This is going to be a really difficult question to ask. Multiplying and dividing by

  • Thread starter flyingpig
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In summary: I believe a definite integral needs to be defined on the open interval between the two endpoints, and x/x is defined on the interval (0,1). What you're saying, which is true, is that it's not defined on the interval [0,1]. But I believe that definite integrals only require the open interval. In your example, if x/x is undefined within the intervals [0,1] and [0,2], then that integral isn't a definite integral.
  • #1
flyingpig
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Homework Statement



I thought I understood what it means to multipyl and divide something, but lately I think it's wrong.

Mostly in Calculus, there are functions that can't be integrated unless we multiply and divide by something. Usually trig integrals need that, but even a simple integral like

[tex]\int_{0}^{1} dx[/tex]

Can go wrong if we multiply and divide by the same thing. Also most of the time we aren't just multiplying and dividing by a number, we do a variable.

For instance, to the above integral, if I go ahead and multiply and divide x (x/x) then that integral becomes undoable.

Often we take granted to use this trick, but sometimes it doesn't work.

How does one recognize it if the function we are "integrating" creates a problem(especially if it is very very complicated)? How do we catch the mistake?
 
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  • #2


flyingpig said:

Homework Statement



I thought I understood what it means to multipyl and divide something, but lately I think it's wrong.

Mostly in Calculus, there are functions that can't be integrated unless we multiply and divide by something. Usually trig integrals need that, but even a simple integral like

[tex]\int_{0}^{1} dx[/tex]

Can go wrong if we multiply and divide by the same thing. Also most of the time we aren't just multiplying and dividing by a number, we do a variable.

For instance, to the above integral, if I go ahead and multiply and divide x (x/x) then that integral becomes undoable.
How so?
If the integrand is now x/x, that function is continuous everywhere except at 0, and is equal to 1 for all x other than zero. Both integrals (the original one and the new one) have identical values, although you need to use a limit to evaluate the new one.
flyingpig said:
Often we take granted to use this trick, but sometimes it doesn't work.

How does one recognize it if the function we are "integrating" creates a problem(especially if it is very very complicated)? How do we catch the mistake?
I don't see what the problem is. The only thing to be concerned about is if you multiply by 1 in the form of some expression over itself, and the denominator expression is zero for some value in the interval of integration.
 
  • #3


No but that's the point, x/x changes the function completely because we can't even integrate it
 
  • #4


No, "multiplying by x/x" doesn't change it at all:
[tex]\int\frac{x}{x}dx= \int dx= x+ C[/tex]
 
  • #5


HallsofIvy said:
No, "multiplying by x/x" doesn't change it at all:
[tex]\int\frac{x}{x}dx= \int dx= x+ C[/tex]

But I referring to the definite integral
 
  • #6


flyingpig said:
But I referring to the definite integral
I covered this in post #2. The two integrals below have exactly the same value, 1.
[tex]\int_{0}^{1} dx[/tex]
[tex]\int_{0}^{1} \frac{x}{x}dx[/tex]
 
  • #7


In the strict sense, the Riemann integral doesn't exist because the integrand isn't defined at x=0, but that's part of the reason why other types of integrals were developed. So in your example, the answer would be that if you're evaluating a Riemann integral, you don't multiply by x/x because you're not allowed to divide by 0.
flyingpig said:
How does one recognize it if the function we are "integrating" creates a problem(especially if it is very very complicated)? How do we catch the mistake?
To put it simply, you have to think carefully. You have to check for and be aware of these possibilities. That's why math teachers like to show you fake proofs of non-sensical results, like 0=1. They're typically based on dividing or multiplying by zero in a non-obvious way. Your teachers want to point out that you always need to be careful to avoid that mistake.

As far as complicated integrands go, you need to learn about how functions behave to avoid running into mistakes. There's a reason why you study curve drawing and other topics which depend on analyzing a function. It's often not good enough to just learn how to turn the crank to obtain an answer.

If you do make a mistake, what happens is usually you get a result that doesn't make sense (e.g., 0=1). It won't usually be as obvious as that example, which is why you should always check your work and check your results to see if they make sense.
 
  • #8


By L'Hospital's rule the function = 1 when x = 0.
 
  • #9


Mark44 said:
I covered this in post #2. The two integrals below have exactly the same value, 1.
[tex]\int_{0}^{1} dx[/tex]
[tex]\int_{0}^{1} \frac{x}{x}dx[/tex]

How can they both be 1 when x = 0 isn't even defined. The area of the rectangles aren't the same.
 
  • #10


flyingpig said:
How can they both be 1 when x = 0 isn't even defined. The area of the rectangles aren't the same.

