Mendicant_Bias
  • Mendicant_Bias
Calculus II question regarding arc length. I'm using Paul's Online Notes and am having trouble understanding why the limits of integration in the first example don't change after u-substitution is utilized, and I have a general question about u-substitution as well. http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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blockcolder
  • blockcolder
In the first place, there was no u-substitution in the first example.
Mendicant_Bias
  • Mendicant_Bias
There was. At least in the way it was implied to have been done. Did you actually try to solve the integral for sec(x)? It requires u-substitution.
Mendicant_Bias
  • Mendicant_Bias
(If there's another way to do it, go ahead, please.)

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blockcolder
  • blockcolder
Well, he already knew that \[\int \sec x\ dx=\ln|\sec x+\tan x| +C\]so there was no need to solve for the integral. He just used Fundamental Theorem of Calculus part 2 after that.
Mendicant_Bias
  • Mendicant_Bias
How did he "already know it"? Just by stating it? While he can state it and be correct and all, not showing work is a little axiomatic and a little ridiculous if this was an actual problem, and if he were to actually write out how to integrate sec(x), it would require u-substitution, and I can't think of a different way to do it, and wolfram|alpha does the same thing. If this was an actual problem that was needed to be solved, which I'm treating it that way, work would have to be shown for integration, and the actual work is where that u-substitution comes up:
Mendicant_Bias
  • Mendicant_Bias
\[\int\limits_{0}^{\frac{ \pi }{ 4 }}\sec(x) = \frac{ \sec(x)\tan(x) + \sec ^{2}(x) }{ \sec(x) + \tan(x) }\]\[u = \sec(x) + \tan(x), du = \sec(x)\tan(x) + \sec ^{2}(x)\]\[\int\limits_{0}^{\frac{ \pi }{ 4 }}\sec(x) = \int\limits_{a}^{b}\frac{ 1 }{ u }du\]Where "a" and "b" are the lower and upper bounds of integration, respectively, which, when you're using u-substitution, you have to take the original bounds and put them through whatever u as a function is. @blockcolder
blockcolder
  • blockcolder
If I recall, he had already proved this in a prior page. http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx
blockcolder
  • blockcolder
It's Example 9 there in the page I just posted, btw.
Mendicant_Bias
  • Mendicant_Bias
^Everything on that page was indefinite integrals, so the limits will never change, how does that involve the limits potentially being changed by u-substitution? PS, I don't regularly use Paul's notes, so I have no idea of the continuity of it. I'll take a look at example nine, give me a minute. Thank you for helping, by the way.
blockcolder
  • blockcolder
By Fundamental Theorem of Calculus part 2, if f is continuous on [a,b], then \[\int_a^b f(x)\ dx=F(b)-F(a),\]where \(F'(x)=f(x),\)or alternatively, \(\int f(x)\ dx=F(x)+C\). Once you know the general antiderivative of f, you can just plug in the limits into F so that you don't have to go through the trouble of solving the antiderivative again.
Mendicant_Bias
  • Mendicant_Bias
^I'm thinking that this whole mess was caused by an issue with W|A, because it explicitly said in step-by-step directions to use u-substitution at that part rather than taking the method of multiplying the integrand by sec(x)+tan(x)/sec(x)+tan(x) and came to the exact same result, and that doesn't seem right. Let me pull up the W|A calculation I'm talking about, because I think there's something inherently wrong with how it's being done, but the odds of that seem really low, so I'm a little confused by the outcome.
blockcolder
  • blockcolder
Besides, if you didn't know that \(\int \sec x\ dx=\ln|\sec x+\tan x|+C\), you can solve it the usual way: \[\int_0^\frac{\pi}{4} \sec x\ dx=\int_0^\frac{\pi}{4} \frac{\sec x(\sec x+\tan x)}{(\sec x+\tan x)}\ dx=\int_1^{1+\sqrt{2}} \frac{1}{u}\ du\]
Mendicant_Bias
  • Mendicant_Bias
^Yeah, that's what I'm confused by, why the limits didn't change like that. I should've actually written it out, but I don't understand why that didn't happen in the problem, why the limits of integration didn't change and why that didn't transfer over into evaluating the antiderivative with those limits.
Mendicant_Bias
  • Mendicant_Bias
Okay, I think I see what happened in that he just approached it in a different way that seemed really, really odd to me.
Mendicant_Bias
  • Mendicant_Bias
(Both u-substitution and just using the fundamental theorem of calculus, part II, given that you know the antiderivative of sec(x) are equally valid methods and I put myself in a sort of-what would you call a false dilemma(?)-position where I only thought u-substitution was a workable method.)
Mendicant_Bias
  • Mendicant_Bias
*a false dilemma with only one available option rather than two. Whatever that would be called.*
blockcolder
  • blockcolder
Yeah, that's kinda what happened.
Mendicant_Bias
  • Mendicant_Bias
Thanks!
blockcolder
  • blockcolder
No problem. :D

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