Proof of Formula for Advanced Calculus: Int. from a(t) to b(t)

In summary, the author is looking for a book or source that can help him understand the Fundamental Theorem of Calculus. This theorem states that, for a function that is continuous on the integration range, the integral of the function over that range is equal to the limit of the function as h approaches 0. He mentions that it can be helpful to visualize where the terms in the derivative come from, and that there is a book that he learned this information from many years ago.
  • #1
alfredblase
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I'm looking for a book/paper/online source where the following formula is shown to be true:

[tex]\frac d{dt} \int_{a(t)}^{b(t)} f(t,\tau)d\tau =\int_{a(t)}^{b(t)} \frac {\partial f(t,\tau)}{\partial t} d\tau\,+\, \frac {da(t)}{dt}f(t,a(t))\,-\, \frac {db(t)}{dt}f(t,b(t))[/tex]

I know one should post attempts at solving the problem but to be honest I wouldn't know where to start. That's one of the reasons why I asked for a reference that would explain/elucidate the proof rather than the proof itself (as well as such a source probably being very useful to me in general). I will say that I can show the Fundamental Theorem of Calculus to be true where the integrand is a function of only one variable if that helps..

Thanks for taking the time to read. Any help you offer will be very much appreciated.
 
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  • #2
When the bounds are independent of t, you have:

[tex]\frac{d}{dt} \int_a^b f(t,\tau)dt = \lim_{h \rightarrow 0} \frac{1}{h} \left( \int_a^b f(t+h,\tau)d\tau-\int_a^b f(t,\tau)d\tau \right)[/tex]

[tex]=\lim_{h \rightarrow 0} \int_a^b \frac{f(t+h,\tau)-f(t,\tau)}{h} d\tau[/tex]

If you could justify exchanging the limit and integral, you'd get that this is equal to:

[tex]=\int_a^b \lim_{h \rightarrow 0} \frac{f(t+h,\tau)-f(t,\tau)}{h} d\tau = \int_a^b \frac{\partial f(t,\tau)}{\partial t}d \tau[/tex]

Then you can get the formula you mentioned by treating the integral as a function F(a,b,t) and using the chain rule and the fundamental theorm of calculus.

One time you can justify switching the limit and the integral is when the derivative is bounded on the integration range, which would be true, eg, if the function was continuously differentiable and the integral was finite.
 
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  • #3
alfredblase said:
I'm looking for a book/paper/online source where the following formula is shown to be true:

[tex]\frac d{dt} \int_{a(t)}^{b(t)} f(t,\tau)d\tau =\int_{a(t)}^{b(t)} \frac {\partial f(t,\tau)}{\partial t} d\tau\,+\, \frac {da(t)}{dt}f(t,a(t))\,-\, \frac {db(t)}{dt}f(t,b(t))[/tex]

I know one should post attempts at solving the problem but to be honest I wouldn't know where to start. That's one of the reasons why I asked for a reference that would explain/elucidate the proof rather than the proof itself (as well as such a source probably being very useful to me in general). I will say that I can show the Fundamental Theorem of Calculus to be true where the integrand is a function of only one variable if that helps..

Thanks for taking the time to read. Any help you offer will be very much appreciated.

Perhaps it will help if you can visualize where the terms in the derivative come from in terms of the area under the curve.
 

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  • #4
"Perhaps it will help if you can visualize where the terms in the derivative come from in terms of the area under the curve."

Indeed it will. It's more useful to learn how to do that than try to remember "the formula". Learning that skill will also tell you that your equation is wrong, just by looking at it - the da/dt term should be minus and the db/dt term should be plus.

I don't have a book recommendation (I learned this stuff a very long time ago) but try and find a book that explains the principles with pictures, rather than just proving things formally.

Learning to think these things out for yourself is like learning to ride a bike, once you have "got it" you will never forget how to do it. You can then get by without remembering formulas, or having to look them up in books.
 
  • #5
thankyou so much OlderDan for your awesome and pedagogical rendition of the problem and AlephZero for that inspirational and, in my humble opinion, profound message :smile:

I'll be sure to apply the methods you described in the future and hopefuly, as in this case, with much success.

:D
 

1. What is a "Proof of Formula for Advanced Calculus: Int. from a(t) to b(t)"?

A "Proof of Formula for Advanced Calculus: Int. from a(t) to b(t)" is a mathematical proof that demonstrates the validity of a formula used in advanced calculus to find the integral of a function from a specific starting point (a) to a specific ending point (b) on a graph.

2. Why is it important to have a proof for this formula?

Having a proof for this formula is important because it ensures the accuracy and reliability of the formula. It allows mathematicians and scientists to confidently use the formula in their calculations and research, knowing that it has been proven to be correct.

3. What are the steps involved in the proof of this formula?

The proof of this formula involves breaking down the integral into smaller, more manageable parts using the fundamental theorem of calculus. This involves finding the derivative of the function and substituting it into the integral. Then, the limits of integration are substituted, and the resulting expression is simplified using algebraic techniques.

4. How is this formula used in real-world applications?

This formula is used in many real-world applications, such as physics, economics, and engineering. It allows scientists and engineers to calculate quantities such as displacement, velocity, and acceleration over a specific time period, which is essential in understanding and predicting the behavior of various systems.

5. Are there any limitations to this formula?

Like any mathematical formula, there are limitations to this formula. It may not be applicable to all functions, and there may be certain cases where the proof does not hold. It is important to understand the assumptions and conditions under which this formula can be used to ensure its accuracy.

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