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Hi, I was wondering how to tell whether to use a Substitution or Integration by Parts rule in equations. What are the main things to look for and decide?
Thank you guys )
Just to clarify, I ment "U-Substitution or Integration by Parts"
Hi Makaros;
Generally it is a matter of experience but here is some genral guidelines. When I see an integrand like sin(f(x)) or e^(f(x)) or (f(x))^n, for instance:
I will think of substitution for the 3x+3 or x+5.
When I see an integrand where I have f(x)*g(x) like
then I will try IBP. But even here a possible substitution for that x-4 is possible.
For difficult integrands, most of the time you end up trying both and more until you find some idea that works.
Try here for some practice on substitution:
http://www.mathisfunforum.com/viewtopic.php?id=4208
Last edited by bobbym (2009-08-03 12:50:36)
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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O thank you
Can you you please help me with this Problem
2
∫ 1/(x-1)^1/3 dx (To Clarify: Integral top = 2 to bottom= 0
0
In this equation I need to find whether the functuion is convergent or divergent?
Please help me integrate 1/(x-1)^1/3
I integrated it and got : ln ( 3/4 (x-1)^4/3 ) Is this correct????
Hi Makaros;
Please help me integrate 1/(x-1)^1/3
I integrated it and got : ln ( 3/4 (x-1)^4/3 ) Is this correct????
That answer is not correct, try to do it this way.
This is a substitution problem; Say:
Last edited by bobbym (2009-08-03 13:17:26)
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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What is an Antiderivative for
9
∫ x/x-3 dx How to integrate it?
5
Hi Makaros;
Please remember when posting to use parentheses.
x/x-3 dx this means
For:
Say:
Substituting back for u:
Now using the intervals of integration 5 and 9.
Last edited by bobbym (2009-08-04 05:40:27)
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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What is an Antiderivative for
9
∫ x/x-3 dx How to integrate it?
5
Hi, here an other way:
9
lets do this way ∫ x/x-3 dx=9
5
9
∫ ( x-3+3)/x-3 dx
5
9
=∫ ( x-3+3)/x-3 dx
5
9
=∫ (( x-3)/(x-3)+(3)/x-3 )dx
5
9 9
=∫ (1dx+∫(3/x-3) dx
5 5
9
=1(9-5)+3ln(x-3)]
5
=4+3(ln(9-3)-ln(5-3))
=4+3(ln6-ln2)
=4+3ln(6/2)
=4+3ln3=3ln3+4
W.B.W
Riad Zaidan
Hi Makaros;
You might want to check out this video for some more info.
http://www.youtube.com/watch?v=qclrs-1rpKI
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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I integrated just the equation and got
∫ x/(x-3) dx
U= x-3
x= x+3
Du= 1
∫ (x+3)/u
= ∫ (x+3)/(x-3) Final Answer
Is this correct?
Hi Makaros;
I am afraid that is not correct. Try to go through my post #7 one step at a time. Post when you don't understand what I did. You are closer to understanding this than you think. I will get you through it 1 step at a time.
Last edited by bobbym (2009-08-04 04:13:33)
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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The problem is taht I dont Understand how to Integrate it.
I'll Show step by step what i understand, after that I am done, Correct me what I did wrong
∫ x/(x-3)
U= x-3
x= u+3
Du= dx
∫ (u+3)/U dx
∫ (x-3+3)/U dx
(x-3+3)/(x-3) +C
+ I belive I should Integrate (x-3+3)/(x-3) But I dont Know how to.
DONE!
Is this Correct? If not, show me what I did wrong and continue from that point
Also, how to Intgrate this Example?
∞
∫ x/(√x+1) (Square root of Full Functiuon x+1
0
I got 2/3(x+1)^3/2
Is this Correct?
Hi Makaros;
∞
∫ x/(√x+1) (Square root of Full Functiuon x+1
0I got 2/3(x+1)^3/2
This answer is not correct. Lets do one thing at a time. For the other problem.
∫ x/(x-3)
u= x-3
x= u+3
du= dx
That is correct.
∫ (u+3)/u dx
This is correct. Now replace the dx with du.
Last edited by bobbym (2009-08-04 05:09:50)
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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Hi,
Yes I understand everythign up to that point.
Is there anything else to it after I substitute into U
Hi Makaros;
Is there anything else to it after I substitute into U
Be careful, in math precision is required. u and U aren't the same.
Last edited by bobbym (2009-08-04 05:31:04)
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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Ok, I just understood up to this point, you can continue.
Hi Makaros;
Are we cool up to here?
Last edited by bobbym (2009-08-04 05:43:35)
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.
Offline
As I undetstand, u/u= 1 and Antiderivative of 1 is u+C
??
If so You may continue, I got it.
Hi;
If you understood the second and third lines then remember that u=x-3 so
We are done with the indefinite integral or antiderivative. But remember that you had a definite integral (one with limits of integration the 9 and 5)
Last edited by bobbym (2009-08-04 05:55:08)
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.
Offline
Ok, I understtod that question,
How do you do this question,
0
∫ 2/(1+x^2)
-1
Please explain me this question.
Whoa! we are not done with the first question.
Using the intervals of integration in the problem:
Is this understandable to you?
Last edited by bobbym (2009-08-04 06:16:33)
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.
Offline
Yeah i understood this question,
thanks
can you continue to other one that I've asked.
Hi Makaros;
For this one we will have to use IBP. There is a substitution but it would be difficult to follow.
We need 4 things we have 2:
We get:
Can you do the above integration?
Last edited by bobbym (2009-08-04 07:35:12)
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.
Offline
Yeah. I understnad up to a point you can continue?