Find Area Under Curve y=x^3 from 0 to 1: Riemann Sum Limit

Click For Summary
SUMMARY

The discussion focuses on calculating the area under the curve defined by the function y=x^3 from 0 to 1 using Riemann sums. Participants derive the limit expression as lim_{n \to \infty} Σ_{i=1}^{n} (i/n)^{3} (1/n) and clarify the summation formula for cubes, (n(n+1)/2)^{2}. The correct evaluation leads to the conclusion that the area is 1/4, confirming the result through both Riemann sums and definite integrals.

PREREQUISITES
  • Understanding of Riemann sums and their applications
  • Familiarity with limits and summation notation
  • Knowledge of the formula for the sum of cubes: Σ i^3 = (n(n+1)/2)^{2}
  • Basic calculus concepts, including definite integrals
NEXT STEPS
  • Learn about Riemann sums and their derivations in calculus
  • Study the properties and applications of the sum of cubes formula
  • Explore the concept of limits in calculus, particularly in relation to summation
  • Practice evaluating definite integrals using various methods
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques and Riemann sums, as well as educators looking for examples of limit evaluations in mathematical discussions.

QuarkCharmer
Messages
1,049
Reaction score
3

Homework Statement


a.) Use definition 2 to find an expression for the area under the curve y=x^3 from 0 to 1 as a limit.
b.)Evaluate the (above) limit using the sum of the cubes of the n integers.

Homework Equations


(\frac{n(n+1)}{2})^{2}

The Attempt at a Solution


For part a.) I wrote my limit like this:
\lim_{n \to \infty} \Sigma_{i=1}^{n}(\frac{i}{n})^{3}\frac{1}{n}The "Definition 2" they have listed just says:
A=\lim_{n \to \infty} R_{n} = \lim_{n \to \infty}(f(x_{1})\Delta x + f(x_{2})\Delta x + . . . + f(x_{n})\Delta x)

Now for part b, I understand that the formula is the sum of all cubes and so on. So I am thinking that the limit should look like this?

\lim_{n \to \infty}(\frac{n(n+1)}{2})^{2}
That should handle the limit and the sum of cubes, now I need each one to multiply by delta x right? So that it comes out to:
\lim_{n \to \infty}(\frac{n(n+1)}{2})^{2}\frac{1}{n}
..because the integral is from 0 to 1, so \Delta x = \frac{1-0}{n}

But I am not sure how to write it in this manner and take the limit from here yet. Is this correct so far? I should just simplify the expression after the limit and then take the limit?
 
Last edited:
Physics news on Phys.org
No, no. The sum of i^3 for i=0 to n is (n(n+1)/2))^2. That's only a PART of the sum of (i/n)^3*(1/n) for i=0 to n. Do some algebra in that sum to separate out the i^3 part and then substitute your summation formula and take the limit.
 
I understand now. Sorry, this is the first time I have ever actually used a summation, other than just knowing what it is.
73p1z5.jpg


I checked with the definite integral and this is correct as far as I can tell.
I'm going to go learn the rules of summation and find some problems to do. Thanks, I didn't realize that this was how it worked.
 
Yes, the integral from x=0 to x=1 is 1/4. And I think you did it correctly through the limit.
 
Thanks for the help!
 
Hey can someone explain to me how did he jumped from lim n-> Ʃ i^3/n * 1/n to lim n->1/n^4 Ʃ i^3??
 
ScienceGeek24 said:
Hey can someone explain to me how did he jumped from lim n-> Ʃ i^3/n * 1/n to lim n->1/n^4 Ʃ i^3??

He didn't jump anywhere. He used a formula the sum of i^3 for i=1 to n.
 
I see... but he committed a mistake in the evaluation at lim n-> Ʃ i^3/n * 1/n is not lim n-> Ʃ i^3/n * 1/n is lim n-> Ʃ i^3/n^3 * 1/n
 
I was having a hard time finding how did he got that 1/n^4... LOL it was an ibvious mistake from part (a) even part (a) is wrong.
 
  • #10
deltax= 1-0/n=1/n than according to definition 2 the R endpoints formula is (a+ideltax)deltax so is i/n*1/n than substituting from x^3 we have (i/n)^3*1/n=i^3/n^3*1/n=i^3/n^4 which makes more sense. Now i just get rid of the 1/n^4 and put it in the other side of the summation and ta chan! problem solved LOL
 
  • #11
ScienceGeek24 said:
I see... but he committed a mistake in the evaluation at lim n-> Ʃ i^3/n * 1/n is not lim n-> Ʃ i^3/n * 1/n is lim n-> Ʃ i^3/n^3 * 1/n

It's pretty clear he meant (i/n)^3 not i^3/n since the n^4 appears in the next line. Why are you resurrecting posts over to a year old to complain about notational mistakes?
 
  • #12
is not clear in part (a).
 
  • #13
ScienceGeek24 said:
is not clear in part (a).

I'll admit the way it was written isn't clear. Using an extra parentheses would have helped.
 
  • #14
Dick said:
I'll admit the way it was written isn't clear. Using an extra parentheses would have helped.

Thank you.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
17
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K