Good. My point - you didn't have numbers and yet you were able to provide the answer. Do you see that using variables in calculations is perfectly normal?Okay.

Area = A x B

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Good. My point - you didn't have numbers and yet you were able to provide the answer. Do you see that using variables in calculations is perfectly normal?Okay.

Area = A x B

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Since the left bound is 0 it is 0 + 1/N = 1/NCan you figure out the right end point of the first interval of length 1/N?

So endpoint number two would just be 1/N + 1/N = 2/N?

Yes, etc.Since the left bound is 0 it is 0 + 1/N = 1/N

So endpoint number two would just be 1/N + 1/N = 2/N?

What about the heights?

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You plug in those x values (the bases) into the equation.Yes, etc.

What about the heights?

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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].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?

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No, this is not correct. 7/N is NOT the left endpoint.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].

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So it would be 7 + 1/NNo, 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.

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I am not sure what you mean. The left most endpoint will be 7, the next one is 7 + 1/N...So it would be 7 + 1/N

Please continue until you get to 10.

7 + 1/N or 7 + 3/N?7, 7+1/N, 7+2/N, 7+3/N, .... 9, 9+1/N, 9+2/N, ..., 9+(N-1)/N, 10

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Into the corner I go. Yep, 7+3/N7 + 1/N or 7 + 3/N?

\(\displaystyle \left \{ \sum_{j=1}^n f \left (a + j * \dfrac{b - a}{n} \right ) * \dfrac{n}{b - a} \right \}.\)

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

\(\displaystyle 1 - 0 = 1.\)

Any problem with that?

Now we are going to divide it into n

Therefore, each sub-interval has a width of \(\displaystyle \dfrac{b - a}{n}.\)

When we want to divide a number into equal parts, we use division, right?

In the case of your problem, \(\displaystyle \dfrac{b - a}{n} = \dfrac{1}{n}.\)

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?

\(\displaystyle \text {Left } a = a + 0 * \dfrac{(b - a)}{n} \text { and Right } = a + 1 * \dfrac{(b - a)}{n}.\)

Make sense? How about the second sub-interval?

\(\displaystyle \text {Left } a = a + 1 * \dfrac{(b - a)}{n} = a + (2 - 1) * \dfrac{(b - a)}{n} \text { and Right } = a + 2 * \dfrac{(b - a)}{n}.\)

How about the third sub-interval?

\(\displaystyle \text {Left } a = a + 2 * \dfrac{(b - a)}{n} = a + (3 - 1) * \dfrac{(b - a)}{n} \text { and Right } = a + 3 * \dfrac{(b - a)}{n}.\)

How about the n

\(\displaystyle \text {Left } a + (n - 1) * \dfrac{(b - a)}{n} \text { and Right } = b = a + n * \dfrac{(b - a)}{n}.\)

I see a pattern here. The endpoints of the j

\(\displaystyle \text {Left } a + (j - 1) * \dfrac{(b - a)}{n} \text { and Right } = b = a + j * \dfrac{(b - a)}{n}.\)

In your problem, because a = 0 and b = 1, b - a reduces to 1 and the endpoints reduce to

\(\displaystyle \text {Left } \dfrac{j - 1}{n} \text { and Right } = \dfrac{j}{n}.\).

Now let's think about the j

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

\(\displaystyle f \left ( \dfrac{j}{n} \right ) * \dfrac{1}{n}.\)

But \(\displaystyle f(x) = x^2 + 1 \implies f \left ( \dfrac{j}{n} \right ) = \dfrac{j^2}{n^2} + 1. \)

So the area of the j

\(\displaystyle \left ( \dfrac{j^2}{n^2} + 1 \right ) * \dfrac{1}{n} = \dfrac{j^2}{n^3} + \dfrac{1}{n}.\)

Now we just add those areas up.

\(\displaystyle \left ( \sum_{j=1}^n \dfrac{j^2}{n^3} + \dfrac{1}{n} \right ) = \left ( \sum_{j=1}^n \dfrac{j^2}{n^3} \right ) + \left ( \sum_{j=1}^n \dfrac{1}{n}\right ) = \)

\(\displaystyle \dfrac{n(n + 1)(2n + 1)}{6n^3} + \dfrac{n}{n} = 1 + \dfrac{2n^3 + 3n^2 + n}{6n^3} = \dfrac{4}{3} + \dfrac{3n^2 + n}{6n^3}.\)

Take the limit of that as n approaches infinity, and you end up with 4/3.

