A Closed Form Expression for the Nth right Riemann sum: ∫712−3x+4dx

[This Cannot Be Left Blank]

In the question
Consider the definite integral ∫712−3x+4dx

Find a closed form expression for the nth right Riemann sum of this integral.
n
I would write this into the form Rn=∑ 3(7+12-7/n⋅i)+4⋅12-7/n
which would simplify to i=1
From now on, I'm refering to the sum as just ∑.
Rn=
5/n* ∑ 3(7+5/n⋅i)+4 (

Then I'd take the 7, add it to the inverse of the 4, and times the 3 by the 5 to get 15/n

=5/n ∑ 3+15n⋅i
Then I'd separate the two into
=5/n ∑ 3+∑ 15/n⋅i
With that I'd make the 5/n 15/n by factoring the 3 out of the first sum and into the 5/n.
=15/n ∑ 1+∑ 15/n⋅i
Then I'd square the denominator and numerator of the second sum's fraction, and take it out of the sum to get
15/n ∑(1)+225/n^2⋅∑(i)
Giving that the sum of 1 in this situation is N and the sum of I is n⋅(n+1)2, I'd convert these. I would get


=15/n⋅n+225/n^2⋅n⋅(n+1)/2

Then I would factor out the Ns on the left of the plus sign, and multiply the equation by 2
=30+450/n^2* n⋅(n+1)


I have the feeling I'm doing something wrong here but i can't tell what.

 

[stapel]

Consider the definite integral ∫712−3x+4dx

Is there any particular reason for not starting by combining the 712 and the 4 to get 716? Also, why does this "definite" integral not have any limits?

Find a closed form expression for the nth right Riemann sum of this integral.
n

Is the "n" above a typo?

I would write this into the form Rn=∑ 3(7+12-7/n⋅i)+4⋅12-7/n

How did you obtain this? For instance, where did the "-7/n" come from? Where did the "4*12" come from?

which would simplify to i=1

How? :shock:

 

[This Cannot Be Left Blank]

Is there any particular reason for not starting by combining the 712 and the 4 to get 716? Also, why does this "definite" integral not have any limits?

I had the that's the intergral with top 12 and bottom 7.
Is the "n" above a typo?
It's supposed to be above the ∑.
How did you obtain this? For instance, where did the "-7/n" come from? Where did the "4*12" come from?



How? :shock:

I had the that's the intergral with top 12 and bottom 7. When I typed in all the symbols it seemed to revert to that.

It's supposed to be above the ∑.

it's supposed to be "4)*12"

I had put spaces there so that the i=1 would be under the ∑, but they seem to have been deleted.

 

[stapel]

I had the that's the intergral with top 12 and bottom 7. When I typed in all the symbols it seemed to revert to that.

It's supposed to be above the ∑.

it's supposed to be "4)*12"

I had put spaces there so that the i=1 would be under the ∑, but they seem to have been deleted.

Okay, it looks like you've edited your first post. Unforunately, I don't know which parts you've changed. Also, your new post seems disconnected from the questions I think it's meant to answer. So I'll have to guess at what you've meant. (This includes trying to decode the spacebar-based formatting, which this forum doesn't support.)

I think the integral is meant to be as follows:

. . . . .$\displaystyle \displaystyle \int_7^{12}\, (-3x\, +\, 4)\, dx$

(To learn how to format math as text, try here.)

You are directed to find a closed-form expression of the Riemann sum; that is, to replace the summation with formulas, and arrive at an algebraic expression which can be simplified and evaluated. I think you are supposed to refer to the Riemann sum with n partitions as "R_n".

The interval has a length of 12 - 7 = 5 units. Dividing this into n (equal) partitions gives n partitions with width 5/n.

The first interval's x-value will be the right-hand endpoint of the first interval, being the lower limit of the integral, plus one partition's width:

. . . . .$\displaystyle 7\, +\, \dfrac{5}{n}$

Each following x-value will be another 5/n down the number line, so the i-th x-value will be:

. . . . .$\displaystyle 7\, +\, \dfrac{5i}{n}$

The corresponding y-values will then be:

. . . . .$\displaystyle -3\left(7\, +\, \dfrac{5i}{n}\right)\, +\, 4$

. . . . .$\displaystyle -21\, -\, \dfrac{15i}{n}\, +\, 4$

. . . . .$\displaystyle -17\, -\, \dfrac{15i}{n}$

This represents the height of each of the partitions. Then the area of each partition, being 5/n in width, is given by:

. . . . .$\displaystyle \left(\dfrac{5}{n}\right)\, \left(-17\, -\, \dfrac{15i}{n}\right)$

. . . . .$\displaystyle -\dfrac{85}{n}\, -\, \dfrac{75i}{n^2}$

The integral is approximated by the sum of these areas:

. . . . .$\displaystyle \displaystyle \sum_{i=1}^n\, \left(-\dfrac{85}{n}\, -\, \dfrac{75i}{n^2}\right)$

Properly, of course, the integral is equal to the limit of the above, as n gets arbitrarily large (that is, as "n goes to infinity"). The limit may be taken after the above is reduced to an algebraic expression, etc, etc. But my question is, how does this relate to what you've posted?

Please be specific. Thank you!

 

[HallsofIvy]

Once you have $\displaystyle \displaystyle \sum_{i=1}^n\, \left(-\dfrac{85}{n}\, -\, \dfrac{75i}{n^2}\right)$ as staple showed, write that as
$\displaystyle -\frac{85}{n}\sum_{i= 1}^n 1- \frac{75}{n^2}\sum_{i=1}^n i$.

