Importance of adding the constant of integration.


by shayaan_musta
Tags: adding, constant, importance, integration
shayaan_musta
shayaan_musta is offline
#1
Oct10-11, 07:36 AM
P: 164
Hello experts!
I have a question below.

Why is it important to add the constant of integration immediately when the integration is performed?

Thanks in advance.
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daveb
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#2
Oct10-11, 08:26 AM
P: 927
If you're wondering why there is a constant of integration, it is because when you have a function f(x) such that F(x) = f'(x), then the derivative of g(x) = f(x) + C for any constant also equals F(x) (i.e., g'(x) = f'(x)), so when you integrate F(x), you need to capture that constant in the solution.

Now, if you are asking why it has to be done immediately I'm not sure what you're asking since adding the constant is done as the last step.
dextercioby
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#3
Oct10-11, 09:26 AM
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To the OP: Assume you must do a double integration. For example:

[tex] \frac{d^2 f(x)}{d x^2} = x^3 + 5 [/tex]

What is then f(x) equal to ?

shayaan_musta
shayaan_musta is offline
#4
Oct11-11, 09:05 AM
P: 164

Importance of adding the constant of integration.


Quote Quote by dextercioby View Post
To the OP: Assume you must do a double integration. For example:

[tex] \frac{d^2 f(x)}{d x^2} = x^3 + 5 [/tex]

What is then f(x) equal to ?
f(x)=[itex]\frac{x^{5}}{5}[/itex] + [itex]\frac{5x^{2}}{2}[/itex] + c

where "c" is the integration constant.

Now what?? :s
shayaan_musta
shayaan_musta is offline
#5
Oct11-11, 09:08 AM
P: 164
Quote Quote by daveb View Post
If you're wondering why there is a constant of integration, it is because when you have a function f(x) such that F(x) = f'(x), then the derivative of g(x) = f(x) + C for any constant also equals F(x) (i.e., g'(x) = f'(x)), so when you integrate F(x), you need to capture that constant in the solution.

Now, if you are asking why it has to be done immediately I'm not sure what you're asking since adding the constant is done as the last step.
Ok daveb thank you. I think you have answered my question.
Thank you very much.
shayaan_musta
shayaan_musta is offline
#6
Oct11-11, 09:09 AM
P: 164
Thank you dextercioby. You also made me to think the answer of my question.
cpt_carrot
cpt_carrot is offline
#7
Oct11-11, 10:59 AM
P: 24
Quote Quote by shayaan_musta View Post
f(x)=[itex]\frac{x^{5}}{5}[/itex] + [itex]\frac{5x^{2}}{2}[/itex] + c

where "c" is the integration constant.

Now what?? :s
Not quite, this is why the constant of integration is important. The first integration gives

[tex] \frac{df}{dx}=\frac{x^4}{4}+5x+c_1 [/tex]

and the second integration gives

[tex] f(x)=\frac{x^5}{20}+\frac{5x^2}{2}+c_1x+c_2 [/tex]

Which is why we need one constant of integration for each integral
HallsofIvy
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#8
Oct11-11, 02:22 PM
Math
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Thanks
PF Gold
P: 38,877
Note that dextercioby's example is a linear second order non-homogenous differential equation which means that the set of all solutions is a two dimensional "linear manifold". That is why you need two undetermined coefficients.
shayaan_musta
shayaan_musta is offline
#9
Oct12-11, 01:55 AM
P: 164
Quote Quote by cpt_carrot View Post
Not quite, this is why the constant of integration is important. The first integration gives

[tex] \frac{df}{dx}=\frac{x^4}{4}+5x+c_1 [/tex]

and the second integration gives

[tex] f(x)=\frac{x^5}{20}+\frac{5x^2}{2}+c_1x+c_2 [/tex]

Which is why we need one constant of integration for each integral
Ok cpt_carrot I got you. Thanks man.
Now I have understood that for each integration a constant is important, as Hallsofivy said.

So help guys on Physics Forums


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