I believe a definite integral needs to be defined on the open interval between the two endpoints, and x/x is defined on the interval (0,1). What you're saying, which is true, is that it's not defined on the interval [0,1]. But I believe that definite integrals only require the open interval.
 
  • #11


That integral is perfectly well defined even when taken from -1 to 1.

At zero, you evaluate the integrand as [d(x)/dx]/[d(x)/dx] = 1/1 = 1.
 
  • #12


The function x/x is Riemann integrable over [0,1]. The point x = 0 doesn't create any problems.

A set E has measure if for every [tex] \epsilon > 0 [/tex] there is a countable covering of E by closed intervals [tex] [a_i, b_i] [/tex] such that [tex] \sum_{i = 0}^\infty (b_i - a_i) < \epsilon [/tex] as [tex] b_i - a_i [/tex] is the length of the interval [tex] [a_i, b_i] [/tex].

A function is Riemann integrable if and only if it is bounded and its set E of points at which it is discontinuous has measure zero.

Since the only point at which the function x/x is discontinuous is at 0, then E = {0}. For any [tex] \epsilon > 0 [/tex] the closed interval [tex] [-\epsilon / 4, \epsilon / 4] [/tex] covers E = {0}, and the interval has length [tex] \epsilon / 2 < \epsilon [/tex]. Thus E has measure zero and x/x is integrable over [0,1]
 
  • #13


Using integration by parts on:
[tex]
\int_{0}^{1}{\frac{x}{x} \, dx}
[/tex]
with [itex]u = 1/x \Rightarrow du = - dx/x^{2}[/itex] and [itex]dv = x \, dx \Rightarrow v = x^{2}/2[/itex], we get:
[tex]
\int_{0}^{1}{\frac{x}{x} \, dx} = \left.\frac{1}{x} \, \frac{x^{2}}{2}\right|^{1}_{0} + \frac{1}{2} \, \int_{0}^{1}{\frac{x^{2}}{x^{2}} \, dx} = \frac{1}{2} + \frac{1}{2} \, \int_{0}^{1}{\frac{x^{2}}{x^{2}} \, dx}
[/tex]
Continuing inductively:
[tex]
\int_{0}^{1}{\frac{x^{n}}{x^{n}} \, dx} = \left.\frac{1}{x^{n}} \, \frac{x^{n + 1}}{n + 1}\right|^{1}_{0} + \frac{n}{n + 1} \, \int_{0}^{1}{\frac{x^{n+1}}{x^{n + 1}} \, dx} = \frac{1}{n + 1} + \frac{n}{n + 1} \, \int_{0}^{1}{\frac{x^{n+1}}{x^{n + 1}} \, dx}
[/tex]
we get:
[tex]
\int^{1}_{0}{\frac{x}{x} \, dx} = \frac{1}{2} + \frac{1}{2} \left[\frac{1}{3} + \frac{2}{3} \, \int_{0}^{1}{\frac{x^{3}}{x^{3}} \, dx}\right] = \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{3} \, \int_{0}^{1}{\frac{x^{3}}{x^{3}} \, dx}
[/tex]
[tex]
=\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \ldots + \frac{1}{n \cdot (n + 1)} + \frac{1}{n + 1} \, \int_{0}^{1}{\frac{x^{n + 1}}{x^{n + 1}} \, dx}
[/tex]
The integral is actually always the same. Assuming it is finite, when we take the limit [itex]n \rightarrow \infty[/itex], the integral on the r.h.s. is multiplied by a factor [itex]1/(n + 1)[/itex], so that term tends to zero and we have the following identity:
[tex]
\int_{0}^{1}{dx} = \sum_{n = 1}^{\infty}{\frac{1}{n \, (n + 1)}}
[/tex]
If you use partial fraction decomposition:
[tex]
\frac{1}{n \, (n + 1)} = \frac{1}{n} - \frac{1}{n + 1}
[/tex]
you will see that the sum is actually:
[tex]
\sum_{n = 1}^{\infty}{\frac{1}{n \, (n + 1)}} = \sum_{n = 1}^{\infty}{\left[\frac{1}{n} - \frac{1}{n + 1}\right]} = 1
[/tex]
as is the value of the integral.

So, nothing is wrong. You just found a method for evaluating some sums.
 
  • #14


Mark44 said:
I covered this in post #2. The two integrals below have exactly the same value, 1.
[tex]\int_{0}^{1} dx[/tex]
[tex]\int_{0}^{1} \frac{x}{x}dx[/tex]

flyingpig said:
How can they both be 1 when x = 0 isn't even defined. The area of the rectangles aren't the same.

x = 0 is defined, but x/x is not defined at 0. I'm sure that's what you meant, but it wasn't what you said.

