Right endpoint approximation part II

Can you figure out the right end point of the first interval of length 1/N?
 
1/N, 2/N, 3/N,....N/N=1 = right endpoint.

Try this one. Suppose the interval is from 7 to 10. Can you label the endpoints?
 
Jomo said:
1/N, 2/N, 3/N,....N/N=1 = right endpoint.

Try this one. Suppose the interval is from 7 to 10. Can you label the endpoints?
Click to expand...

So I am just adding 1/N to the previous endpoint to find the new right endpoint? If 7/N is the left endpoint, then you would have 8/N, 9/N and 10/N as the sub intervals over the interval [7,10].
 
Hckyplayer8 said:
So I am just adding 1/N to the previous endpoint to find the new right endpoint? If 7/N is the left endpoint, then you would have 8/N, 9/N and 10/N as the sub intervals over the interval [7,10].
Click to expand...
No, this is not correct. 7/N is NOT the left endpoint. 7 is the left endpoint. And the base width is 1/N. Please try again to label the endpoints. you need to get this down before we proceed to the next step.
 
Jomo said:
No, this is not correct. 7/N is NOT the left endpoint. 7 is the left endpoint. And the base width is 1/N. Please try again to label the endpoints. you need to get this down before we proceed to the next step.
Click to expand...

So it would be 7 + 1/N
 
7, 7+1/N, 7+2/N, 7+3/N, .... 9, 9+1/N, 9+2/N, ..., 9+(N-1)/N, 10

Now suppose \(\displaystyle f(x) = x^2 + x\). Approximate the area under the curve of f(x) from x=3 to x=7 using N rectangles.
 
I'd probably find it easier to describe the endpoints as [MATH]7, 7 + \frac{3}{N}, 7 + \frac{6}{N}, \dots, 7 + \frac{3k}{N}, \dots, 7 + \frac{3(N-1)}{N}, 7 + \frac{3N}{N} = 10[/MATH], with k ranging from 0 to N (the first right endpoint being [MATH]7 + \frac{3}{N}[/MATH]).
 
Let's start at the very beginning. The right Riemann Sum is defined as follows:

[MATH]\left \{ \sum_{j=1}^n f \left (a + j * \dfrac{b - a}{n} \right ) * \dfrac{n}{b - a} \right \}.[/MATH]
This formula tells you exactly what to do, each step of the way.

What is b - a? It is the length of the interval. In the case of your problem, it is

[MATH]1 - 0 = 1.[/MATH]
Any problem with that?

Now we are going to divide it into n equal sub-intervals.

Therefore, each sub-interval has a width of [MATH]\dfrac{b - a}{n}.[/MATH]
When we want to divide a number into equal parts, we use division, right?

In the case of your problem, [MATH]\dfrac{b - a}{n} = \dfrac{1}{n}.[/MATH]
Let's number the sub-intervals from left to right.

What are the boundaries of n sub-intervals along the x-axis over [a, b]? How about the very first sub-interval?

[MATH]\text {Left } a = a + 0 * \dfrac{(b - a)}{n} \text { and Right } = a + 1 * \dfrac{(b - a)}{n}.[/MATH]
Make sense? How about the second sub-interval?

[MATH]\text {Left } a = a + 1 * \dfrac{(b - a)}{n} = a + (2 - 1) * \dfrac{(b - a)}{n} \text { and Right } = a + 2 * \dfrac{(b - a)}{n}.[/MATH]
How about the third sub-interval?

[MATH]\text {Left } a = a + 2 * \dfrac{(b - a)}{n} = a + (3 - 1) * \dfrac{(b - a)}{n} \text { and Right } = a + 3 * \dfrac{(b - a)}{n}.[/MATH]
How about the nth sub-interval?

[MATH]\text {Left } a + (n - 1) * \dfrac{(b - a)}{n} \text { and Right } = b = a + n * \dfrac{(b - a)}{n}.[/MATH]
I see a pattern here. The endpoints of the jth sub-interval are

[MATH]\text {Left } a + (j - 1) * \dfrac{(b - a)}{n} \text { and Right } = b = a + j * \dfrac{(b - a)}{n}.[/MATH]
In your problem, because a = 0 and b = 1, b - a reduces to 1 and the endpoints reduce to

[MATH]\text {Left } \dfrac{j - 1}{n} \text { and Right } = \dfrac{j}{n}.[/MATH].

Now let's think about the jth rectangle

The height is simply f(the right end point of the sub-interval) and the width is 1 / n so the area is their product

[MATH]f \left ( \dfrac{j}{n} \right ) * \dfrac{1}{n}.[/MATH]
But [MATH]f(x) = x^2 + 1 \implies f \left ( \dfrac{j}{n} \right ) = \dfrac{j^2}{n^2} + 1. [/MATH]
So the area of the jth rectangle is

[MATH]\left ( \dfrac{j^2}{n^2} + 1 \right ) * \dfrac{1}{n} = \dfrac{j^2}{n^3} + \dfrac{1}{n}.[/MATH]
Now we just add those areas up.

[MATH]\left ( \sum_{j=1}^n \dfrac{j^2}{n^3} + \dfrac{1}{n} \right ) = \left ( \sum_{j=1}^n \dfrac{j^2}{n^3} \right ) + \left ( \sum_{j=1}^n \dfrac{1}{n}\right ) = [/MATH]
[MATH]\dfrac{n(n + 1)(2n + 1)}{6n^3} + \dfrac{n}{n} = 1 + \dfrac{2n^3 + 3n^2 + n}{6n^3} = \dfrac{4}{3} + \dfrac{3n^2 + n}{6n^3}.[/MATH]
Take the limit of that as n approaches infinity, and you end up with 4/3.

