, but by the 5th our (i.e. t=5) I may have shoveled 12 untis of snow.
So in that case you have $R(t)=\int_0^tr(x)\,dx$, where r(t) is the rate of snow shoveling.

Quote:
 Originally Posted by mfetch22 What does the amount of snow shoveled multiplied by the infinitismal time increment $\Delta t$, what does this value represent? In mathematical terms, and in terms of the snow shoveled.
Let's suppose for a moment that you shoveled at a constant rate of 3 units of snow per hour for the first two hours, then slowed to a rate of 2 units of snow per hour afterward. It's easy to see that this gives 3 units of snow at time 1 and 12 units of snow at time 5.

A first approximation of the amount of snow you shoveled (if we didn't know that it was 12 units) would be to take a sample point in the range, say at time 2.5, and multiply that rate by the amount of time, 5 hours. That would give 2 units per hour * 5 hours = 10 units. A better approximation would split it in two parts centered around, say, time 1.25 and time 3.75, with rates 3 units per hour and 2 units per hour; this would give 3 units per hour * 2.5 hours + 2 units per hour * 2.5 hours = 12.5 units. Splitting into finer and finer intervals will eventually get close to the true answer, 12 units.

In these examples, Delta-t was 5 and 2.5, respectively. If you split the range from x = a to b into k parts, Delta-x will be (b - a)/k.
This helps, but lets say we just have some function $f(x)$ and I want to know what:

$\int_a^b[f(x)]dx$

stands for mathematically without referencing to anything geometric or anything about its being the opposite of the derivitive. What I mean is what is this value mathematically, what does it mean to the function $f(x)$ in no geometric terms? Or is there only a geomtric definition? My whole point of this question is to better understand how one can use integrals to continuosly add up cetain things. One thing where I am confused is the line integral over a vector field, being:

$\int_C\;W(\psi(t)) \cdot \frac{d\psi}{dt}dt$

I understand that this integral adds up all the vectors along this path of curve C, but wouldn't that just be:

$\int_C\;W(\psi(t))dt$???

I know it isnt, but I'm just confused as to where or what the $\frac{d\psi}{dt}$. I'm fimmiliar with vectors and their operations, and the dot and cross product so thats not neccesarily the problem. Is the line integral over a vector field to be considered as adding up all the vectors along the curve continuously? Or is there a better interpretation?

June 16th, 2010, 10:17 AM   #6
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Re: Integral as continous summation vs. area computation??

Quote:
 Originally Posted by mfetch22 This helps, but lets say we just have some function $f(x)$ and I want to know what: $\int_a^b[f(x)]dx$ stands for mathematically without referencing to anything geometric or anything about its being the opposite of the derivitive.
$\int_a^bf(x)\,dx=\lim_{n\to\infty}\sum_{k=1}^nf\le ft(a+k\frac{b-a}{n}\right)\frac{b-a}{n}$, if you like. There are many equivalent definitions.

The geometric interpretation is just there to help explain where this formula comes from.

 June 16th, 2010, 03:55 PM #7 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: Integral as continous summation vs. area computation?? CRGreathouse's definition (actually, an equivalent, but slightly more general looking) is the standard one you'll see. The (geometric) intuition for the formula is that the product $f\left(a+k\frac{b-a}{n}\right)\frac{b-a}{n}$ is the area under the curve with small, uniform-width rectangles. As the number of rectangles goes to infinity (and hence, the width of each goes to 0), the value of that sum approaches the integral. If you're still confused about the "dx" at the end, just think of this as denoting what function you're integrating with respect to. You can actually integrate f(x) with respect to g(x). $\int_a^b[f(x)]d(g(x))$ And the definition of the integral that CRG gave changes a bit... Modifying the exact formula CRG gave is difficult, but instead of the width of each rectangle, you use a "weighted width", whose weight is dependent on the values of g at the boundaries of the rectangle.

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