Improper integral question
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Improper integral question
The problem:
[math]\int_{2}^{14} \! \frac{14}{\sqrt[4]{x+2}}\, \mathrm{d}x[/math]
What I get, unless I am wrong, is
[math]\lim_{t\to2}14\ln{\sqrt[4]{x+2}}\Biggr]_t^{14}[/math]
...Which, as far as I know, should be divergent, since ln 0 is undefined, approaching negative infinity. However, (since this is online and I can attempt this problem multiple times) I know that it converges, but I don't understand how (unless I am right and webassign is wrong).
I actually do know the answer, which is 448/3 (which I arrived at with a calculator and the ability to try a problem almost exactly the same, whose answer I could see after not getting it rightnot the most enlightening process), but I don't understand how it is 448/3. So, if anybody understand why it is that...please help me see why.
[math]\int_{2}^{14} \! \frac{14}{\sqrt[4]{x+2}}\, \mathrm{d}x[/math]
What I get, unless I am wrong, is
[math]\lim_{t\to2}14\ln{\sqrt[4]{x+2}}\Biggr]_t^{14}[/math]
...Which, as far as I know, should be divergent, since ln 0 is undefined, approaching negative infinity. However, (since this is online and I can attempt this problem multiple times) I know that it converges, but I don't understand how (unless I am right and webassign is wrong).
I actually do know the answer, which is 448/3 (which I arrived at with a calculator and the ability to try a problem almost exactly the same, whose answer I could see after not getting it rightnot the most enlightening process), but I don't understand how it is 448/3. So, if anybody understand why it is that...please help me see why.
Re: Improper integral question
Your integral is wrong. Note that you can write your expression as 14(x+2)^(1/4)
double epsilon = .0000001;

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Re: Improper integral question
Oh...right.
/Double Facepalm
Thank you.
/Double Facepalm
Thank you.
Re: Improper integral question
I teach calculus. Both you and my students know the formula
[math]\int \frac1x dx = \ln x[/math]
which, casually paraphrased, could be said as "integral of 1 over something is natural log of something."
I often tell my students to be careful with that casual paraphrase. Notice that, for example,
[math]\int \frac1{x^2+1} dx = \arctan x[/math]
which has absolutely nothing to do with logarithms.
It is true that
[math]\int \frac1{x^2+1} 2x dx = \ln(x^2+1)[/math]
but that's a totally different integral.
Perhaps the above casual paraphrase would be better expressed as "integral of 1 over something times d something is natural log of something".
[math]\int \frac1x dx = \ln x[/math]
which, casually paraphrased, could be said as "integral of 1 over something is natural log of something."
I often tell my students to be careful with that casual paraphrase. Notice that, for example,
[math]\int \frac1{x^2+1} dx = \arctan x[/math]
which has absolutely nothing to do with logarithms.
It is true that
[math]\int \frac1{x^2+1} 2x dx = \ln(x^2+1)[/math]
but that's a totally different integral.
Perhaps the above casual paraphrase would be better expressed as "integral of 1 over something times d something is natural log of something".
 Talith
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Re: Improper integral question
I find it much easier to remember that the integral of f'(x)/f(x) is log(f(x)).
Re: Improper integral question
skullturf wrote:I often tell my students to be careful with that casual paraphrase. Notice that, for example,
[math]\int \frac1{x^2+1} dx = \arctan x[/math]
which has absolutely nothing to do with logarithms.
Well, not nothing. 1/(x^{2}+1) = i/2 (1/(i+x) + 1/(ix)). So its integral is i/2 (log(i+x)  log(ix)) = i/2 log((i+x)/(ix)). You can see that this is the same thing as arctan x, since when x is real it is the imaginary part of log(1+ix).
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Re: Improper integral question
skullturf wrote:I teach calculus. Both you and my students know the formula
[math]\int \frac1x dx = \ln x[/math]
which, casually paraphrased, could be said as "integral of 1 over something is natural log of something."
This is why I beat people over the head for saying those kinds of things, even if it's a casual paraphrase.
I find it better to say "integral of 1 over x plus something is is natural log of x plus something".
It's not much harder to say, but you won't get tripped up on something like 1 / (x^2 + 1).
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Re: Improper integral question
Substitution, people!
The problem is:
[math]\int_{2}^{14} \! \frac{14}{\sqrt[4]{x+2}}\, \mathrm{d}x[/math]
Now, let: [math]u=x+2[/math]
[math]\frac{du}{dx}\ = 1[/math]
[math]du=dx[/math]
When: [math]x=14, u = 16[/math]
[math]x=2, u = 0[/math]
So the new integral problem is:
[math]=\int_{0}^{16} \! \frac{14}{\sqrt[4]{u}}\, \mathrm{d}u[/math]
[math]=\int_{0}^{16} \! 14u^\frac{1}{4}\, \mathrm{d}u[/math]
[math]=\frac{4}{3}*14u^\frac{3}{4}\Biggr]_0^{16}[/math]
[math]=(\frac{56}{3}16^\frac{3}{4})(0)[/math]
[math]=\frac{448}{3}\[/math]
Can I ask where you got this problem from? Homework or otherwise  where did you find it? What are you doing it for?
The problem is:
[math]\int_{2}^{14} \! \frac{14}{\sqrt[4]{x+2}}\, \mathrm{d}x[/math]
Now, let: [math]u=x+2[/math]
[math]\frac{du}{dx}\ = 1[/math]
[math]du=dx[/math]
When: [math]x=14, u = 16[/math]
[math]x=2, u = 0[/math]
So the new integral problem is:
[math]=\int_{0}^{16} \! \frac{14}{\sqrt[4]{u}}\, \mathrm{d}u[/math]
[math]=\int_{0}^{16} \! 14u^\frac{1}{4}\, \mathrm{d}u[/math]
[math]=\frac{4}{3}*14u^\frac{3}{4}\Biggr]_0^{16}[/math]
[math]=(\frac{56}{3}16^\frac{3}{4})(0)[/math]
[math]=\frac{448}{3}\[/math]
Can I ask where you got this problem from? Homework or otherwise  where did you find it? What are you doing it for?
Last edited by Annihilist on Mon Apr 02, 2012 1:02 am UTC, edited 2 times in total.

