- #1

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On number 18...

1) What is the logic behind separating x^7 into x^4 x^3

2) In the translation from dv to v, how did x^3 become (1/12)? I fail to see what is happening here.

Thanks for the help guys.

- Thread starter yUNeeC
- Start date

- #1

- 34

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On number 18...

1) What is the logic behind separating x^7 into x^4 x^3

2) In the translation from dv to v, how did x^3 become (1/12)? I fail to see what is happening here.

Thanks for the help guys.

- #2

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...I'm sorry but I don't think you understood my questions.

- #3

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1) I don't see what your question is. x^7 = x^4*x^3 because 3 + 4 = 7. This is a basic property of exponents.

2) Do the substitution w = x^2. Then dw = 2x dx, therefore x dx = 1/2 dw. Using this,

x^3 * (5 + 3*x^4)^1/2 dx = x^2 * x * (5 + 3*x^4)^1/2 dx = w * x * (5 + 3*w^2)^1/2 dx = w * (5 + 3*w^2)^1/2 x dx = 1/2 w * (5 + 3*w^2)^1/2 du. Then use integral (31) found here: http://www.integral-table.com/" [Broken].

2) Do the substitution w = x^2. Then dw = 2x dx, therefore x dx = 1/2 dw. Using this,

x^3 * (5 + 3*x^4)^1/2 dx = x^2 * x * (5 + 3*x^4)^1/2 dx = w * x * (5 + 3*w^2)^1/2 dx = w * (5 + 3*w^2)^1/2 x dx = 1/2 w * (5 + 3*w^2)^1/2 du. Then use integral (31) found here: http://www.integral-table.com/" [Broken].

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- #4

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For the expression they split out:

dv = x^{3}(5+3x^{4})^{1/2}

You need to find an integral of this. An integral of this would look something along the lines of

(5 + 3x^{4})^{3/2}

Differentiating (5 + 3x^{4})^{3/2} would give you

12x^{3} * 3/2 (5 + 3x^{4})^{1/2}

or

18x^{3}(5 + 3x^{4})^{1/2}

so to bring it in line with the expression you have, the integral needs a constant in the form of 1/18 meaning

x^{3}(5+3x^{4})^{1/2}

integrates to

1/18(5+3x^{4})^{3/2}

If you don't follow the logic of this, revisit some more basic integration.

This also explains the logic behind splitting x^{7} into x^{4} and x^{3}. It gives you the correct power of x to multiply the bracket with in order to make an integration of the bracket easier (without doing this, you would have:

u = x^{7}

dv = (5 + 3x^{4})^{1/2}

v = 1/18 * x^{-3}(5 + 3x^{4})^{3/2}

which just isn't as nice to work with (and the factor of x^{-3} should end up cancelling out in the next step anyway)

dv = x

You need to find an integral of this. An integral of this would look something along the lines of

(5 + 3x

Differentiating (5 + 3x

12x

or

18x

so to bring it in line with the expression you have, the integral needs a constant in the form of 1/18 meaning

x

integrates to

1/18(5+3x

If you don't follow the logic of this, revisit some more basic integration.

This also explains the logic behind splitting x

u = x

dv = (5 + 3x

v = 1/18 * x

which just isn't as nice to work with (and the factor of x

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- #5

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I appreciate the help. I'm trying to learn Calc II by myself, and some of these forms of integration elude me. I need to see one example, and understand one example of each type before I "get it," and this is the first time I've had to integrate something like this. Just yesterday I began grasping the integration of the "brothers" (for lack of a better word) of e and lnx...now I'm trying to understand roots.

A couple of questions:

^{3}? Where does the /(3/2) go to?

Basically, why isn't the integral of x^{3}(5+3x)^{1/2}.....x^{4} / 4 (2/3)(5+3x)^{3/2}?

And when you differentiate again...how does the 1/12 suddenly get reintroduced?

So...once you see that x^{3} = (1/12), you use that to simplify all x^{3}?

Gracias

Edit:

I think I just saw something. Do you take the derivative of what's under the square root and multiply it by whats outside...resulting in x^{3} / 12x^{3}...equalling 1/12? If this is the case, how does the 1/12 disappear and then reappear?

A couple of questions:

Do you just drop the xFor the expression they split out:

dv = x^{3}(5+3x^{4})^{1/2}

You need to find an integral of this. An integral of this would look something along the lines of

(5 + 3x^{4})^{3/2}

Basically, why isn't the integral of x

And when you differentiate again...how does the 1/12 suddenly get reintroduced?

