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Upper Bounds Integration

  1. Apr 25, 2008 #1
    [SOLVED] Upper Bounds Integration

    1. The problem statement, all variables and given/known data
    Integrate y=4x from 2 to 5 using the limit with circumscribed rectangles.


    2. Relevant equations

    A=lim(n to inf.) Summation of f(xsubi) times delta (xsubi)

    3. The attempt at a solution

    A=lim(4/n)(4/n)(4)(2+3+4+......+(n+1))
    =64/n^2((n^2+3n)/2))= 32lim((n+3)/n)) =32. But from integration the answer is obviously 48. What am I doing wrong? (Sorry about lack of typo skills-newbie)
     
  2. jcsd
  3. Apr 25, 2008 #2
    why do you think the answer is obviously 48?
    Your answer isn't correct again thout..
     
  4. Apr 25, 2008 #3
    well i got 42 as my answer, either by directly integrating

    [tex]\int_2^5 4xdx[/tex] and also by using Rieman sums.

    I'll try to post my work, on my next post.
     
  5. Apr 25, 2008 #4
    Sorry-wrong limits!

    Sorry! The limits were 1 to 5, not 2 to 5!
     
  6. Apr 25, 2008 #5
    we want to calculate

    [tex]\lim_{n\to\infty}\sum_{i=1}^{n}f(\epsilon_i)\delta x_i[/tex]

    now let us create n mini segments on the segment [2,5]

    that is let the points be

    [tex]x_0=2,x_1,x_2,......x_i_-_1,x_i,....,x_n=5[/tex]

    Now our concern is to determine what our function will be.
    First let's notice certian facts:

    [tex]\delta x_i=x_i-x_i_-_1[/tex] also let [tex]\epsilon_i=x_i[/tex]

    this way we have:

    [tex]\epsilon_i=\delta x_i+x_i_-_1[/tex]

    also: [tex]\delta x_i=\frac{5-2}{n}=\frac{3}{n}[/tex]

    Now, for to determine our function lets try some values for i=1,2,3,...

    [tex]f(x_1)=4\left(\frac{3}{n}+2\right),f(x_2)=4(\frac{6}{n}+2),f(x_3)=4(\frac{9}{n}+2),......, f(x_i)=4(\frac{3i}{n}+2)[/tex]


    Hence:

    [tex]\int_2^54xdx=\lim_{n\to\infty}\sum_{i=1}^{n}4\left(\frac{3i}{n}+2\right)\frac{3}{n}=...=42[/tex]
     
    Last edited: Apr 25, 2008
  7. Apr 25, 2008 #6
    Well, then do the same thing as i did here, just take into consideration that you have the lower limit 1, in this case. I am not gonna troube to go the same route again, i think you can do it now. If you can't ask again.

    cheers!
     
  8. Apr 25, 2008 #7
    Well it doesn't change a lot by the way, the difference is that now you'll have

    [tex]\delta x_i=\frac{4}{n}[/tex] and

    [tex]f(x_i)=4\left(\frac{4i}{n}+1\right)[/tex]

    and the answer will be 48.
     
  9. Apr 25, 2008 #8
    Thanks very much for your replies- I'm still stuck expanding the summation- will attempt another query when I have time, and can clarify.
     
  10. Apr 25, 2008 #9
    Got it- I wasn't adding the 1 to the 4/n. Thanks again for your answer.
     
  11. Apr 25, 2008 #10
    I tried to post a detailed answer, including how the summation expanded and all that stuff, but after i typed it all, i don't know for some crappy reason it did not show up. Anyways, i'm glad you got it !
     
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