\(\displaystyle \int \ x^2 + 1 \ dx = \dfrac{x^3}{3} + x + C \implies\)

\(\displaystyle \int_0^1 \ x^2 + 1 \ dx = \left ( \dfrac{1^3}{3} + 1 + C \right ) - \left ( \dfrac{0^3}{3} + 0 + C \right) = \dfrac{4}{3}.\)

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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 jobdefinedas follows:

\(\displaystyle \left \{ \sum_{j=1}^n f \left (a + j * \dfrac{b - a}{n} \right ) * \dfrac{n}{b - a} \right \}.\)

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

\(\displaystyle 1 - 0 = 1.\)

Any problem with that?

Now we are going to divide it into nequalsub-intervals.

Therefore, each sub-interval has a width of \(\displaystyle \dfrac{b - a}{n}.\)

When we want to divide a number into equal parts, we use division, right?

In the case of your problem, \(\displaystyle \dfrac{b - a}{n} = \dfrac{1}{n}.\)

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?

\(\displaystyle \text {Left } a = a + 0 * \dfrac{(b - a)}{n} \text { and Right } = a + 1 * \dfrac{(b - a)}{n}.\)

Make sense? How about the second sub-interval?

\(\displaystyle \text {Left } a = a + 1 * \dfrac{(b - a)}{n} = a + (2 - 1) * \dfrac{(b - a)}{n} \text { and Right } = a + 2 * \dfrac{(b - a)}{n}.\)

How about the third sub-interval?

\(\displaystyle \text {Left } a = a + 2 * \dfrac{(b - a)}{n} = a + (3 - 1) * \dfrac{(b - a)}{n} \text { and Right } = a + 3 * \dfrac{(b - a)}{n}.\)

How about the n^{th}sub-interval?

\(\displaystyle \text {Left } a + (n - 1) * \dfrac{(b - a)}{n} \text { and Right } = b = a + n * \dfrac{(b - a)}{n}.\)

I see a pattern here. The endpoints of the j^{th}sub-interval are

\(\displaystyle \text {Left } a + (j - 1) * \dfrac{(b - a)}{n} \text { and Right } = b = a + j * \dfrac{(b - a)}{n}.\)

In your problem, because a = 0 and b = 1, b - a reduces to 1 and the endpoints reduce to

\(\displaystyle \text {Left } \dfrac{j - 1}{n} \text { and Right } = \dfrac{j}{n}.\).

Now let's think about the j^{th}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

\(\displaystyle f \left ( \dfrac{j}{n} \right ) * \dfrac{1}{n}.\)

But \(\displaystyle f(x) = x^2 + 1 \implies f \left ( \dfrac{j}{n} \right ) = \dfrac{j^2}{n^2} + 1. \)

So the area of the j^{th}rectangle is

\(\displaystyle \left ( \dfrac{j^2}{n^2} + 1 \right ) * \dfrac{1}{n} = \dfrac{j^2}{n^3} + \dfrac{1}{n}.\)

Now we just add those areas up.

\(\displaystyle \left ( \sum_{j=1}^n \dfrac{j^2}{n^3} + \dfrac{1}{n} \right ) = \left ( \sum_{j=1}^n \dfrac{j^2}{n^3} \right ) + \left ( \sum_{j=1}^n \dfrac{1}{n}\right ) = \)

\(\displaystyle \dfrac{n(n + 1)(2n + 1)}{6n^3} + \dfrac{n}{n} = 1 + \dfrac{2n^3 + 3n^2 + n}{6n^3} = \dfrac{4}{3} + \dfrac{3n^2 + n}{6n^3}.\)

Take the limit of that as n approaches infinity, and you end up with 4/3.

\(\displaystyle \int \ x^2 + 1 \ dx = \dfrac{x^3}{3} + x + C \implies\)

\(\displaystyle \int_0^1 \ x^2 + 1 \ dx = \left ( \dfrac{1^3}{3} + 1 + C \right ) - \left ( \dfrac{0^3}{3} + 0 + C \right) = \dfrac{4}{3}.\)

JomoNicely 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

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

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

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How did I miss that one!Posr #35 has a confusing typo. In the definition, the second factor should be (b - a)/n.

Well, I wrote it and still missed it. The problem with formulas is that we know what they ought to be and so miss typos.How did I miss that one!