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awibroe1-Visitor
1-Visitor
March 20, 2017
Solved

Integration as Summation

  • March 20, 2017
  • 2 replies
  • 3051 views

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.JPG

2.JPG

Best answer by Werner_E

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:

2 replies

W
Werner_E25-Diamond I
25-Diamond I
March 20, 2017

Notsure which problem you experience.

Maybe you should rather attach your worksheet.

A
awibroe1-VisitorAuthor
1-Visitor
March 20, 2017

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

W
Werner_E25-Diamond I
25-Diamond I
March 20, 2017

> 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

V
vlehner1-Visitor
1-Visitor
March 20, 2017

Hi Andy,

your textbook is right.

See attachement please.

What we have here is a gaussian summation:

Hope it helps you

best regards, Volker

A
awibroe1-VisitorAuthor
1-Visitor
March 20, 2017

‌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?

V
vlehner1-Visitor
1-Visitor
March 20, 2017

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

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