Integration from Summation

  • Thread starter Crystal037
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  • #1
Crystal037
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Homework Statement:
we know that integration is continuous summation
If we take a graph of x=y and let pts A and B on x-axis represent x=a and x=b
and try to find the area under the curve between a and b
We divide the length AB in N equal intervals.
Then the length of each interval is Δx=(b-a)/N.
The height of the first shaded bar is y=x=a, of second bar is y=x=a+Δx, that of third bar is y=x=a+2Δx. The height of the Nth bar is y=x=a+(N-1)Δx. The width of each bar is Δx, then total area of all bars is
I=aΔx+(a+Δx)Δx+(a+2Δx)Δx+.........+[a+(N-1)Δx]Δx
={a+(a+Δx)+(a+2Δx)+.........+[a+(N-1)Δx]}Δx ....(1)
=∑xi(Δx) where 1<=i<N
As Δx tends to 0 the total area of the bars becomes the area under the curve.
thus the required area is I=lim Δx tends to 0 ∑xi(Δx) where 1<=i<N
Now the terms making the series in curly bracket in eq(1) are in arithmetic progression
hence this series may be summed up using formulae S=n/2(a+l).
Thus, I=N/2{a+[a+(N-1)Δx]}Δx
=NΔx/2{2a+NΔx-Δx}
=b-a/2{2a+b-a-Δx}
=b-a/2{a+b-Δx}
Thus the area under curve is
I=lim Δx tends to 0 (b-a)/2(a+b-Δx)
=(b-a)/2(a+b)
1/2(b^2-a^2)
Hence integral from a to b = 1/2(b^2-a^2)
Similarly using summation find the integral of y=x^2 from a to b
Relevant Equations:
y=x^2
Here, width of first bar, y=x^2=a^2
y=x^2=(a+Δx)^2
height of nth bar=y=(a+(N-1)Δx)^2
Total area,I={a^2+(a+Δx)^2+(a+2Δx^2)+...+[a+(N-1)Δx]^2}Δx
I={Na^2 + 2aΔx +...}
I can't seem to get forward to get the required result which is 1/3(b^3-a^3)
 

Answers and Replies

  • #2
Andrew Mason
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Try using: [tex]Area = \lim_{N\rightarrow\infty}\sum_{i=1}^{N} f(x_i)\Delta x[/tex]

where ##f(x_i) = x_i^2 = (a + i(\Delta x))^2## and ##\Delta x = \frac{b-a}{N}##

AM
 
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  • #3
Crystal037
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N
Area= lim ∑ xi(Δx)
Δx→0 i=1

Both of these are same but I am not getting the steps to reach the result 1/3(b^3-a^3)
 
  • #4
Crystal037
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Also can you give me an idea on how to take limits on summation
 
  • #5
PeroK
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Also can you give me an idea on how to take limits on summation

This is really just an exercise is some suprisingly complicated algebra:

1) Write down the sum and expand the terms.

2) Calculate (or look up) the formula for ##\sum_{1}^{N}i## and ##\sum_{1}^{N}i^2##.

3) Take the limit as ##N \rightarrow \infty##, which simplifies things.

4) Simplify to the final expression.
 
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  • #6
Andrew Mason
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N
Area= lim ∑ xi(Δx)
Δx→0 i=1

Both of these are same but I am not getting the steps to reach the result 1/3(b^3-a^3)
I think you will need to work out:
[tex]\sum_{i=1}i^{2} [/tex]

using a clever summation technique similar to

[tex]\sum_{i=1}^{N} i = N(N+1)/2[/tex] which approaches [tex]N^2/2\text{ as }N\rightarrow\infty[/tex]

which you used in the first part. Showing how that is done requires either a fair degree of cleverness or moderate googling skills...

AM
 
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  • #7
WWGD
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You will likely just get an expression for the sum of squares that is a "polynomial" ( on n) of degree 3 , divided by 3, and then you can ignore the lower degree terms. Similar to the way that ##\Sigma i = n(n+1)/2 =(n^2+n)/2 ##
 
  • #8
Crystal037
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Please give me some kind of proper solution or a direction on how to get to the result
 
  • #9
WWGD
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Please see my above. Look for a closed-form formula for the sum of squares in a book or online. It will be an expression in powers of n. The highest power will dominate the others , which you can then ignore, as you approach infinity.
 
