Definite integral example: don't understand

[zeion]

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.

 

[Dick]

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.

 

[zeion]

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]

 

[Dick]

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.

 

[zeion]

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)

 

[Dick]

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?

 

[zeion]

.. 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.

 

[Mark44]

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 -- x_{i - 1}, and the largest function value comes at the right end -- x_i.

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 x_{i -1} < x_i, x_{i - 1}² < x_i². A number that's between x_{i - 1}² and x_i² is x_{i - 1}x_i.

Now if you add these three expressions together, x_{i - 1}² + x_{i - 1}x_i + x_i², 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.

 

[Dick]

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.

 

[zeion]

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?

 

[Dick]

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".