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Integration help

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data
    The interval [0,4] is partitioned into n equal subintervals and ti, is chosen in [Xi-1, Xi] for each i.
    What is:

    Lim........ inf
    x->inf......Ʃ e^(sqrt ti)/N(sqrt ti)
    ............i=1

    3. The attempt at a solution
    It was problem we had to do in class as a group, teacher wasnt there to provide us an answer, and some people in my group have conflicting answers


    What i did:

    I wrote it as integral

    4
    ∫ (e^(sqrt X))/(N(sqrt X))
    0

    I then use the substitution method:
    U= sqrt X
    du= 1/(2 sqrtX) dx
    2du= 1/(sqrt X)

    .......4
    2/n ∫ e^u = (2e^(sqrt 4))-(2e/n^(sqrt 0))
    .......0

    = 2e^2-2
    My answer = 2/n(e^2 - 1)



    Can someone help me out, i dont really understand what we are suppose to do, and i apologize for the bad input. Appreciate it.
     
  2. jcsd
  3. Nov 14, 2012 #2

    SammyS

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    Hello smiggles. Welcome to PF !

    I assume that N and n are the same thing.

    What quantity represents Δx in the sum ?
     
  4. Nov 14, 2012 #3
    Hey sammy, thanks for the reply!!!

    Yes N and n are the same thing.

    im not sure if i under stand what youre asking me, but would Δx = (b-a)/n = (4-0)/n = 4/n
     
  5. Nov 14, 2012 #4

    SammyS

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    Yes, Δx = 4/n .

    Your integral does not have a dx in it. The quantity in the sum that would correspond to dx is Δx . Can you figure out how to have Δx, i.e. 4/n, in your sum?
     
  6. Nov 14, 2012 #5
    4
    ∫ (e^(sqrt x))/(n(sqrt x)) * (4/n)
    0

    or could i just multiply the (4/n) to my answer:

    = (8/n^2)(e^2 - 1)
     
    Last edited: Nov 14, 2012
  7. Nov 14, 2012 #6

    SammyS

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    There's still no dx there ... but that's not what I asked you to do anyway.

    How can you get 4/n to be in the sum ?

    [itex]\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}[/itex]
     
  8. Nov 14, 2012 #7
    dx= 4/n

    can i solve for n, so it would be n=(dx/4), and i would plug it in?
    :(

    edit:

    or can i just tack on a dx
    [itex]\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}[/itex] DX

    2nd edit:
    to get 4/n to be in the sum, can i put it in front of the sigma?

    4/n[itex]\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}[/itex]
     
    Last edited: Nov 14, 2012
  9. Nov 14, 2012 #8

    SammyS

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    No. It's like this:

    [itex]\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}=\frac{1}{4}\sum_{i=1}^{\infty} \frac{4e^\sqrt{t_i}}{n\sqrt{t_i}}[/itex]

    By the way, maybe we should be looking at [itex]\displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\ .[/itex]
     
  10. Nov 14, 2012 #9
    Nice, one question, where did the 1/4 come from?

    Looking back at [itex]\displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\ .[/itex]

    My first instict would be to change
    [itex]\frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\ .[/itex]
    to

    4
    ∫ fx dx = [itex]\frac{e^\sqrt{x}}{n\sqrt{x}}\ dx .[/itex]
    0

    Is there something im overlooking, or a step im skipping?

    I really appreciate your help, and your great patience!!
     
  11. Nov 14, 2012 #10

    SammyS

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    The use of ti instead of some xi (I know, it's because xi has a different use.) may be making this a little less straight forward.

    You need something like:

    [itex]\displaystyle \sum_{i=1}^{n} (f(t_i)\Delta x)[/itex]

    to become

    [itex]\displaystyle \int_{a}^{b} f(x)\,dx\ .[/itex]

    The sum [itex]\displaystyle \ \ \frac{1}{4}\sum_{ i=1}^{\infty} \frac{4e^\sqrt{t_i}}{n\sqrt{t_i}}\ \ [/itex] does that.

