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Express e^x from 1 to 8 as a Riemann Sum. Please, check my work?

  1. Jan 18, 2013 #1
    Express e^x from 1 to 8 as a Riemann Sum. Please, check my work? :)

    1. Express ∫1 to 8 of e^xdx as a limit of a Riemann Sum.

    (Please ignore the __ behind the n's. The format is not kept without it...)

    _____n
    2. lim Ʃ f(xi)(Δx)dx
    x→∞ i=1

    Δx= (b-a)/n = 8-1/n = 7/n

    xi= 1 + 7i/n

    ____n
    lim Ʃ (1 + 7i/n)(7/n)
    x→∞

    It also says to keep it in terms of i and n. So... is that the answer then?
     
  2. jcsd
  3. Jan 18, 2013 #2

    jbunniii

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    Re: Express e^x from 1 to 8 as a Riemann Sum. Please, check my work? :)

    What happened to your function (e^x)? Surely it must appear in the answer. Also, what does [itex]\lim_{x\rightarrow\infty}[/itex] refer to, when the rest of the expression doesn't depend on [itex]x[/itex]? Also please indicate exactly what values of [itex]i[/itex] you are summing over.
     
  4. Jan 18, 2013 #3

    Ray Vickson

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    Re: Express e^x from 1 to 8 as a Riemann Sum. Please, check my work? :)

    You cannot have Δx= 8-1/n, because this → 8 as n → ∞. However, you *can* have Δx= (8-1)/n, which → 0 as n → ∞. Do you see what a difference it makes to use parentheses?

    Your expression ## \sum f(x_i)\, \Delta x \, dx ## makes no sense: the *integral* has dx in it, but the finite sum should not. Your final expression is also incorrect: the quantity
    [tex] \sum_{i=1}^n \left( 1 + \frac{7i}{n} \right) \left( \frac{7}{n} \right)[/tex] does not contain 'x' anywhere, so you cannot have ##\lim_{x \to \infty}## of it! Finally, your sum is a Riemann sum approximation to ##\int f(x) \,dx## for some function f(x); what f(x) is that? It certainly is not ##e^x##.
     
  5. Jan 18, 2013 #4

    Mark44

    Staff: Mentor

    Re: Express e^x from 1 to 8 as a Riemann Sum. Please, check my work? :)

    You can use LaTeX to present integrals, limits, and sums in a nice way.

    The above is
    $$\int_1^8 e^x dx$$
    The LaTeX for this is
    \int_1^8 e^x dx (with $$ at the front and at the end).
    The limit should be as n gets large, not x. Also, there should not be a factor of dx. In LaTeX, the correct limit is:
    $$ \lim_{n \to \infty} \sum_{i = 1}^n f(x_i) Δx$$

    My LaTeX script: \lim_{n \to \infty} \sum_{i = 1}^n f(x_i) Δx (again, with a pair of $ at beginning and end.
    You don't mean 8 - (1/n), so use parentheses: (8 - 1)/n.
     
  6. Jan 18, 2013 #5
    Re: Express e^x from 1 to 8 as a Riemann Sum. Please, check my work? :)

    Oh, right.

    I left out the e^x! :/

    So would it be:

    _____n
    lim Ʃ (e^(1 + 7i/n))(7/n)
    n→∞ i=1


    So let me try to say what is going on in words...
    So by the Ʃ, we are adding together the area of all the rectangles inside of the curve.

    Area of a rectangle= L*W. Every rectangle has a length of (e^(1 + (7i/n)) and a width of
    (7/n).

    The number of rectangles inside of the curve start from 1 and go until n.
    n is a number that approaches infinity.

    So we are saying that there are infinity rectangles inside of the curve and that each have a length of (e^(1 + 7i/n)) and a width of (7/n).
    We need to add up all of the many tiny rectangles to approximate the area under the curve.

    ... i does mean the number rectangle you start out on, right...? Like if it's i=1, that means you are summing rectangle #1 all the way through rectangle #n?

    If it's i=7, then you are summing rectangle #7 through rectangle #n?

    Am I thinking about somthing incorrectly here? Thanks.
     
  7. Jan 18, 2013 #6

    jbunniii

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    Re: Express e^x from 1 to 8 as a Riemann Sum. Please, check my work? :)

    Yes, this looks right.
    Correct.

    I would rephrase it slightly: the number of rectangles inside the curve is n. We number these rectangles from i=1 to i=n. As n grows, we are increasing the number of rectangles and decreasing their width. As n approaches infinity, the sum of the areas of the rectangles becomes as close as we like to the area under the curve.

    The reason I rephrased it like that is that there aren't really infinitely many rectangles. If you had infinitely many of them, then each one would have zero width (and therefore zero area), so our attempt to calculate the area would result in infinity times zero, an undefined operation. Instead, we use finitely many, n rectangles, to compute an area A(n), and then we take the limit of A(n) as n approaches infinity.

    Yes.

    Right, if the sum went from i = 7 to i = n, you would be adding the areas of rectangles #7 through #n. (This assumes that n >= 7.)

    No, I think you pretty much understand it correctly.
     
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