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Definite inegral - please help

  1. Nov 28, 2009 #1
    Use the definiton of the definite inegral (with right hand rule) to evaluate the following integral. Show work please

    Can NOT use shortcut method.. but be the long process


    1
    S (3x^2 - 5x - 6) dx
    -4
     
  2. jcsd
  3. Nov 29, 2009 #2
    [tex]\int_{-4}^{1}(3x^2-5x-6)dx=\int_{-4}^{1}3x^2dx-\int_{-4}^{1}5x-\int_{-4}^{1}6dx=3\frac{x^3}{3}-5\frac{x^2}{2}-6x]_{-4}^{1}[/tex]

    =[tex]\frac{145}{2}[/tex]

    But i don't understand "with right hand rule".What do you mean?
     
  4. Nov 29, 2009 #3

    HallsofIvy

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    I suspect DemiMike means using the Riemann sum definition, using the right end point of each subinterval as the height of the rectangle.

    And, of course, requiring that it be done in a particular way makes it sound an awful lot like homework, so I am moving it. DemiMike, please post school work in the appropriate place.

    Also, whether homework or not, you must show some effort to do the problem yourself. This particular problem is a bit tedious but a straightforward calculation.
     
  5. Nov 29, 2009 #4
    this is what i got ./.i want to see what other people get first

    ∫[-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
     
  6. Nov 29, 2009 #5
    so any 1 ?
     
  7. Nov 30, 2009 #6

    Redbelly98

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    Looks good :smile:
     
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