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Re: Improper integral question
skullturf wrote:I teach calculus. Both you and my students know the formula
[math]\int \frac1x dx = \ln x[/math]
which, casually paraphrased, could be said as "integral of 1 over something is natural log of something."
I often tell my students to be careful with that casual paraphrase. Notice that, for example,
[math]\int \frac1{x^2+1} dx = \arctan x[/math]
which has absolutely nothing to do with logarithms.
It is true that
[math]\int \frac1{x^2+1} 2x dx = \ln(x^2+1)[/math]
but that's a totally different integral.
Perhaps the above casual paraphrase would be better expressed as "integral of 1 over something times d something is natural log of something".
[math]\int \frac{f'(x)}{f(x)} dx = \ln f(x)+c[/math]
(Edited the +c in)
Last edited by Annihilist on Mon Apr 02, 2012 7:17 am UTC, edited 1 time in total.
 jestingrabbit
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Re: Improper integral question
Annihilist wrote:Substitution, people!
Meh. Anytime that you get du/dx = 1, you should just write out the antiderivative imo.
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Re: Improper integral question
Yeah thats what I did. Just slower.jestingrabbit wrote:Annihilist wrote:Substitution, people!
Meh. Anytime that you get du/dx = 1, you should just write out the antiderivative imo.
 Proginoskes
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Re: Improper integral question
Annihilist wrote:[math]\int \frac{f'(x)}{f(x)} dx = \ln f(x)[/math]
Sorry, you lose a point. The correct answer is
[math]\int \frac{f'(x)}{f(x)} dx = \ln f(x) + C[/math]

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Re: Improper integral question
Ah shit. Thanks.Proginoskes wrote:Annihilist wrote:[math]\int \frac{f'(x)}{f(x)} dx = \ln f(x)[/math]
Sorry, you lose a point. The correct answer is
[math]\int \frac{f'(x)}{f(x)} dx = \ln f(x) + C[/math]
No one else was using 'c's so I forgot to add it.
EDIT: "c = 0" is a valid answer though, unless you have been given a specific point which the function passes through.
Re: Improper integral question
Proginoskes wrote:Annihilist wrote:[math]\int \frac{f'(x)}{f(x)} dx = \ln f(x)[/math]
Sorry, you lose a point. The correct answer is
[math]\int \frac{f'(x)}{f(x)} dx = \ln f(x) + C[/math]
Well if you want to play that game:
[imath]\displaystyle \int \frac{f'(x)}{f(x)} dx = \lnf(x) + C[/imath]
where f(x) ≠ 0
Re: Improper integral question
Annihilist wrote:Ah shit. Thanks.Proginoskes wrote:Annihilist wrote:[math]\int \frac{f'(x)}{f(x)} dx = \ln f(x)[/math]
Sorry, you lose a point. The correct answer is
[math]\int \frac{f'(x)}{f(x)} dx = \ln f(x) + C[/math]
No one else was using 'c's so I forgot to add it.
EDIT: "c = 0" is a valid answer though, unless you have been given a specific point which the function passes through.
Sure, we'll give you partial credit for getting one point.
+1/infinity points.
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