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## Integration as Summation  14-Alexandrite

## Integration as Summation

Dear All,

I am again coming to you with an issue I have related to work I am doing from a text book this time on the topic of integration as summation. The text book is trying to prove that integration is essentially the summation of an infinite number of rectangles. I have inserted images of the text in question as this is easier (sorry for this being sideways).

The example looks at a simple graph of y=x as seen in the below images. The problem I am having is I understand the concept, indeed I know that to find the area under the line y=x can be derived algebraically by integrating x to give x^2/2 and calculating for x=4 and x=2 and taking one from the other to get an area of 6 units^2.

However, following the example in the text book in the second image I can't see how the penultimate line of lim n to infin 4/n *n + lim n to infin 4/n^2*n(n+1)/2 gives 4 + 2 = 6. I can see that 4/n*n will give 4 as the ns cancel but on the other side I see that 4/n^2*n(n+1)/2 should give 4n^2+n/2n^2 would leave 2+n i.e. the n^2s cancel 4/2 is 2 and there is a remaining n.

I have used Mathcad to confirm my assumptions but Mathcad's symbolics indicate that the top expression in the second image below does give 6 but the bottom expression does not it gives 4 + 2 * lim n to infin n(n+1)/n^2.

Can anyone tell me if I have misunderstood something or if the text book is wrong? I presume this is the former not the latter.

Cheers,

Andy  1 ACCEPTED SOLUTION

Accepted Solutions  24-Ruby V
(To:awibroe)
 4/n^2*(n*(n+1))/2 how is this simplified to 2?

Its not. As already written above: If you cancel 2 and one n, split it into two fraction and cancel one n again It simplifies to 2/n+2

Its the limit for n->infinity which yields the 2.

 (4*n^2+n)/(2*n^2)

Thats wrong! You lost a factor 4 (distributive law!) It would have been a better idea to cancel before expanding the expression: 8 REPLIES 8  24-Ruby V
(To:awibroe)

Notsure which problem you experience.

Maybe you should rather attach your worksheet.   14-Alexandrite
(To:Werner_E)

Hi,

I see what I was doing wrong. I didn't put a * between n and (n and assumed Mathcad would know that n(n+1) meant n * (n+1).

However, can you explain mathematically how this results in 2?

I was of the understanding that n*(n+1) would give n^2+n.

Then 4/n^2 * n^2+n/2 = 4n^2+n/2n^2

Which results in 2+n i.e. 4/2 is 2 and n^2/n^2 is 1?

Where am I going wrong in my understanding?

Andy  24-Ruby V
(To:awibroe)

> I didn't put a * between n and (n and assumed Mathcad would know that n(n+1) meant n * (n+1).

n(n+1) would mean a function n with argument n+1

>  I was of the understanding that n*(n+1) would give n^2+n.

correct!

> Then 4/n^2 * n^2+n/2 = 4n^2+n/2n^2

Oops! You rather should say that  4/n^2 * (n^2+n)/2=2+2/n   13-Aquamarine
(To:awibroe)

Hi Andy,

What we have here is a gaussian summation: Hope it helps you

best regards, Volker

Volker  14-Alexandrite
(To:vlehner)

‌Sorry guys I'm not sure I've explained where I am getting stuck. In the second expression which is simplified to:

4/n^2*(n*(n+1))/2

how is this simplified to 2?

Am I correct in saying this equates to:

(4*n*(n+1))/(2*n^2)

Then

(4*n^2+n)/(2*n^2)

I'm then a bit lost as how this ends in 2?  13-Aquamarine
(To:awibroe)

Andy,

look here: with n---> infinity you get 2 as result of a limit value.

 Am I correct in saying this equates to: (4*n*(n+1))/(2*n^2)

Yes.

 Then (4*n^2+n)/(2*n^2) I'm then a bit lost as how this ends in 2?

it's a limit value as said.

regards, Volker

Volker  24-Ruby V
(To:awibroe)
 4/n^2*(n*(n+1))/2 how is this simplified to 2?

Its not. As already written above: If you cancel 2 and one n, split it into two fraction and cancel one n again It simplifies to 2/n+2

Its the limit for n->infinity which yields the 2.

 (4*n^2+n)/(2*n^2)

Thats wrong! You lost a factor 4 (distributive law!) It would have been a better idea to cancel before expanding the expression:   14-Alexandrite
(To:Werner_E)

‌the penny has dropped.

thanks all! 