Now can you do those two sums? The first should be easy: $\displaystyle \sum_{i= 1}^3 i= 1+1+ 1= 3$. etc.

The other is only a little harder: $\displaystyle \sum{i=1}^3 i=1+ 2+ 3= 6$, $\displaystyle \sum_{i=1}^4= 6+ 4= 10$, etc. Do you see that those are "triangular numbers"?

 

[This Cannot Be Left Blank]

Okay, it looks like you've edited your first post. Unforunately, I don't know which parts you've changed. Also, your new post seems disconnected from the questions I think it's meant to answer. So I'll have to guess at what you've meant. (This includes trying to decode the spacebar-based formatting, which this forum doesn't support.)

I think the integral is meant to be as follows:

. . . . .$\displaystyle \displaystyle \int_7^{12}\, (-3x\, +\, 4)\, dx$

(To learn how to format math as text, try here.)

You are directed to find a closed-form expression of the Riemann sum; that is, to replace the summation with formulas, and arrive at an algebraic expression which can be simplified and evaluated. I think you are supposed to refer to the Riemann sum with n partitions as "R_n".

The interval has a length of 12 - 7 = 5 units. Dividing this into n (equal) partitions gives n partitions with width 5/n.

The first interval's x-value will be the right-hand endpoint of the first interval, being the lower limit of the integral, plus one partition's width:

. . . . .$\displaystyle 7\, +\, \dfrac{5}{n}$

Each following x-value will be another 5/n down the number line, so the i-th x-value will be:

. . . . .$\displaystyle 7\, +\, \dfrac{5i}{n}$

The corresponding y-values will then be:

. . . . .$\displaystyle -3\left(7\, +\, \dfrac{5i}{n}\right)\, +\, 4$

. . . . .$\displaystyle -21\, -\, \dfrac{15i}{n}\, +\, 4$

. . . . .$\displaystyle -17\, -\, \dfrac{15i}{n}$

This represents the height of each of the partitions. Then the area of each partition, being 5/n in width, is given by:

. . . . .$\displaystyle \left(\dfrac{5}{n}\right)\, \left(-17\, -\, \dfrac{15i}{n}\right)$

. . . . .$\displaystyle -\dfrac{85}{n}\, -\, \dfrac{75i}{n^2}$

The integral is approximated by the sum of these areas:

. . . . .$\displaystyle \displaystyle \sum_{i=1}^n\, \left(-\dfrac{85}{n}\, -\, \dfrac{75i}{n^2}\right)$

Properly, of course, the integral is equal to the limit of the above, as n gets arbitrarily large (that is, as "n goes to infinity"). The limit may be taken after the above is reduced to an algebraic expression, etc, etc. But my question is, how does this relate to what you've posted?

Please be specific. Thank you!

It said on the question to "Find a closed form expression for the nth right Riemann sum of this integral.", in this case, $\displaystyle \displaystyle \int_7^{12}\, (-3x\, +\, 4)\, dx$ (well not the exact equation, I changed the numbers a bit)
Also, It might help, but when I typed in the 225/n^2⋅n⋅(n+1)/2 in the =15/n⋅n+225/n^2⋅n⋅(n+1)/2 step, it seemed to simplify to 15+ 225/2(n+1/n)

 

[pka]

In the question
Consider the definite integral $\displaystyle \displaystyle\int_7^{12} {\left( { - 3x + 4} \right)dx}$
Find a closed form expression for the nth right Riemann sum of this integral.

This is what I have found so very frustrating about this thread: The term nth right Riemann sum​has a precise historically agreed upon meaning. You must start with a uniform division( or net) of $\displaystyle [7,12]$, then the evaluation points $\displaystyle x^*$ are the right-endpoints of the net.
The length of each subinterval of the net is $\displaystyle \delta x=\frac{5}{n}$.
Therefore, each evaluation point is $\displaystyle x^*_k=7+k(\delta x),~k=1,\cdots,~n$, a right endpoint of a subinterval in the standard partition.
Now that we have the gobbledygook out of the way, lets apply it all.
$\displaystyle \sum\limits_{k = 1}^n {f\left( {\mathop x\nolimits_k^* } \right)} \delta x = \frac{5}{n}\sum\limits_{k = 1}^n {\left( { - 3\left[ {7 + \frac{{5k}}{n}} \right] + 4} \right)}$

 

[This Cannot Be Left Blank]

This is what I have found so very frustrating about this thread: The term nth right Riemann sum​has a precise historically agreed upon meaning. You must start with a uniform division( or net) of $\displaystyle [7,12]$, then the evaluation points $\displaystyle x^*$ are the right-endpoints of the net.
The length of each subinterval of the net is $\displaystyle \delta x=\frac{5}{n}$.
Therefore, each evaluation point is $\displaystyle x^*_k=7+k(\delta x),~k=1,\cdots,~n$, a right endpoint of a subinterval in the standard partition.
Now that we have the gobbledygook out of the way, lets apply it all.
$\displaystyle \sum\limits_{k = 1}^n {f\left( {\mathop x\nolimits_k^* } \right)} \delta x = \frac{5}{n}\sum\limits_{k = 1}^n {\left( { - 3\left[ {7 + \frac{{5k}}{n}} \right] + 4} \right)}$

There's some similar questions on a practice quiz for it, and in the online course, if you get one of the practice quizzes wrong, it not only gives you the answer, it shows the steps. Am I allowed to show one of these so you can get a general idea of what the course wants me to do? Because in those, the final answer, the one the course apparently wants, did not involve sums at all. There were sums in the steps of course, but the final answer didn't involve sums.