Area of the region? Help please!
- Thread starter bobopedic
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Your so-called "improper" integrals are fun! The usual plan is to look at a finite version and see what happens as it wanders out somewhere, unbounded.
With your constraints, first do this:
For some b, where 0<b<3
\(\displaystyle \int_{b}^{3}\frac{1}{(x-1)^{\frac{2}{3}}}\;dx\;=\;3*2^{\frac{1}{3}}\;-\;3\cdot(b-1)^{\frac{1}{3}}\)
Now, move that b out toward zero and see if you get a finite value.
b = 1 gives about 3.78
b = 1/2 gives about ... hmmm... That's weird.
I guess we should point out that your calculator or computer-based math system may give different answers for what is shown above and this:
\(\displaystyle 3*\sqrt[3]{2}\;-\;3\cdot\sqrt[3]{(b-1)}\)
Okay, now b = 1/2 gives about 6.161
b = 1/4 gives about 6.505 - That didn't move much. Maybe it will converge.
b = 1/8 gives about 6.649
That's a cute little arithmetical discovery. Can we prove that the limit exists and is finite? This is your task.
With your constraints, first do this:
For some b, where 0<b<3
\(\displaystyle \int_{b}^{3}\frac{1}{(x-1)^{\frac{2}{3}}}\;dx\;=\;3*2^{\frac{1}{3}}\;-\;3\cdot(b-1)^{\frac{1}{3}}\)
Now, move that b out toward zero and see if you get a finite value.
b = 1 gives about 3.78
b = 1/2 gives about ... hmmm... That's weird.
I guess we should point out that your calculator or computer-based math system may give different answers for what is shown above and this:
\(\displaystyle 3*\sqrt[3]{2}\;-\;3\cdot\sqrt[3]{(b-1)}\)
Okay, now b = 1/2 gives about 6.161
b = 1/4 gives about 6.505 - That didn't move much. Maybe it will converge.
b = 1/8 gives about 6.649
That's a cute little arithmetical discovery. Can we prove that the limit exists and is finite? This is your task.
I'm not sure how to reply to your message. Was my answer close to yours? What I did was take two integrals, one from 0 to 1 and the other from 1 to 3 of dx/(x-1)^2/3, for the former I plugged in k which approached one from the left and for the next one I plugged in k for 1 from the right...got -3 for the first and 3*cuberootof 2 for the second. Is that right?
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That's funny. Since I seem to be working the entirely wrong problem, I'll give up the irritating inquisition. I'm not sure why I thought the asymptotic behavior was at x=0. I think I need to go stand in the corner for a few moments.
Very good. You seem to have the idea, excepting that "-3" out front. There is no ordinate less than zero on [0,1)
Very good. You seem to have the idea, excepting that "-3" out front. There is no ordinate less than zero on [0,1)
I have a question. How come on an example in the book (not this problem), one step is:
lim [3(k-2)^1/3 - 3(1-2)^1/3] = 3
k--->2 (from the left)
I thought for directional limits that you don't touch the number...2 in this example, I know the second part gives 3, so the first part must give zero but how? Maybe as k approaches 2 from the left...3(k-2)^1/3 goes to zero? Is that right?
I'll be posting my work from the original problem...if you can check it for me that'd be great!
lim [3(k-2)^1/3 - 3(1-2)^1/3] = 3
k--->2 (from the left)
I thought for directional limits that you don't touch the number...2 in this example, I know the second part gives 3, so the first part must give zero but how? Maybe as k approaches 2 from the left...3(k-2)^1/3 goes to zero? Is that right?
I'll be posting my work from the original problem...if you can check it for me that'd be great!
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You're assuming I've gotten out fo the corner?
"don't touch" - This is not well-defined.
You can get as close as you like.
After some value, call it M, however far away you are at M, you will never be that far away for anything greater than M.
The distance from where you are and where you want to be approaches zero (0).
Also not well-defined, but closer than "don't touch".
In the world of limits, if a function is continuous, and the limit is the same from both sides, then a simple substitution at the point gives the limit.
"don't touch" - This is not well-defined.
You can get as close as you like.
After some value, call it M, however far away you are at M, you will never be that far away for anything greater than M.
The distance from where you are and where you want to be approaches zero (0).
Also not well-defined, but closer than "don't touch".
In the world of limits, if a function is continuous, and the limit is the same from both sides, then a simple substitution at the point gives the limit.
bobopedic said:
Here's my work:
x=0, y=0, x=3, and y=1/(x-1)^(2/3)
I took that to mean:
integral from 3 to 0 of[ dx/(x-1)^(2/3)
I think the problem point is at 1, so I seperated it into two parts: one from 0 to 1 and the other from 1 to 3
Now integral from 0 to 1 of[ dx/(x-1)^(2/3)] = limit k--> 1 from the left of integral from 0 to k of[ dx/(x-1)^(2/3)] and the integral of that is 3(x-1)^1/3
now that becomes limit k-->1 from the left of [3(k-1)^1/3 - 3(0-1)^1/3 = -3
In the second part I skipped the integral part and just came to limit k-->1 from the right of k to 3 of 3(x-1)^1/3 which resulted in:
limit k--> 1 from the right of [3(3-1)^1/3 - 3(k-1)^1/3] = 3*cuberoot of 2
and then I added them to get (3*cube root of 2) + (-3)
Is that correct?
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"integral from 3 to 0"
Which partial symbol do you mean?
\(\displaystyle \int_{0}^{3} or \int_{3}^{0}\)
If you mean the former, then this is where you have confused it.
You use all of these terms:
integral from 3 to 0
one from 0 to 1
the other from 1 to 3
integral from 0 to 1
integral from 0 to k
I'm a little confused by the first one.
You need
\(\displaystyle \int_{0}^{k_{1}} and \int_{k_{2}}^{3}\)
Make sure you keep them in the right order. Maybe don't "skip the integration part" so you can see the difference.
Notice, if an integral exists
\(\displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx\)
This would explain a sign error.
Which partial symbol do you mean?
\(\displaystyle \int_{0}^{3} or \int_{3}^{0}\)
If you mean the former, then this is where you have confused it.
You use all of these terms:
integral from 3 to 0
one from 0 to 1
the other from 1 to 3
integral from 0 to 1
integral from 0 to k
I'm a little confused by the first one.
You need
\(\displaystyle \int_{0}^{k_{1}} and \int_{k_{2}}^{3}\)
Make sure you keep them in the right order. Maybe don't "skip the integration part" so you can see the difference.
Notice, if an integral exists
\(\displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx\)
This would explain a sign error.