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If this correct please look - definite inegral

  1. Nov 29, 2009 #1
    The question:
    Use the definiton of the definite inegral (with right hand rule) to evaluate the following integral. Show work please
    Can NOT use shortcut method.. must be the long process

    S (3x^2 - 5x - 6) dx

    ∫[-4,1] (3x^2 - 5x - 6) dx =
    lim[n-->∞] 5/n ∑[i=1 to n] {3(-4 + 5/n)² - 5(-4 + 5/n) - 6} =
    lim[n-->∞] 5*∑[i=1 to n] (48/n - 120i/n² + 75i²/n³ + 20/n -25i/n² - 6/n) =
    lim[n-->∞] 5*∑[i=1 to n] (62/n - 145i/n² + 75i²/n³) =
    lim[n-->∞] 5[62n/n - 145n(n+1)/(2n²) + 75n(n+1)(2n+1)/(6n³)] =
    5(62 - 145/2 + 25) = 72.5

    ∑[i=1 to n] 1 = n
    ∑[i=1 to n] i = n(n+1)/2
    ∑[i=1 to n] i² = n(n+1)(2n+1)/6

    Please check if this is correct and let me know
  2. jcsd
  3. Nov 30, 2009 #2
    In my opinion it may be,
    Area under the curve
    For 3n^2 [tex][3(n^2)+3((2n)^2)+3((3n)^2)+..]n=3(1^2+2^2+3^2...)n^3=3x_i^2n^3[/tex]

    For 5n [tex]5(n+2n+3n+...)n=5(1+2+3+...)n^2=5x_in^2[/tex]

    For 6 [tex]6n[/tex]

    [tex]\lim_{n\rightarrow 0}\sum_{i=1}^{5/n}(3x_i^2n^3-5x_in^2-6n)=\lim_{n\rightarrow0}\sum_{i=1}^{5/n}3x_i^2n^3-\lim_{n\rightarrow0}\sum_{i=1}^{5/n}5x_in^2-\lim_{n\rightarrow 0}\sum_{i=1}^{5/n}6n[/tex]

    Because we divide [-4,1] interval n pieces and the n pieces are our base of rectangles.Also our heights are the images on function of the little pieces.For example our base is n and our height is 3/n(interval of sum [1,5/n])). So our rectangle's area is [tex](3(3/n)^2n^3-5(3/n)n^2-6n)[/tex].If we sum the all of these areas and base converges to 0 then we get the true area at [-4,1].
    Last edited: Nov 30, 2009
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