# Solution 75: Integrating Limit, Arithmetic Mean, and Trig

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- Published on Jan 17, 2019
- Hmm, we have n approaching infinity and a factor of 1/n residing inside arithmetic means. Can you guess what this limit is equal to?

Congratulations to Bamdad Shamaei, Parth Pawar, PRAKHAR AGARWAL, Kwekinator117, 123 gogo, Nicholas Patel, Jaleb, Hiren Bavaskar, Arun Bharadwaj, and Isaac YIU Math Studio for successfully solving this math challenge question! Bamdad Shamaei was the first person to solve the question.

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Sitanshu Chaudhary2 months ago^{+1}But i have doubt what is the limit n tends infinite sin(2ln(n)) is converge

aaronlabuka6 months agowaw, impressing

trey hanckok8 months agoI tried to integrate by parts however my solution included ln0 which is und. Any thoughts?

Aaron He6 months agoTake the limit and ln0 is -infinity.

Jeffrey Cloete8 months ago^{+2}Sir..whomever you are..you are a brilliant problem poser..and have an ingenious faculty for maths. Thank you for sharing your skill on USclip! !

LetsSolveMathProblems7 months agoThank you for your heartwarming compliment. I'm glad you find my videos enjoyable and educational! =)

민뚜TV8 months ago私は韓国人だよ

君と友達になりたい

君のチャンネルに購読したよ。

私のチャンネルにも来てくれる?

ツGMD Novacraft8 months ago^{+3}like if u dont understand nothing but still liking this vids

Rot8 months agoWhere can I learn about this

Mike Fernandez8 months agoLove your videos, I always learn something new!

Janda1258 months ago^{+8}You put both sin(ln n) and sin(ln i) = A, same with the cosines, you put cos(ln n) and cos(ln i) as B

Janda1258 months ago^{+3}LetsSolveMathProblems No worries, it was a great video, no need to apologize

LetsSolveMathProblems8 months ago^{+3}Oops! You are correct. I apologize for the error.

Janda1258 months ago^{+1}I know it wasn’t the exact expretion, I just wanted to point out to you which ln n and ln i belonged to which cos and sin. I think you ment ln n = A and ln i = B, right?

Furious Fthefics8 months ago^{+1}I can do it easily but how I can post my solution picture...here???

Felipe Lorenzzon8 months ago^{+1}I think you can post it elsewhere as a picture or with latex and put the link to it in your comment. But I would recommend you to give at least a brief description of your solution and also your final answer

LetsSolveMathProblems8 months agoThe solution to this weekly math challenge had to be submitted on the "Problem 75" video by yesterday (January 15, 2019) at the latest, so it is too late to turn in any solution to this specific problem. If you are interested, here is a link to this week's challenge problem: usclip.net/video/znl2l2nEjLk/video.html.

Ben Burdick8 months ago^{+1}That was a really cool solution, but It's unclear to me how we determined the limits of integration. Why was it 0 to 1?

LetsSolveMathProblems8 months ago^{+3}You are right (but make sure you also have 5i/n inside the function, along with the factor 5/n outside the summation). Here is the actual limit definition of the definite integral from 0 to 5 of sin(ln(x))dx using right Riemann Sum:

Limit as n -> infinity of (5/n) * Sum from i=1 to n of ( sin(ln( 5i/n )) ).

Ben Burdick8 months ago^{+2}@LetsSolveMathProblems Ah, ok this makes sense. So if I wanted to integrate from 0 to 5, the summation would have been in the form 5/n?

LetsSolveMathProblems8 months ago^{+3}Excellent question. To see why the limits of integration were chosen as in the video, I will give two examples with different limits of integration.

1. Integral from 0 to 2: In this case, the width of each rectangle is 2/n, NOT 1/n, (because we are dividing an interval of length 2 into n subintervals) and, more importantly, we should evaluate our function at 2i/n, NOT i/n.

2. Integral from 1 to 2: In this case, we should evaluate of our function at 1+i/n (that is, 1+1/n, 1+2/n, ..., 1+n/n = 2), NOT i/n.

In our example, our function is being evaluated at 0+i/n, so the bounds should be from 0 to 1.

James Wilson8 months agoSweet problem! And very good solution Bamdad Shamaei!

adandap8 months agoI failed on this one last week, though got it out when I had another look at it this morning. But I was quite pleased with one aspect of my failed attempt, which at least let let me bound the answer. I took a complex combination c_n = a_n + i b_n. Then it's not hard to see that the average of c_n = 1/n sum(j=1,n) exp(i ln(j) ), which has modulus

Andrés Robles8 months ago^{+2}Tus retos son únicos 👏.