Proving Calculus:Invariant w.r.t. Integral function?

In summary, In high school we are taught one semantic of x and u, while in this problem there are two. The equation can still be solved using either semantics, but it is more clear and easier to do so using the first semantics. However, one problem with this solution is that it might not be mathematically correct to assume that u=x.
  • #1
cyt91
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Hi.

How do we solve this? I tried using the hint,but I still couldn't prove it.

Thank you for your help.
 
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  • #2
cyt91 said:
I tried using the hint,but I still couldn't prove it.

Well, can you at least show what you got?
 
  • #4
You're almost done, do you see it?

It might help to change your dummy variables back to x, so that what you have more closely resembles the thing you're trying to prove, and maybe to rewrite in full the equation you have proven with your manipulations.
 
  • #5
I got the solution by replacing u with x.

However,one thing that bothers me is that,can we assume u=pi-x and then later assume u=x?
Doesn't that make things mathematically incorrect? Since pi-x =/= x?

Is this what we call invariance w.r.t. integral function? What is invariance w.r.t. integral function anyway?
 
  • #6
cyt91 said:
What is invariance w.r.t. integral function anyway?
I'm not familiar with the phrase. Where did you hear it? Can you use it in a sentence or a paragraph?


However,one thing that bothers me is that,can we assume u=pi-x and then later assume u=x?
There are two perspectives that lead to the same result.

The semantics you're taught in elementary calculus, all of the occurrences of x and u in this problem are dummy variables.

It's not like solving a geometry problem where you have a hypothetical rectangle of and you define L to mean its length and W to mean its width and they have a fixed and relevant meaning throughout the lifetime of the problem.

Instead, their only meaning is "I am the variable this definite integral is integrating over" and they have no meaning outside of the integral -- its only purpose is to make it easier to write the function inside the integral. The expressions [itex]\int_a^b f(p) \, dp[/itex] and [itex]\int_a^b f(q) \, dq[/itex] literally mean exactly the same thing. Tnd the notation [itex]\int_a^b f[/itex] is sometimes used when feasible, which further emphasizes the irrelevance of dummy variables.




In the other semantics that's often used (but rarely stated explicitly at your level), the integrals [itex]\int_a^b f(p) \, dp[/itex] and [itex]\int_a^b f(q) \, dq[/itex] might mean different things, but the definition of the integral immediately implies that the two integrals evaluate to the same real number -- you have an identity:
[tex]\int_a^b f(p) \, dp = \int_a^b f(q) \, dq[/tex]​
which is just another integral law.
 
  • #7
Hi.
Thank you for your helpful explanation.
I understand now.

The term invariant w.r.t. integral function is basically what you've explained. That is:

[itex]\int_a^b f(p) \, dp = \int_a^b f(q) \, dq[/itex]

My tutor said that we can change x to any variable we like( in your words,dummy variables) so long as the other notation/symbols are unchanged. Hence the phrase 'invariant w.r.t to integral function'. Anyway,I have not encountered the term in high school before and I don't think it's very widely used.

I feel that your explanation is more useful and simpler.

Thank you.
 

FAQ: Proving Calculus:Invariant w.r.t. Integral function?

1. What is the basic concept of proving calculus: invariant w.r.t. integral function?

The basic concept is to show that the integral function does not change under certain transformations, such as change of variables or limits of integration. This is important in calculus because it allows us to simplify complex integrals and make use of known properties of the integral function.

2. What is the importance of proving calculus: invariant w.r.t. integral function?

The importance lies in the fact that it allows us to solve a wide range of problems in calculus with greater ease and efficiency. It also provides a deeper understanding of the fundamental concepts of calculus.

3. What are some common techniques used to prove calculus: invariant w.r.t. integral function?

Some common techniques include substitution, integration by parts, u-substitution, and trigonometric identities. These techniques involve manipulating the integral function to make it more easily solvable while keeping it invariant.

4. Can you provide an example of proving calculus: invariant w.r.t. integral function?

Sure, one example is proving the invariance of the integral function under substitution. This involves substituting a new variable in place of the original variable in the integral and showing that the result is the same. For example, if we have the integral of x^2 dx, we can substitute u = x^2 and show that the integral becomes 1/2 u du, which is equal to 1/3 x^3 + C, the original integral.

5. How does proving calculus: invariant w.r.t. integral function relate to real-world applications?

The concept of invariance of the integral function is used in a wide range of real-world applications, such as physics, engineering, and economics. For example, in physics, it is used to calculate the work done by a variable force, and in economics, it is used to calculate consumer surplus. By proving the invariance of the integral function, we can accurately solve these real-world problems and make meaningful predictions.

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