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Homework Help: Definite integral example: don't understand

  1. Mar 4, 2010 #1
    1. The problem statement, all variables and given/known data

    This is am example from my textbook

    [tex] \int_{1}^{3} x^2 dx = 26/3 [/tex]

    It then goes to define the arbitrary partition in [1,3] and maximum / minimum for each subinterval blah blah.. then it says

    For each index i, 1 <= i <= n,

    [tex] 3x_{i-1}^2 \leq x_{i-1}^2 + x_{i-1}x_i + x_i^2 \leq 3x_i^2 [/tex]

    2. Relevant equations



    3. The attempt at a solution

    I'm totally lost about this.. why are the min / max of the interval multiplied by 3? How did the middle part get like that?
    I've seen the last example of the identity function f(x) = x they can get a midpoint between the min / max then work with that.. but this is not the same.
     
  2. jcsd
  3. Mar 4, 2010 #2

    Dick

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    x_(i-1)<x_i. Put every x in your expression x_(i-1)^2+x_i*x_(i-1)+x_i^2 to be x_(i-1). That gives you a lower bound. Now put them all to be x_i. That gives you an upper bound.
     
  4. Mar 4, 2010 #3
    Huh..
    I still don't get it.. how did that expression get there in the first place?
    I put them to be [tex]x_i[/tex] and [tex]x_{i-1}[/tex] then I get [tex]3x_i^2[/tex] and [tex]3x_{i-1}^2[/tex]
     
  5. Mar 4, 2010 #4

    Dick

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    How the expression got there is probably in the "blah, blah" part. I don't know how they are trying to work this problem. I'm just telling you where the 3's came from.
     
  6. Mar 4, 2010 #5
    Ok this is exactly what it says:

    Let P = [tex]{x_0, x_1, ... , x_n}[/tex] be an arbitrary partition of [1, 3]. On each subinterval [tex][x_{i-1}, x_i][/tex] the function [tex] f(x) = x^2[/tex] has a maximum [tex]M_i = x_i^2[/tex] and minimum [tex]m_i = x_{i-1}^2.[/tex]
    It follows that [tex]U_f(P) = x_1^2\Delta x_1 + ... + x_n^2\Delta x_n[/tex]
    and
    [tex]L_f(P) = x_0^2\Delta x_1 + ... + x_{n-1}^2\Delta x_n[/tex]
    and then it pops to the part I was asking about

    For each index i, 1 <= i <= n,
    [tex]
    3x_{i-1}^2 \leq x_{i-1}^2 + x_{i-1}x_i + x_i^2 \leq 3x_i^2
    [/tex]
    (verify this)
     
  7. Mar 4, 2010 #6

    Dick

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    Sorry, but I'm really not sure where they are going with that. But as far as the (verify this) part, I think you are ok. Where do they go from there?
     
  8. Mar 4, 2010 #7
    .. then they just start to simplify it, they divide by 3 to all parts and then they multiply everything by delta x_i so the middle is bounded by the lower and upper sums.. and then the middle part simplifies to (1/3)(x_i^3 - x_(i-1)^3).. then they sum up all the index i 1 to n and the left part and right part becomes the lower and upper sums of the whole interval and the middle stuff becomes 26/3.. which is I guess the area under the curve from 1 to 3.
     
  9. Mar 5, 2010 #8

    Mark44

    Staff: Mentor

    I'm not sure whether you understand this part. The function f is strictly increasing for x >= 0, so on each subinterval, the smallest function value comes at the left end -- xi - 1, and the largest function value comes at the right end -- xi.
    Since xi -1 < xi, xi - 12 < xi2. A number that's between xi - 12 and xi2 is xi - 1xi.

    Now if you add these three expressions together, xi - 12 + xi - 1xi + xi2, you get an expression that is greater than or equal to 3 times the smallest number, and less than or equal to 3 times the larger number.

    Evidently they're going to do something with this fact, but what that is isn't apparent just yet.
     
  10. Mar 5, 2010 #9

    Dick

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    Ohhh. I see where they are going with it, Mark44. If you multiply that expression by the interval lengths (x_i-x_(i-1)) you get an approximation to 3*I where I is the integral you are after. But if you multiply it out you get (x_i)^3-(x_(i-1))^3. This last expression is very easy to sum over i.
     
  11. Mar 5, 2010 #10
    Okay so,
    if a < b, then
    a2 < b 2
    a2 < ab < b2

    and then

    3a < a2 + ab + b2 < 3b ?

    Is this some general inequality that I should know? Or are they just pulling this out of nowhere so to make things work out nicely in the end?
     
  12. Mar 5, 2010 #11

    Dick

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    3a^2 < a2 + ab + b2 < 3b^2. It's not a general inequality you should know. They do know the integral of x^2 should come out to be. x^3/3. That expression happens to be handy because b^3-a^3=(b-a)*(a^2+ab+b^2). So, yes, in some sense they are "pulling this out of nowhere so to make things work out nicely in the end".
     
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