  • #10
Crystal037
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The expression for sum of squares is N(N+1)(2N+1)/6 =(2N^3+3N^2+N)/6
Taking the highest power of N we'll get 2N^3/6=N^3/3
But this formula is when n belongs to natural numbers
But here why are we taking n to be a natural number
 
  • #11
WWGD
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The expression for sum of squares is N(N+1)(2N+1)/6 =(2N^3+3N^2+N)/6
Taking the highest power of N we'll get 2N^3/6=N^3/3
But this formula is when n belongs to natural numbers
But here why are we taking n to be a natural number
As you go to infinity with natural numbers, your numbers remain natural numbers: n, n+1,..., n+k,...
 
  • #12
Crystal037
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But why are we taking N which is the number of intervals to be a natural number
 
  • #13
WWGD
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But why are we taking N which is the number of intervals to be a natural number
The number of intervals must be a natural number. You cannot have, e.g., 3.7 or ##\pi## intervals. It must be a whole number. And this is what the formula tells you, asks you to do.
 
  • #14
Crystal037
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But it can be any rational number or real numer why not irrational number
 
  • #15
WWGD
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But it can be any rational number or real numer why not irrational number
How can you have 3.7 intervals? How can it be a Rational that is not an integer. How do you take 1.7 times the interval [0,1]? How do you divide the interval [a,b] into 1.7 or ## \mathbb e ## parts?
 
  • #16
Crystal037
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Can you explain me the interval part with some example
I am still not getting why?
 
  • #17
Crystal037
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Try using: [tex]Area = \lim_{N\rightarrow\infty}\sum_{i=1}^{N} f(x_i)\Delta x[/tex]

where ##f(x_i) = x_i^2 = (a + i(\Delta x))^2## and ##\Delta x = \frac{b-a}{N}##

AM
shouldn't lower bound of summation i=0 rather than i=1 since you have kept the function f(xi)=(a+i(Δx))^2
as for the area under the function we had a terms (a+a+Δx+a+2Δx+...+a+9(N-1)Δx)Δx
 
  • #18
PeroK
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But why are we taking N which is the number of intervals to be a natural number

Why don't you divide ##[0, 1]## into 1.5 intervals for us? Show us how it's done.
 
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  • #19
WWGD
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Why don't you divide ##[0, 1]## into 1.5 intervals for us? Show us how it's done.
I'd prefer ##\pi## intervals, or 2.8765213 , but 1.5 will do it too.
 
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  • #20
Andrew Mason
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shouldn't lower bound of summation i=0 rather than i=1 since you have kept the function f(xi)=(a+i(Δx))^2
as for the area under the function we had a terms (a+a+Δx+a+2Δx+...+a+9(N-1)Δx)Δx
The difference is not material. The only reason to start at 1 is that the number of intervals is N. If you start at 0, the number of intervals has to be N-1. But in the end, the limits of N(N-1)/2 and N(N+1)/2 are both ##N^2/2## as ##N \rightarrow \infty##.
 
  • #21
PeroK
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The difference is not material. The only reason to start at 1 is that the number of intervals is N. If you start at 0, the number of intervals has to be N-1. But in the end, the limits of N(N-1)/2 and N(N+1)/2 are both ##N^2/2## as ##N \rightarrow \infty##.

With ##1## to ##N## you get the upper sum, with the function taken at its greatest value in each interval. With ##0## to ##N-1## you get the lower sum, with the function taken at its least value in each interval.

The two sums, of course, converge as ##N \rightarrow \infty##.
 
  • #22
kimbyd
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Can you explain me the interval part with some example
I am still not getting why?
Half an interval isn't a meaningful thing. You could have an interval of a different size, but that's not half an interval: it's just a different-sized interval.

So if, say, you wanted to divide a range into "3.5 intervals", one way of doing that is having 3 intervals of size 1, and 1 interval of size 0.5. You really have four intervals, of course, but "3.5" describes the system in a comprehensible way.

You can certainly do something like that. But why would you want to? It adds unnecessary complexity to the calculation and provides a mechanism for errors to creep into the calculation that you might never expect. I definitely wouldn't even consider trying anything that looks like a non-integer number of intervals without a compelling reason to do so.

Aside: using intervals of varying size is a really useful technique for numeric integration, it turns out, because functions tend to vary a lot in how many samples are required to get a good integral approximation at different locations. Good numerical integration libraries tend to use a small number of samples to start, then subdivide regions if the estimated error in those regions is too high.
 
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  • #23
WWGD
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Problem Statement: we know that integration is continuous summation
Not quite. It is a form of sampling, taking a very large sample for a population of function values. A sum over a continuous index with more than countably-many non-zero terms is not defined, i.e., will not converge.
 
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  • #24
jambaugh
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I think there are several reasons we don't worry about the possibility of these residual fractional intervals. One is the numbers here are fundamentally counts of things that are not naturally subdivided in a unique way and the other is that since the count is going to infinity in the limit (which is the more important point) the effect of any fractional piece on the, intentionally, diverging limit of the count will necessarily, in the limit, become inconsequential.

