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Homework Help: Determine the value of expression (Calc-II)

  1. May 19, 2014 #1
    Please look picture.
    I represented the expression with this integral :

    integral of (1-x)^3 from 0 to 1.

    I got 0, which makes no sense...

    How should I proceed?

    Attached Files:

  2. jcsd
  3. May 19, 2014 #2

    I represented this using:

    sum of (1-i/n)^3(1/n), i=1..infinity.

    I expand it, and apply summation rules.

    Problem solved?
  4. May 19, 2014 #3


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    It took me a moment to understand what you were doing! The problem you gave "Find the integral of (1- x)^3, from 0 to 1" can be done with a simple integration. But what you are doing is using the basic "Riemann sums" definition. Yes, if you divide the interval from 0 to 1 into n parts each part has length 1/n. And, at the right endpoint of the ith interval the function value is (1- i/n)^3 so the "area" of the thin rectangle there is (1- i/n)^3(1/n) and the entire area, and so the integral, is approximated by the sum.

    No, i does not go from 1 to infinity- it goes from 1 to n and then, to get the actual integral, you take the limit as n goes to infinity.
    You should have
    [tex]\frac{1}{n}\sum_{i=1}^n \left( 1- 3\frac{i}{n}+ 3\frac{i^2}{n^2}- \frac{i^3}{n^3}\right)[/tex]
    [tex]\frac{1}{n}\sum_{i= 1}^n 1- \frac{3}{n^2}\sum_{i= 1}^n i+ \frac{3}{n^3}\sum_{i= 1}^n i^2-\frac{1}{n^4}\sum_{i= 1}^n i^3[/tex]
  5. May 19, 2014 #4
    I don't see how you get that. Do you see that it should be:
    $$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^3$$

    EDIT: Ok, your form is correct too. Now just convert it to a definite integral which is:
    $$\int_0^1 (1-x)^3\,dx$$
    There is no need to expand it.
    Last edited: May 19, 2014
  6. May 19, 2014 #5
    Hmm... If I evaluate the integral, I get 1/4. But that answer is just right if I assume n goes to infinity. If am asked to determine the value of this expression, should I not use hallsofivy's last expression and the summation formulas to expand the i's into n?

    I think my final answer should contain n's.

    Ultimately, the integral is irrelevant in this problem?
  7. May 19, 2014 #6


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    Yes, use Halls's last expression.

    Then use the summation formulas, the results of which will contain n, but not i .
  8. May 19, 2014 #7
    What about the integral? Is it "useless" in this specific problem?
  9. May 19, 2014 #8


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    I think Halls has the correct interpretation of what's expected. Use the basic "Riemann sums" definition of the integral - taking the limit as n → ∞ .

    To check your answer, find the value of the following integral in the usual way, by finding the anti-derivative of the integrand, etc.
    ##\displaystyle \int_0^1 (1-x)^3\,dx##​
  10. May 20, 2014 #9
    Why would it be take the infinite sum? The question is determine the value of this expression without saying as n goes to infinity.
  11. May 20, 2014 #10
    We can do that too but this is easily done using definite integrals. If you insist, here's how. I will be using the form I shown before because that doesn't requires expanding the terms as HallsOfIvy has done.

    The sum under consideration is:
    $$\frac{1}{n}\sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^3=\frac{1}{n^4}\sum_{i=1}^{n-1}i^3$$
    Since ##\displaystyle \sum_{i=1}^{n-1}i^3=\left(\frac{(n-1)n}{2}\right)^2##, we have the following limit:
    \lim_{n\rightarrow \infty} \frac{1}{n^4}\left(\frac{(n-1)n}{2}\right)^2 &= \lim_{n\rightarrow \infty} \frac{(n-1)^2}{4n^2}\\
    &= \lim_{n\rightarrow \infty} \frac{1}{4}\left(1-\frac{1}{n}\right)^2 \\
    &= \boxed{\dfrac{1}{4}}\\
  12. May 20, 2014 #11
    I'm sorry. Perhaps I was not clear. My question is: the way you just did it supposes that n->infinity. But the original question is : determine the value of this expression in the picture. And in the picture, there is NO indication that n->infinity.
    Do you understand what I'm trying to convey?
  13. May 20, 2014 #12
    Ah ok, I thought this was another question related to Riemann sums. In that case, you leave it till the following step:
    I hope your issue is addressed now. :)
  14. May 20, 2014 #13
    Excellent! This forum is filled of hardworking people!
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