    [itex]\displaystyle \frac{e^\sqrt{t_i}}{\sqrt{t_i}}[/itex] is f(ti).

    [itex] \displaystyle \frac{4}{n}\ \ [/itex] is Δx .
     
  12. Nov 15, 2012 #11
    so i should have
    1/4[itex]\displaystyle \\\int_{0}^{4} \ \frac{4e^\sqrt{x}}{n\sqrt{x}}\ \ [/itex]

    sorry to ask again, but where is the 1/4 coming from?
     
  13. Nov 15, 2012 #12

    SammyS

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    No.

    The 4/n, which is Δx, becomes dx in for the integral version.

    Added in Edit:

    The 1/4 comes about because you need 4/n for Δx, so that you have dx to properly write tour integral.

    Have you ever seen in your textbook, a case where an integral is written without a differential such as dx ?
     
    Last edited: Nov 15, 2012
  14. Nov 15, 2012 #13

    SammyS

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    What you have is close to the right answer, but never quite there.

    OK

    Let's say we want to write a right Riemann sum to approximate the integral, [itex]\displaystyle \int_{0}^{4}f(x)\,dx \ .[/itex]

    Let's partition the interval [0, 4] into 100 equal-width intervals.

    The width of each interval is Δx = 4/100 = 0.04 .

    The height of each interval is f(xk), where xk is the right hand x value of the k-th interval.

    We would then approximate the integral as:

    [itex]\displaystyle f(0.04)\left(0.04\right)+f(0.08)\left(0.04\right)+f(0.12)\left(0.04\right)
    +\dots[/itex]
    [itex]\displaystyle =\sum_{k=1}^{100}f(x_k)\frac{4}{100}\ .
    [/itex]​
    If you use n intervals instead, you have

    [itex]\displaystyle \sum_{k=1}^{n}f(x_k)\frac{4}{n}\ .[/itex]


    The function you're using already has the n in it, but not the 4 .
     
  15. Nov 15, 2012 #14

    HallsofIvy

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    You cannot take the limit as "x goes to infinity"- there is NO "x" in your formula. I think you meant "n goes to infinity".

    The integral is the limit of the sums as n goes to infinity. There should be no "n" (or "N") in this.

     
  16. Nov 16, 2012 #15
    Thank you very much for your help, greatly appreciate it!

    Thanks for clearing that up for me, apologize about the mistakes.

    Glad there is still great people around to help others appreciate it!
     
  17. Nov 16, 2012 #16

    SammyS

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    So, what integral did you finally come up with ?
     
  18. Nov 17, 2012 #17
    The professor put up the answer, does it seem correct, how did the integral go from 0 to 4 TO 0 to 2. My original answer is very close to that, but i multiplied by 2 instead of multiplying by 1/2.

    and how are you able to multiply 4/n by n/4, and what is the purpose to doing this?
     
    Last edited: Nov 17, 2012
  19. Nov 17, 2012 #18

    SammyS

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    I agree with your professor's answer.

    The limits of integration change due to doing integration by substitution.

    Added in Edit:

    So, why did you edit your post, erasing the image?


    As for your question about 4/n and n/4:

    Your professor took the expression [itex]\displaystyle \ \ \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\ \ [/itex] and multiplied that by [itex]\displaystyle \ \ \frac{n}{4}\frac{4}{n}\ .\ [/itex] Why can he do that ?

    His motivation for doing that was that he needed [itex]\ \Delta x\ [/itex] in his sum and [itex]\displaystyle \ \ \Delta x=\frac{4}{n}\ .[/itex] He then kept the [itex]\ \Delta x\ [/itex] isolated and combined the [itex]\displaystyle \ \ \frac{n}{4}\ \ [/itex] with the given expression, giving:
    [itex]\displaystyle \left(\frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\frac{n}{4} \right)\frac{4}{n}=\frac{e^\sqrt{t_i}}{4\sqrt{t_i}}\Delta x[/itex]​

    You didn't have the 4 in the denominator.

    (1/4)2 = 1/2 . So, there's a 1/2 rather than the 2 you had.
     
    Last edited: Nov 17, 2012
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