You are experimenting with what happens to a camel as you add a number, n of inch long straws to the pack on it's back. You find that in the limit as n goes to infinity the camel's back breaks. This limit is unaffected if you use n - 1 inch long straws and one inch and a quarter long straw.

I think the worry here is because in the definition of a Riemann sum there are actually two limits going on here. To define a (more general) Riemann sum you subdivide the interval into, say ##n## subintervals and you also measure the longest sub-interval an give that, say ##d_max##.

For the limit of the Riemann sums to converge to a unique, well defined value (when that's possible) you need, most specifically that the maximum length converge to zero ## d_\max= \Delta x_\max \to 0##. Not this necessarily implies that for non-zero-lengthed intervals that the number of subdivisions ##n## diverge to infinity. That's an inference and not the main premise in the definition of a Riemann sum.

I'm coming to this discussion late but hope this is to the point and helpful in the discussion.
 
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  • #25
WWGD
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I think there are several reasons we don't worry about the possibility of these residual fractional intervals. One is the numbers here are fundamentally counts of things that are not naturally subdivided in a unique way and the other is that since the count is going to infinity in the limit (which is the more important point) the effect of any fractional piece on the, intentionally, diverging limit of the count will necessarily, in the limit, become inconsequential.

You are experimenting with what happens to a camel as you add a number, n of inch long straws to the pack on it's back. You find that in the limit as n goes to infinity the camel's back breaks. This limit is unaffected if you use n - 1 inch long straws and one inch and a quarter long straw.

I think the worry here is because in the definition of a Riemann sum there are actually two limits going on here. To define a (more general) Riemann sum you subdivide the interval into, say ##n## subintervals and you also measure the longest sub-interval an give that, say ##d_max##.

For the limit of the Riemann sums to converge to a unique, well defined value (when that's possible) you need, most specifically that the maximum length converge to zero ## d_\max= \Delta x_\max \to 0##. Not this necessarily implies that for non-zero-lengthed intervals that the number of subdivisions ##n## diverge to infinity. That's an inference and not the main premise in the definition of a Riemann sum.

I'm coming to this discussion late but hope this is to the point and helpful in the discussion.
I think you misunderstood the situation. I completely agree with you; the person I referred to seems not to see it this way.
 
  • #26
scottdave
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So if you have a string of a certain length (say 1 meter) and then you make a cut - no matter where you make the cut, you now have 2 pieces of string. If you make another cut, then you have 3 pieces of string, and so on. If you did not cut into equal lengths, then that means each piece will have a different value for ## \Delta x##
 
  • #27
Stephen Tashi
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Total area,I={a^2+(a+Δx)^2+(a+2Δx^2)+...+[a+(N-1)Δx]^2}Δx

It should say ##(a +2 \triangle x)^2## instead of ##(a + 2\triangle x^2)##.


## = \{ a^2 + ( a^2 + 2\triangle x + \triangle x^2) + (a^2 + (2)(2)\triangle x + 2^2 \triangle x^2) + (a^2 + (2)(3)\triangle x + 3^2 \triangle x^2) + ... \} \triangle x ##

## = \{ ( a^2 + a^2 + a^2 + ...) + (2\triangle x + (2)(2)\triangle x + (2)(3)\triangle x + ...) + (\triangle x^2 + 2^2 \triangle x^2 + 3^2 \triangle x^2 + ...) \} \triangle x ##

## = \{ \sum_{i=1}^N a^2 + \sum_{i=1}^{N-1} 2 i \triangle x + \sum_{i=1}^{N-1} i^2 \triangle x^2\} \triangle x ##

## = \{ N a^2 + \triangle x \sum_{i=1}^{N-1} i + \triangle x^2 \sum_{i=1}^{N-1} i^2 )\} \triangle x ##

So the sums you need are sums of natural numbers. The "closed form" forumulas for the sums are functions of ##N##.

To examine the limit of the above sums as ##\triangle x \rightarrow 0##, you must use the fact that ##N = (b-a)/ \triangle x ##
 
  • #28
vanhees71
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This is a convincing example for, why nothing is more practical than abstract thought in mathematics. Instead of bothering with calculating the (Riemann) integral directly from its definition just develop the theory further until the point of the fundamental theorem of calculus, i.e., that for a continuous function,
$$\int_a^x \mathrm{d} x' f(x')=F(x)$$
one has
$$F'(x)=f(x).$$
Then it's very easy to calculate the integral for ##f(x)=x^2##.
 

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