Sorry, how to integrate this function from step1 to step 2

In summary, integrating a function is a mathematical process used to find the area under a curve between two given points. The method of integration used depends on the type of function, and there are various methods such as substitution, integration by parts, and trigonometric substitution. Calculators and software programs can also be used, but it is important to have a solid understanding and be able to check for accuracy. The basic steps for integrating a function involve identifying the limits of integration, choosing an appropriate method, integrating the function, and evaluating the definite integral. Special cases and exceptions, such as functions with vertical asymptotes or improper integrals, require specific techniques for integration.
  • #1
garylau
70
3
Sorry how to integrate this function from step1 to step 2(see the circle inside the picture)

it looks too complicated
i can't help with it
i cannot use partial fraction to do because it looks hard to decompose it
thank
 

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  • #2
I thought partial fractional decomposition required both numerator and denominator to be polynomials?
 
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  • #3
We're trying to integrate
[itex]\frac{z-Ru}{(R^2 + z^2 - 2Rzu)^{\frac{3}{2}}}[/itex]

We can rewrite this as:

[itex]\frac{R}{(2Rz)^{\frac{3}{2}}} \frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}[/itex]

Now [itex]\frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}[/itex] is of the form

[itex](A-u)(C-u)^\alpha[/itex]

with [itex]A = \frac{z}{R}[/itex], [itex]C = \frac{R^2 + z^2}{2Rz}[/itex], [itex]\alpha = \frac{-3}{2}[/itex]

Here's the general way to integrate such expressions:

Rewrite this as:
[itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex]

We can easily integrate these terms, to get:

[itex]-\frac{A-C}{\alpha+1} (C-u)^{\alpha+1} - \frac{1}{\alpha+2} (C-u)^{\alpha+2}[/itex]
[itex]= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} ((A-C)(\alpha+2) + (C-u)(\alpha+1))[/itex]
[itex]= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} (A (\alpha+2) - C - u(\alpha+1))[/itex]

So if you plug in our known values for [itex]A, C, \alpha[/itex], you should get the desired result (after some simplification).
 
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  • #4
stevendaryl said:
We're trying to integrate
[itex]\frac{z-Ru}{(R^2 + z^2 - 2Rzu)^{\frac{3}{2}}}[/itex]

We can rewrite this as:

[itex]\frac{R}{(2Rz)^{\frac{3}{2}}} \frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}[/itex]

Now [itex]\frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}[/itex] is of the form

[itex](A-u)(C-u)^\alpha[/itex]

with [itex]A = \frac{z}{R}[/itex], [itex]C = \frac{R^2 + z^2}{2Rz}[/itex], [itex]\alpha = \frac{-3}{2}[/itex]

Here's the general way to integrate such expressions:

Rewrite this as:
[itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex]

We can easily integrate these terms, to get:

[itex]-\frac{A-C}{\alpha+1} (C-u)^{\alpha+1} - \frac{1}{\alpha+2} (C-u)^{\alpha+2}[/itex]
[itex]= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} ((A-C)(\alpha+2) + (C-u)(\alpha+1))[/itex]
[itex]= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} (A (\alpha+2) - C - u(\alpha+1))[/itex]

So if you plug in our known values for [itex]A, C, \alpha[/itex], you should get the desired result (after some simplification).

where is this form
[itex](A-u)(C-u)^\alpha[/itex]
coming from

and Where do you got [itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex] this equation from?

thank
 
Last edited:
  • #5
garylau said:
where is this form
[itex](A-u)(C-u)^\alpha[/itex]
coming from

He's just substituting there. A and C are the two terms that have r and z in it. In the first step he just rearranged it to get u by itself in both the numerator and deniminator, so that he could make the substitution.

The second part is just adding and subtracting C, and then using the associstive property. You'll note that A - u = A + C - C - u.
 
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  • #6
Honestly that's a really clever trick and I think I'm going to commit this one to memory.
 
  • #7
garylau said:
where is this form
[itex](A-u)(C-u)^\alpha[/itex]
coming from

and Where do you got [itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex] this equation from?

thank
I suspect both of those relationships came from a book of integrals.
 
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  • #8
garylau said:
where is this form
[itex](A-u)(C-u)^\alpha[/itex]
coming from

The integrand is (after factor out the factor [itex]\frac{2R}{(2Rz)^{3/2}}[/itex]):

[itex]\frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^{3/2}}[/itex]

That's the same as:

[itex](\frac{z}{R} - u)(\frac{R^2 + z^2}{2Rz} - u)^{-3/2}[/itex]

Now, let [itex]A = \frac{z}{R}, C = \frac{R^2 + z^2}{2Rz}, \alpha = -3/2[/itex], so you get

[itex](A - u)(C - u)^{\alpha}[/itex]

...and Where do you got [itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex] this equation from?

Battlemage! explained it above.
 
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  • #9
But the temptation to integrate with respect to ##z## (since ##(z^2-Rzu)'=2(z-Ru)##), then integrate with respect to ##u##, and then differentiate with respect to ##z## is just irresistible...
 
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  • #10
Dragon27 said:
But the temptation to integrate with respect to ##z## (since ##(z^2-Rzu)'=2(z-Ru)##), then integrate with respect to ##u##, and then differentiate with respect to ##z## is just irresistible...
Ye gads, is this not cheating? Would that actually work, and if so is it easier?
 
  • #11
Battlemage! said:
Ye gads, is this not cheating?
It feels so good, that it probably is...
Battlemage! said:
Would that actually work, and if so is it easier?
The integrals in the intermediate steps are much easier, in my opinion (or, maybe, more obvious). In the end you just differentiate a simple fraction, it's not too hard.
 
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1. What is the purpose of integrating a function?

Integrating a function is a mathematical process used to find the area under a curve between two given points. It is often used in physics, engineering, and other scientific fields to calculate things such as displacement, velocity, and acceleration.

2. How do I know which method of integration to use?

The method of integration used depends on the type of function being integrated. Some common methods include substitution, integration by parts, and trigonometric substitution. It is important to understand the properties of each method and determine which one is best suited for the given function.

3. Can I use a calculator to integrate a function?

Yes, there are many online calculators and software programs that can integrate functions for you. However, it is important to have a solid understanding of the integration process and the ability to check the accuracy of the results.

4. What are the steps involved in integrating a function?

The basic steps for integrating a function include identifying the limits of integration, choosing an appropriate method, integrating the function, and evaluating the definite integral using the limits of integration. It is also important to simplify the result and check for any errors.

5. Are there any special cases or exceptions when integrating a function?

Yes, there are some special cases and exceptions when integrating a function. These include functions with vertical asymptotes, discontinuities, and improper integrals. It is important to be aware of these cases and use proper techniques to integrate them.

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