Question

integral question
will post with editor

Answer 1

\[\int\limits_{0}^{6}(6x+1)dx \] compute as a limit of a Riemann square

Answer 2

if it is two different equations not sure how to get different values for a nad b

Answer 3

\[\int\limits_{0}^{b}f(x)-\int\limits_{0}^{a}f(x)\]

Answer 4

c=3?

Answer 5

Although I don't have the energy to stick with you throughout the entire solution of this problem, perhaps I can give you info to help you get started. First of all, you don't need\[\int\limits\limits_{0}^{b}f(x)-\int\limits\limits_{0}^{a}f(x)\]

Answer 6

ok

Answer 7

You are integrating from x=0 to =6. Thus, your a = 0 and your b=6.

Supposing that you subdivide the interval [0,6] into n subdivisions, how wide is each subdivision? Call it delta x:\[\Delta x=\frac{ b-a }{ n }\]

With your known b values, and the general divisor n, what is the width of each subinterval?

Answer 8

You will be doing this calculation again and again when working with Riemann Sums.

Answer 9

That's fine, 'though I'd prefer you label it and call it delta x:\[\Delta x=\frac{ b-a }{ n }=\frac{ 6-0 }{ n }=\frac{ 6 }{ n }\]

Answer 10

Next, DH, you need to develop a formula for the n x-values that you're going to use.
Have you developed such formulas before?

Answer 11

\[\Delta x=6/n\]

Answer 12

Great. Now a formula for \[x _{i}\]might be ... what ??

Answer 13

that letter " i " is just a counter; it begins with 1 and ends with n.

Answer 14

f(ck)=(ck)^2 = (ka/n)^2 = k^2a^2/n^2

Answer 15

ive done loops in another class

Answer 16

I would like to present the formula I normally use for these x values. Remembering that i is just a counter, \[x _{i}=a+(n-1)*\Delta x\]with i beginning at 1 and ending at n. Have you seen this formula before in the context of Riemann Sums?

Answer 17

yes, with a different notation

Answer 18

6 [ (6^2/2) - (0^2/2) ] = 108
i know that is wrong

Answer 19

All right, DH. Since our a = 0, the formula for the n x-values then becomes\[x _{i }=(i-1)\frac{ 6 }{ n }\] with i starting at 1 and ending at n.

Answer 20

Takes a while to get through this kind of problem...please let's do it step by step.

Answer 21

ok, thank you

Answer 22

so Xo=-6/n ?

Answer 23

For example, if i = 1, 2, 3, we get the x-values 6/n, 2(6/n), 3(6/n), and so on. OK with that?
Actually, DH, I'm using "right end points," so there won't be an Xo at all. Using right end points is often the easiest approach.

Answer 24

ok

Answer 25

OK. We're almost ready to add up the areas of lots and lots of vertical strips of width 6/n and height \[f(x _{i})=6x _{i}+1.\] That's just your original function, f(x)=6x+1, with \[x _{i}\]substituted for x. OK?

Answer 26

ok

Answer 27

Remembering that each rectangle has area (height)*(width), we want the sum of billions of little rectangles:\[\sum_{i=1}^{n}[6((i-1)*(6/n)+1)]*(6/n)\] .

Answer 28

The expression inside [ ] represents the heights of the rectangles and the final 6/n represents the width of each. OK?

Answer 29

yes

Answer 30

We could simplify this by factoring two of the three 6's out:\[6*6*\sum_{i=1}^{n}[(i-1)*(6/n)+1](1/n)\]

Answer 31

ok

Answer 32

At this point I'm going to ask you whether you know what needs to be done next. We need to be able to apply two special summation formulas which I imagine you have seen before. One of them is \[\sum_{i=1}^{n}i=\frac{ n(n+1) }{ 2 }\]

Answer 33

and the other is \[\sum_{i=1}^{n}1=n\]

Answer 34

Familiar with these?

Answer 35

yes

Answer 36

So, time to pick your brains. What's our next step?

Answer 37

= n(n+1)(2n+1)/6

Answer 38

That's the next formula, and is fine. But I was asking what the next step in the Riemann Sum approach to solving your posted math problem would be. We have\[6*6*\sum_{i=1}^{n}[(i-1)*(6/n)+1](1/n)....\].... Now what?

Answer 39

sorry lost connection

Answer 40

start loop

Answer 41

Not quite. We have to multiply out the expression to the right of the summation symbol so as to obtain forms that look like either \[\sum_{?}^{?}i ~or~\sum_{?}^{?}1\]

Answer 42

so that we can take advantage of the simplifying formulas we discussed 3 minutes ago.

Answer 43

ok

Answer 44

What I'm getting looks like this:\[6*6[\sum_{i=1}^{n}\frac{ 6 }{ n }i-\sum_{i=1}^{n}\frac{ 6 }{ n }+\sum_{i=1}^{n}1]\frac{ 1 }{ n }\]

Answer 45

ok

Answer 46

Pretty awful, no???

