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Definite integral example: don't understand

  • Thread starter zeion
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  • #1
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Homework Statement



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]

Homework Equations





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.
 

Answers and Replies

  • #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.
 
  • #3
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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]
 
  • #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.
 
  • #5
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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)
 
  • #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?
 
  • #7
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.. 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.
 
  • #8
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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]
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.
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)
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.
 
  • #9
Dick
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Evidently they're going to do something with this fact, but what that is isn't apparent just yet.
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.
 
  • #10
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0
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?
 
  • #11
Dick
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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?
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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