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Integration Question

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Jefferson :
Hi guys, please help me with the attachment below.
thanks!

jamonwindeyer:

--- Quote from: Jefferson  on January 29, 2019, 06:28:45 pm ---Hi guys, please help me with the attachment below.
thanks!

--- End quote ---

Hey Jefferson! Interesting question this one. Take your substitution:



Essentially, you now have two ways of writing \(u\):



Basically, as long as these line up, you are safe. For example, if you let \(\sqrt{4-x}=u\), then you must let \(u=2\), because \(\sqrt{4-x}\neq-2\), and similarly, \(-\sqrt{4-x}\neq2\), the signs need to line up!! ;D

Jefferson :

--- Quote from: jamonwindeyer on January 29, 2019, 07:13:03 pm ---Hey Jefferson! Interesting question this one. Take your substitution:



Essentially, you now have two ways of writing \(u\):



Basically, as long as these line up, you are safe. For example, if you let \(\sqrt{4-x}=u\), then you must let \(u=2\), because \(\sqrt{4-x}\neq-2\), and similarly, \(-\sqrt{4-x}\neq2\), the signs need to line up!! ;D

--- End quote ---

That makes a lot of sense.
Thank you so much!

RuiAce:
Already addressed of course but gonna buzz in here.

Usually, when we substitute \( \boxed{x\text{ is some function of }u} \) instead of other way around, the problems more or less lie in that \(u\) is no longer necessarily a function of \(x\). (It's sneakily kinda equivalent to how some functions don't have inverse functions.)

At times like these, usually when we write the substitution (here \(x=4-u^2\) which would give \(dx = -2u\,du\)), we impose what's essentially a "domain" restriction on \(u\). For these quadratic substitutions, we treat \(4-u^2\) as a quadratic in terms of \(u\), which has axis of symmetry (or equivalently its stationary point) at \(u=0\). So to formally work around this issue, we would prefer to say "let \(x = 4-u^2\) (\(u\geq 0\))".

That way, we can now rearrange the equation without ambiguity.

Why the positive case? Just a convention - it's usually the easier one because it avoids a lot of dumb negative signs, since now \(u = +\sqrt{4-x}\). You could've picked \(u \leq 0\) in which you'd get \(u=-\sqrt{4-x}\) and it will still work, but it might be a tiny bit uglier.

Jefferson :

--- Quote from: RuiAce on January 30, 2019, 12:15:37 am ---Already addressed of course but gonna buzz in here.

Usually, when we substitute \( \boxed{x\text{ is some function of }u} \) instead of other way around, the problems more or less lie in that \(u\) is no longer necessarily a function of \(x\). (It's sneakily kinda equivalent to how some functions don't have inverse functions.)

At times like these, usually when we write the substitution (here \(x=4-u^2\) which would give \(dx = -2u\,du\)), we impose what's essentially a "domain" restriction on \(u\). For these quadratic substitutions, we treat \(4-u^2\) as a quadratic in terms of \(u\), which has axis of symmetry (or equivalently its stationary point) at \(u=0\). So to formally work around this issue, we would prefer to say "let \(x = 4-u^2\) (\(u\geq 0\))".

That way, we can now rearrange the equation without ambiguity.

Why the positive case? Just a convention - it's usually the easier one because it avoids a lot of dumb negative signs, since now \(u = +\sqrt{4-x}\). You could've picked \(u \leq 0\) in which you'd get \(u=-\sqrt{4-x}\) and it will still work, but it might be a tiny bit uglier.

--- End quote ---

Thank you for taking the extra time to clarify the issue. It really helped me with my understanding of integration.
:)

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