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

In summary, the homework statement states that the area under the curve y=x^3 from 0 to 1 as a limit is given by \lim_{n \to \infty} \Sigma_{i=1}^{n}(\frac{i}{n})^{3}\frac{1}{n}
  • #1
QuarkCharmer
1,051
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


[tex](\frac{n(n+1)}{2})^{2}[/tex]

The Attempt at a Solution


For part a.) I wrote my limit like this:
[tex]\lim_{n \to \infty} \Sigma_{i=1}^{n}(\frac{i}{n})^{3}\frac{1}{n}[/tex]The "Definition 2" they have listed just says:
[tex]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)[/tex]

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?

[tex]\lim_{n \to \infty}(\frac{n(n+1)}{2})^{2}[/tex]
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:
[tex]\lim_{n \to \infty}(\frac{n(n+1)}{2})^{2}\frac{1}{n}[/tex]
..because the integral is from 0 to 1, so [itex]\Delta x = \frac{1-0}{n}[/itex]

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?
 
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  • #2
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.
 
  • #3
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.
 
  • #4
Yes, the integral from x=0 to x=1 is 1/4. And I think you did it correctly through the limit.
 
  • #5
Thanks for the help!
 
  • #6
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??
 
  • #7
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.
 
  • #8
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
 
  • #9
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.
 

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

What is the concept of area under a curve?

The area under a curve is the measure of the region bounded by the curve, the x-axis, and the vertical lines at the limits of integration. It represents the total value or quantity of a function over a given interval.

What is the Riemann Sum method for finding area under a curve?

The Riemann Sum method is a mathematical technique for approximating the area under a curve by dividing the region into smaller rectangles and summing up their areas. As the number of rectangles increases, the approximation becomes more accurate and approaches the actual area under the curve.

How do you calculate the Riemann Sum for a function?

To calculate the Riemann Sum for a function, you need to divide the interval into smaller subintervals of equal width. Then, evaluate the function at the left or right endpoint of each subinterval and multiply the result by the width of the subinterval. Finally, sum up all the resulting values to get the approximate area under the curve.

What is the limit of the Riemann Sum?

The limit of the Riemann Sum is the exact area under the curve. As the number of rectangles approaches infinity, the width of each rectangle approaches zero, resulting in a more accurate approximation of the area under the curve.

How do you find the area under the curve y=x^3 from 0 to 1 using the Riemann Sum Limit?

To find the area under the curve y=x^3 from 0 to 1 using the Riemann Sum Limit, you need to first divide the interval [0,1] into smaller subintervals of equal width. Then, evaluate the function at the left or right endpoint of each subinterval and multiply the result by the width of the subinterval. Finally, take the limit as the number of rectangles approaches infinity to get the exact area under the curve.

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