Are you going to be around tomorrow at any time? As I said before, I'm not sure I can take you all the way through this problem solution tonight, but if it's not urgent and you have time tomorrow, i could help you further then.

Answer 47

In any case I strongly encourage you to copy down everything you think is important here, for future reference, so we won't have to type it all over again.

Answer 48

sounds good, i work until 3 so will be online from 3 to 5 tomorrow.

I have it all wrote down.

Thank you for taking the time

Answer 49

cent

Answer 50

l

Answer 51

central time

Answer 52

Before we stop, let me suggest something that would simplify this probelm solution:

Treat 6x and 1 as separate functions, finding the Riemann Sum for each; in the end, you just add the results together.

Considerably easier and less confusing.

Answer 53

I'm in California and see that it's 9 : 24 p.m. What time have you?

Answer 54

11:25

Answer 55

pm in oklahoma

Answer 56

So that would mean 2 hours earlier in the day for me; your 3-5 p.m. would be my 1-3 p.m., right?

Answer 57

correct

Answer 58

OK. I'll try my best to be online when you are. I do need to get 3 sets of income tax returns mailed; that's the only obstacle.

Great working with you. Til later, then. Good night.

Answer 59

is that good for you, can be on at this time tomorrow also starting at 7pm your time

Answer 60

thank you.

Answer 61

My pleasure. Actually the 7 p.m. (my time) arrangement would be much better for me than mid-day.

Answer 62

Private message me through OpenStudy when you're on or about to go on...I check my OS messages frequently.

Answer 63

sounds good be on here around 7pm tomorrow night. Will do.

Answer 64

:)

Answer 65

hello

Answer 66

Hello... glad you're back!

Answer 67

you too

Answer 68

Last night we looked at finding the definite integral of f(x) = 6x+1 from 0 to 6 using Riemann Sums.
Have you had an opp to look over any of our previous discussion yet?

Answer 69

briefly

Answer 70

In any case, I would like to start by focusing solely on the 6x part of the integrand, and take care of the +1 part later. Makes this process a bit easier to undrstand.

Answer 71

Have you a textbook or any other resource that presents you with the definition of Riemann Sums?

Answer 72

ok, I think I understand that part has you take the integral of 6x which is 6^2/2 - 0^2/2

Answer 73

What you just did here, DH, was to apply the power rule for integration. Of course you can do that, but I'd thought you wanted to use Riemann Sums.

Answer 74

\[Sn = \sum_{i=1}^{n} F (a + k((b - a)/n)(b-a)/n\]

Answer 75

\[\int\limits_{0}^{6}6x dx=6\frac{ x ^{1+1} }{ 1+1 } \]

Answer 76

i do

Answer 77

evaluated from 0 to 6. But we're supposed to use Riemann sums.

Answer 78

ok

Answer 79

OK. Now looking at the formula you've just typed in, I see you have it correct. This formula is equivalent to the definite integral of your function F(x) from a to b.

Let's look just at the 6x term and find the def integ of that from 0 to 6, using Riemann sums.
From last night, the width of each thin vertical strip is (b-a)/n, or ... what?

Answer 80

6/n

Answer 81

\[\Delta x=\frac{ b-a }{ n }\]

Answer 82

Correct, and so we next want a formula for the n x values. May I borrow what we developed last night, or want to go through the procedure of finding that formul again?

Answer 83

i have it wrote down

Answer 84

i have a question

Answer 85

Correct, and so we next want a formula for the n x values. May I borrow what we developed last night, or want to go through the procedure of finding that formul again?\[x _{i}=a+i \Delta x\]

Answer 86

By all means go ahead with your question.

Answer 87

when going from
Xi=(i-1) to F

Answer 88

oops

Answer 89

when going from
Xi= (i-1) to F(Xi)=6Xi +1
Why did we change the sign before the 1?

Answer 90

just because the given question?

Answer 91

No, our formula for xi is completely independent of the function being addressed.
The formula for xi that I'm giving you tonight is a bit simpler in that I ditched the (i-1) part and am letting i begin at 0 (instead of at 1 like last night) and go to n.

Answer 92

ok

Answer 93

Here, delta x is 6/n, as you said before,
so the formula for the ith x value becomes

\[x _{i}=i \frac{ 6 }{n }\]

Answer 94

with i beginning at zero and going up to n.

Answer 95

If you can accept this, then our function f(x) = 6x becomes f(xi) = 6(i[6/n]), or

Answer 96

\[f(x)=6x \rightarrow f(x _{i})=6(i \frac{ 6 }{ n })\]

Answer 97

I am purposely leaving out the +1 part of your f(x) = 6x + 1.

Answer 98

ok