# Definite integral example: don't understand

• zeion
In summary, the conversation discusses an example from a textbook involving an integral and an arbitrary partition. It explains how the function has a maximum and minimum on each subinterval and how this is used to find upper and lower sums. The conversation then focuses on an expression involving the function and its integral, and how it is simplified to find the area under the curve. The conversation ends with a discussion on a general inequality that is used to make the calculations work out nicely.

## 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.

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 $$3x_{i-1}^2$$

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
$$L_f(P) = x_0^2\Delta x_1 + ... + x_{n-1}^2\Delta x_n$$

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)

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.

zeion said:
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.
zeion said:
It follows that $$U_f(P) = x_1^2\Delta x_1 + ... + x_n^2\Delta x_n$$
and
$$L_f(P) = x_0^2\Delta x_1 + ... + x_{n-1}^2\Delta x_n$$

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.

Mark44 said:
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?

zeion said:
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".

## 1. What is a definite integral?

A definite integral is a mathematical concept that represents the area between a function and the x-axis on a given interval. It is denoted by the symbol ∫ and has two limits, the lower and upper bound, which determine the interval of integration.

## 2. How is a definite integral calculated?

A definite integral can be calculated using the fundamental theorem of calculus, which states that the definite integral of a function can be found by evaluating the antiderivative of the function at the upper bound and subtracting the antiderivative at the lower bound.

## 3. What is the purpose of a definite integral?

The purpose of a definite integral is to find the area under a curve or the net change in a quantity over a given interval. It is also used in many real-world applications such as calculating volumes, work, and displacement.

## 4. What is the difference between a definite integral and an indefinite integral?

A definite integral has specific limits, whereas an indefinite integral does not. This means that a definite integral will give a numerical value, while an indefinite integral will give a function with a constant of integration.

## 5. How can I better understand definite integrals?

To better understand definite integrals, it is helpful to practice with various examples and to have a strong understanding of basic calculus concepts such as derivatives and antiderivatives. It is also beneficial to visualize the area under a curve and how it relates to the definite integral.