Calculating Integrals Using the Fundamental Theorem of Calculus

In summary: You might look at the Riemann-Stieltjes integral, for example.But why are we taking N which is the number of intervals to be a natural numberIf you don't take a natural number of intervals, then there is no way that ##\Delta x## can be a rational number. And if ##\Delta x## is not a rational number, then the "lower sum" and "upper sum" don't make sense, and you have no way to calculate the integral. That is, you cannot add up an infinite number of irrational numbers to get a rational number. (You can't even add a finite number of irrational
  • #1
Crystal037
167
7
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)
 
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  • #2
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
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
Also can you give me an idea on how to take limits on summation
 
  • #5
Crystal037 said:
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
Crystal037 said:
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
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
Please give me some kind of proper solution or a direction on how to get to the result
 
  • #9
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
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
Crystal037 said:
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
But why are we taking N which is the number of intervals to be a natural number
 
  • #13
Crystal037 said:
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.
 
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  • #14
But it can be any rational number or real numer why not irrational number
 
  • #15
Crystal037 said:
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
Can you explain me the interval part with some example
I am still not getting why?
 
  • #17
Andrew Mason said:
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
...
Crystal037 said:
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
PeroK said:
...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
Crystal037 said:
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
Andrew Mason said:
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
Crystal037 said:
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
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
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
jambaugh said:
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
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
Crystal037 said:
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 ##
 
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  • #28
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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Related to Calculating Integrals Using the Fundamental Theorem of Calculus

1. What is integration from summation?

Integration from summation is a mathematical technique used to find the area under a curve by summing up smaller sections of the curve. It is also known as the Riemann sum method.

2. How is integration from summation different from other integration methods?

Integration from summation is different from other integration methods, such as the Trapezoidal rule or Simpson's rule, because it uses discrete data points instead of continuous functions. This makes it more suitable for solving problems with discrete data or irregular curves.

3. What are the steps involved in integration from summation?

The steps involved in integration from summation include dividing the curve into smaller sections, finding the area of each section, and then summing up all the areas to get an approximation of the total area under the curve. As the number of sections increases, the approximation becomes more accurate.

4. What are the applications of integration from summation?

Integration from summation has various applications in fields such as physics, engineering, and economics. It is used to calculate the total distance traveled by an object with varying velocity, to find the total work done by a varying force, and to estimate the total cost of production with varying inputs, among others.

5. Are there any limitations to integration from summation?

Yes, there are limitations to integration from summation. It can only provide an approximation of the total area under a curve, and the accuracy of the approximation depends on the number of sections used. It also requires the curve to be divided into smaller sections, which can be time-consuming for complex curves.

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