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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Hello! I was just working on a review assignment and in the 10% or so left, I've come across a question for which I have no idea on how to begin.

I've looked into the http://en.wikipedia.org/wiki/Leibniz_integral_rule but I'm not sure if that is the method I should be employing, as I've never come across it before. Or it is possible that I have but like I said, this is a review assignment and I haven't been in a math course for over 2 semesters so I may have forgotten it.

Any ideas on how to get started? I tried using substitution to express √(1+t^3) using only x but to no avail as the integral of √(1+t^3) itself looks far too long and complicated to be required for such a straightforward question (using a math engine).

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,930

hi Anakin,

In other words you can substitute into the integral process any alternative variable.

So I cannot see why you should just 'cancel out' the integration with the differentiation and just sub in the limits.

This worked OK with a simple function, but, I confess, I'm not 100% sure on this. As the day moves ahead, hopefully we'll get a second opinion on this.

Bob

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,475

Hi;

Do you mean differentiation under the integral sign?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Bob: If one was to change the variable in the function, wouldn't the intervals on the integral also change?

Bobbym: I mean differentiating the definite integral, with respect to x. Or at least that's what I think the question appears to be asking for.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,475

Hi Anakin;

A Leibniz integral is an differentiation under the integral sign and is something else. Otherwise Bob is right just cancel the integral and the derivative.

Please look here:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Hi,

By looking at the link that you last posted, I was able to get the following solution. Could you tell me if it seems right?

Also, it would seem that simply cancelling the integral and derivative would lead to a different solution - one that is lacking the 4x and sin(x) in front of the square roots.

Which is the right one then?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,475

Hi Anakin;

I am not getting that answer, let me look at it agian.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Hi Bobbym,

Okay, thanks. I appreciate it!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,475

Hi;

This is what I get plugging into that formula:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

That's exactly what I've got above as well. I'm not sure whether you're simply checking if my work was free of errors or whether that is the solution to the question. Could you clarify?

Also, I meant plugging the intervals into sqrt(1 + t^3) by canceling the derivative and integral out would yield a different answer from the one I got.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,475

Whenever I have seen the derivative operator in front of an integral like that they have always meant

differentiate under the integral sign. So, I am going with that.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Alright then, I'll go with that as well. Plus it fits the formula from the page you linked. 5 hours until class, I'm gonna catch some sleep.

Thanks Bobbym (and Bob) for the help, I'm very grateful.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,475

Hi Anakin;

Good luck at class.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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bobbym wrote:

Hi;

This is what I get plugging into that formula:

This is the correct answer. However, for the benefit of future viewers, I demonstrate a quick way to tackle such problems (without the need for memorising formulae). Recall the following theorem:

Let f be locally Riemann integrable over I, and let F: I -> R be an indefinite integral of f. Then:

(i) F is continuous on I;

(ii) F is differentiable at each interior point c ∈ I at which f is continuous, and satisfies F'(c) = f(c);

(iii) If f is continuous on I, then F is a *primitive* of f.

This theorem allows us to perform the following manipulations on Anakin's problem.

We let

, and put. Then:So by the chain rule:

.bob bundy wrote:

hi Anakin,

In other words you can substitute into the integral process any alternative variable.

Not strictly true -- you'd need the 't' to act as the dummy variable.

*Last edited by zetafunc (2014-05-23 21:07:22)*

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