[MATH]\int \ x^2 + 1 \ dx = \dfrac{x^3}{3} + x + C \implies[/MATH]
[MATH]\int_0^1 \ x^2 + 1 \ dx = \left ( \dfrac{1^3}{3} + 1 + C \right ) - \left ( \dfrac{0^3}{3} + 0 + C \right) = \dfrac{4}{3}.[/MATH]
 
JeffM said:
Let's start at the very beginning. The right Riemann Sum is defined as follows:

[MATH]\left \{ \sum_{j=1}^n f \left (a + j * \dfrac{b - a}{n} \right ) * \dfrac{n}{b - a} \right \}.[/MATH]
This formula tells you exactly what to do, each step of the way.

What is b - a? It is the length of the interval. In the case of your problem, it is

[MATH]1 - 0 = 1.[/MATH]
Any problem with that?

Now we are going to divide it into n equal sub-intervals.

Therefore, each sub-interval has a width of [MATH]\dfrac{b - a}{n}.[/MATH]
When we want to divide a number into equal parts, we use division, right?

In the case of your problem, [MATH]\dfrac{b - a}{n} = \dfrac{1}{n}.[/MATH]
Let's number the sub-intervals from left to right.

What are the boundaries of n sub-intervals along the x-axis over [a, b]? How about the very first sub-interval?

[MATH]\text {Left } a = a + 0 * \dfrac{(b - a)}{n} \text { and Right } = a + 1 * \dfrac{(b - a)}{n}.[/MATH]
Make sense? How about the second sub-interval?

[MATH]\text {Left } a = a + 1 * \dfrac{(b - a)}{n} = a + (2 - 1) * \dfrac{(b - a)}{n} \text { and Right } = a + 2 * \dfrac{(b - a)}{n}.[/MATH]
How about the third sub-interval?

[MATH]\text {Left } a = a + 2 * \dfrac{(b - a)}{n} = a + (3 - 1) * \dfrac{(b - a)}{n} \text { and Right } = a + 3 * \dfrac{(b - a)}{n}.[/MATH]
How about the nth sub-interval?

[MATH]\text {Left } a + (n - 1) * \dfrac{(b - a)}{n} \text { and Right } = b = a + n * \dfrac{(b - a)}{n}.[/MATH]
I see a pattern here. The endpoints of the jth sub-interval are

[MATH]\text {Left } a + (j - 1) * \dfrac{(b - a)}{n} \text { and Right } = b = a + j * \dfrac{(b - a)}{n}.[/MATH]
In your problem, because a = 0 and b = 1, b - a reduces to 1 and the endpoints reduce to

[MATH]\text {Left } \dfrac{j - 1}{n} \text { and Right } = \dfrac{j}{n}.[/MATH].

Now let's think about the jth rectangle

The height is simply f(the right end point of the sub-interval) and the width is 1 / n so the area is their product

[MATH]f \left ( \dfrac{j}{n} \right ) * \dfrac{1}{n}.[/MATH]
But [MATH]f(x) = x^2 + 1 \implies f \left ( \dfrac{j}{n} \right ) = \dfrac{j^2}{n^2} + 1. [/MATH]
So the area of the jth rectangle is

[MATH]\left ( \dfrac{j^2}{n^2} + 1 \right ) * \dfrac{1}{n} = \dfrac{j^2}{n^3} + \dfrac{1}{n}.[/MATH]
Now we just add those areas up.

[MATH]\left ( \sum_{j=1}^n \dfrac{j^2}{n^3} + \dfrac{1}{n} \right ) = \left ( \sum_{j=1}^n \dfrac{j^2}{n^3} \right ) + \left ( \sum_{j=1}^n \dfrac{1}{n}\right ) = [/MATH]
[MATH]\dfrac{n(n + 1)(2n + 1)}{6n^3} + \dfrac{n}{n} = 1 + \dfrac{2n^3 + 3n^2 + n}{6n^3} = \dfrac{4}{3} + \dfrac{3n^2 + n}{6n^3}.[/MATH]
Take the limit of that as n approaches infinity, and you end up with 4/3.

[MATH]\int \ x^2 + 1 \ dx = \dfrac{x^3}{3} + x + C \implies[/MATH]
[MATH]\int_0^1 \ x^2 + 1 \ dx = \left ( \dfrac{1^3}{3} + 1 + C \right ) - \left ( \dfrac{0^3}{3} + 0 + C \right) = \dfrac{4}{3}.[/MATH]
Click to expand...
Nicely done. I was think about showing Hckeyplayer8 the formula but chose the longer way. I have been thinking I was wrong mainly because hckeyplayer8 was not getting it and my method always seems to be a little different for different formulas. Nice job
 
Jomo said:
Nicely done. I was think about showing Hckeyplayer8 the formula but chose the longer way. I have been thinking I was wrong mainly because hckeyplayer8 was not getting it and my method always seems to be a little different for different formulas. Nice job
Click to expand...
Jomo

One thought I have frequently had is that trying to teach an abbreviated version of analysis to start calculus is backwards. Most students find it hard and see no purpose in it. Yes, the logical foundation of calculus lies in analysis, but using calculus does not require any understanding of analysis. Nor do we start teaching students algebra by introducing the Peano postulates. Foundations came historically after what sits on those foundations. Psychology mirrors history.

I suspect that the only pedagogically persuasive reason to start with any form of analysis is to convince students that calculus has some logical basis. I further suspect that that could perhaps be done with a far simpler structure than either standard or non-standard analysis.

But if we are going to teach an introduction to analysis in calculus, I'd prefer to base it on the very carefully defined concepts of Newton quotient and Riemann sum and present them as recipes to be followed
 
Posr #35 has a confusing typo. In the definition, the second factor should be (b - a)/n.
 
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