This also explains the logic behind splitting x^{7}into x^{4}and x^{3}. It gives you the correct power of x to multiply the bracket with in order to make an integration of the bracket easier (without doing this, you would have:

u = x^{7}

dv = (5 + 3x^{4})^{1/2}

v = 1/18 * x^{-3}(5 + 3x^{4})^{3/2}

which just isn't as nice to work with (and the factor of x^{-3}should end up cancelling out in the next step anyway)

So...once you see that x

Gracias

Edit:

I think I just saw something. Do you take the derivative of what's under the square root and multiply it by whats outside...resulting in x

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- #6

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Ok, remember that integration is the reverse of differentiation. If you want to differentiate a formula such as:

v = (5 + 3x^{4})^{3/2}

Then the differentiation really needs to be done by substitution as such:

U = 5 + 3x^{4}

v = U^{3/2}

and

dv/dx = (dU/dx)(dv/dU)

Now, differentiating,

dv/dU = 3/2U^{1/2} (a pretty standard differentiation)

dU/dx = 12x^{3}

so

dv/dx = 12x^{3} * 3/2U^{1/2}

and substituting U back in and simplifying

dv/dx = 18x^{3}(5 + 3x^{4})^{1/2}

so, integrating is the reverse of this. Therefore if you have a formula such as

dv = x^{3}(5 + 3x^{4})^{1/2}

you can see it would integrate to something like

v = (5 + 3x^{4})^{3/2}

but in this case, we don't have the factor of 18 at the front. So to get the correct integral (ignoring the constant of integration here) you need to divide your final answer by this factor to correct it, arriving at an integral of:

1/18(5 + 3x^{4})^{3/2}

Now, with your integration by parts, you can either split it along the obvious bounds of

u=x^{7}

dv = (5 + 3x^{4})^{1/2}

and procede simply like this, or you can recognise that you can 'borrow' x^{3} from x^{7} to give you

u=x^{4}

dv=x^{3}(5 + 3x^{4})^{1/2}

and this gives you a simpler v to work with when you integrate it. Either way should produce the same answer in this case, however it is useful to learn to recognise these forms as sometimes a split like that is the only way to correctly integrate, or provides a much easier route to the final answer.

I hope that makes things clearer.

Also, if you are self-teaching and want to learn (one method) of how integration works, I'd suggest going for something like the book here:

http://www.math.wisc.edu/~keisler/calc.html

This is an approach to calculus from first principles using the fairly easily understood 'infinitesimal' approach with hyperreal numbers (basically real numbers with the concept of infinitely small and infinitely big numbers introduced). With a first principles approach like this, all these integration and differentiation forms fall out algebraically which makes them much easier to understand and remember in the future :)

v = (5 + 3x

Then the differentiation really needs to be done by substitution as such:

U = 5 + 3x

v = U

and

dv/dx = (dU/dx)(dv/dU)

Now, differentiating,

dv/dU = 3/2U

dU/dx = 12x

so

dv/dx = 12x

and substituting U back in and simplifying

dv/dx = 18x

so, integrating is the reverse of this. Therefore if you have a formula such as

dv = x

you can see it would integrate to something like

v = (5 + 3x

but in this case, we don't have the factor of 18 at the front. So to get the correct integral (ignoring the constant of integration here) you need to divide your final answer by this factor to correct it, arriving at an integral of:

1/18(5 + 3x

Now, with your integration by parts, you can either split it along the obvious bounds of

u=x

dv = (5 + 3x

and procede simply like this, or you can recognise that you can 'borrow' x

u=x

dv=x

and this gives you a simpler v to work with when you integrate it. Either way should produce the same answer in this case, however it is useful to learn to recognise these forms as sometimes a split like that is the only way to correctly integrate, or provides a much easier route to the final answer.

I hope that makes things clearer.

Also, if you are self-teaching and want to learn (one method) of how integration works, I'd suggest going for something like the book here:

http://www.math.wisc.edu/~keisler/calc.html

This is an approach to calculus from first principles using the fairly easily understood 'infinitesimal' approach with hyperreal numbers (basically real numbers with the concept of infinitely small and infinitely big numbers introduced). With a first principles approach like this, all these integration and differentiation forms fall out algebraically which makes them much easier to understand and remember in the future :)

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- #7

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That was an amazing help. Thanks so much.

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I think part of my problem is just not understanding the background info of the integral. I can do u-substitution, partial integration, and a decent amount of integration by parts...but when you mix the integration by parts with needed u-substitutions I can get lost. But then again, I haven't seen this stuff before...but hopefully this book will familiarize me with things so I know WHY to do _________ instead of just how.

- #11

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I'm not familiar with american courses either though... I remember doing this stuff in my Further Maths A-Level about 8 years ago now... and then my degree where I never touched the stuff and now I'm interested in maths and physics again. That book was a fairly recent find of mine and although I haven't read all of it yet I'm happier with my knowledge of calculus just knowing there are nice algebraic methods to work everything out and rough ideas of what they are :)

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