The areas of the two rectangles are exactly the same.

As I mentioned before, the second integral above is improper, so it needs to be evaluated using a limit.
[tex]\int_{0}^{1} \frac{x}{x}dx = \lim_{a \to 0^+}\int_a^1 \frac{x}{x}dx [/tex]
[tex]= \lim_{a \to 0^+}\int_a^1 dx = 1 [/tex]
 
  • #15


Mark44 said:
x = 0 is defined, but x/x is not defined at 0. I'm sure that's what you meant, but it wasn't what you said.

The areas of the two rectangles are exactly the same.

As I mentioned before, the second integral above is improper, so it needs to be evaluated using a limit.
[tex]\int_{0}^{1} \frac{x}{x}dx = \lim_{a \to 0^+}\int_a^1 \frac{x}{x}dx [/tex]
[tex]= \lim_{a \to 0^+}\int_a^1 dx = 1 [/tex]

Okay this makes more sense to me.

But that's only when we take the limit. The two rectangles still aren't the same, I mean one of them have a hole in the corner. I know I am being really persistent and annoying with this even though the numbers are telling me I am wrong, but I just find it very difficult to accept this
 
  • #16


The "hole in the corner" doesn't affect the area of the rectangle. Nor does it affect the limit.
 
  • #17


HallsofIvy said:
The "hole in the corner" doesn't affect the area of the rectangle. Nor does it affect the limit.

So you are telling me that if I decide to not just poke a hole and just poke a line of holes, essentially wiping out the line,the rectangle's area is still the same? I guess what I am trying to ask is, how many holes can I poke to change the area?
 
  • #18


How many holes? Changing the value of a function at a countable number of points does not change the value of the integral of the function.
 
  • #19


So what if I decide to create another discontinutiy by multiplying it by say (x-2)/(x-2) and x/x together?

Still no change?

What if I do

[tex]\frac{x}{x} \frac{x-1}{x-1} \frac{x-2}{x-2} \frac{x-3}{x-3}...[/tex]
 
  • #20


I answered that in post 18:
Changing the value of a function at a countable number of points does not change the value of the integral of the function.
 
  • #21


HallsofIvy said:
I answered that in post 18:

Is this count as countable?

[tex]\frac{x}{x} \frac{x-1}{x-1} \frac{x-2}{x-2} \frac{x-3}{x-3}...[/tex]

What do you mean countable anyways? Do you mean alternating holes?
 
  • #22


I don't find this idea to be mathematically sound, but it's just a little something you might find useful. If you think of the integral as being the area between the x-axis and the function, if the function isn't defined at a point then the "area" will be a vertical line shooting off to infinite. Lines don't have any area so you can neglect it. So really, the integral will be no different if we have the line there or not - thus, the hole in the function doesn't matter.
 
  • #23


A set is countable if there is a one-to-one mapping between the members of the set and the positive integers. In your example, you are adding a countable number of discontinuities to the integrand, so the value of the integral does not change.
 

Related to This is going to be a really difficult question to ask. Multiplying and dividing by

What does "multiplying and dividing by" mean?

"Multiplying" and "dividing" are mathematical operations used to calculate the total or relative amount of a quantity. When we say "multiplying and dividing by", it means using these operations to manipulate a number or set of numbers in a specific way.

Why is multiplying and dividing by important?

Multiplying and dividing by is important because it allows us to solve complex mathematical problems and make calculations more efficiently. It is also used in many real-life situations, such as calculating prices, measurements, and proportions.

What is the order of operations when multiplying and dividing by?

The order of operations when multiplying and dividing by is the same as the general order of operations in math: parentheses, exponents, multiplication and division (from left to right), and addition and subtraction (from left to right).

How do I know when to multiply and when to divide?

You can determine whether to multiply or divide by looking at the given problem or situation. For example, if you need to find the total cost of 5 items that each cost $2, you would multiply 5 by 2. If you need to find the average of 10 numbers, you would divide the sum of the numbers by 10.

Are there any rules or tricks for multiplying and dividing by certain numbers?

Yes, there are a few rules and tricks for multiplying and dividing by certain numbers. For example, when multiplying by 10, you can simply add a 0 to the end of the number. When dividing by 2, you can divide the number by 2 and then move the decimal point one place to the left. There are also specific rules for multiplying and dividing by negative numbers. However, it is important to understand the concept of multiplying and dividing before relying on these tricks.

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