# Definite integral example: don't understand

## Homework Statement

This is am example from my textbook

$$\int_{1}^{3} x^2 dx = 26/3$$

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,

$$3x_{i-1}^2 \leq x_{i-1}^2 + x_{i-1}x_i + x_i^2 \leq 3x_i^2$$

## 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

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Dick
Science Advisor
Homework Helper
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.

Huh..
I still don't get it.. how did that expression get there in the first place?
I put them to be $$x_i$$ and $$x_{i-1}$$ then I get $$3x_i^2$$ and $$3x_{i-1}^2$$

Dick
Science Advisor
Homework Helper
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.

Ok this is exactly what it says:

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

For each index i, 1 <= i <= n,
$$3x_{i-1}^2 \leq x_{i-1}^2 + x_{i-1}x_i + x_i^2 \leq 3x_i^2$$
(verify this)

Dick
Science Advisor
Homework Helper
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?

.. 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
Mentor
Ok this is exactly what it says:

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

For each index i, 1 <= i <= n,
$$3x_{i-1}^2 \leq x_{i-1}^2 + x_{i-1}x_i + x_i^2 \leq 3x_i^2$$
(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.

Dick
Science Advisor
Homework Helper
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.

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
Science Advisor
Homework Helper
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".