
Topic review (newest first)
irspow wrote:You are correct ryos, it wouldn't really matter if this problem was to be solved by a computer. The two answers are indeed equal. I was thinking however about the scenario in which the result was going to be an integral (pun not intended) part of one's work.
And that, in a nutshell, is the difference between scientists and engineers.
 mikau
 20060219 05:40:26
at any rate, a simpler formula will be quicker to evaluate by the computer. (so we could all save a few billionths of a second)
 irspow
 20060219 00:23:22
You are correct ryos, it wouldn't really matter if this problem was to be solved by a computer. The two answers are indeed equal. I was thinking however about the scenario in which the result was going to be an integral (pun not intended) part of one's work. Remembering and applying their formula would be much harder than would be necessary. I, for one, had until now believed that such programs simplified their answers for our convenience.
Sorry for making such a big thing of this, but I found it really interesting. My level of "trust" in such software has diminished as a result anyway.
As my Calc teacher once said, "Machines sometimes give us some interesting answers..."
Remember that computers have no intuition. I'm sure they've got their algorithms such that if a function can be integrated, they will integrate it somehow. It's doubtless not the best at selecting the proper method for every integral, and for some it probably doesn't have a "best method." But if you have the attention span of a computer, it works just fine.
Or maybe somebody taught the computer about artificial job security...
Anyway, I guess this is a case where a calculator isn't the best tool after all. Though, if you're using computers end to end, then the computer doesn't care how ugly the function you ask it to evaluate is...
 irspow
 20060218 11:22:24
Is the Mathmatica Integrator broken or that limited?
 irspow
 20060216 13:16:42
If you guys want to laugh, Mathmatica's online integrator gives;
(1/27)√(9 + 4/(x^(2/3)))(9x^(2/3) + 4)(x^(1/3)) for the integral I solved above.
I put it into my TI89 and it spit out an even more ridiculous answer.
I checked and my little (x^(2/3) + 4/9)^(3/2) produces the same values as the mess above.
What in the world are these people at Mathmatica and Texas Instruments thinking about?
Sorry ryos, but this seems to be a perfect example of why we might not want to use calculators if we don't have to.
It took me ten minutes just to simplify that to;
√((729x^2 + 972x^(4/3) + 432x^(2/3) + 64) / 729)
I don't know what algorithyms they're using, but I think that they need to hire some more math majors.
 mikau
 20060216 11:36:42
not all the time. But on occasions were it happens, best to be prepared.
 irspow
 20060216 08:11:31
Yeah, mikau, I have never had any trouble with doing that. I don't recall ever being taught to do it specifically, but like I said, I haven't seen it ever getting me into trouble before.
In the example you gave;
u = x²
du = 2x dx
dx = du/2x
But x = √ u
So, yes,
dx = du/2√u
I do that kind of manipulation all of the time. Actually, I don't remember how else you would do it.
Oh, yeah!
Then 2x would have to actually appear in the integral for you to use the "standard" method. Or even x I guess would still leave you with du/2.
I can see how if you were "lucky enough to have the derivative of u in the integral it would make sense to use it, but how often does that happen?
 mikau
 20060215 13:20:49
Very true. Nice observation.
as far as the u substitution, yeah I know you can replace u with the value it stood for after integrating, rather then changing the variable to evaluate it at (in terms of u).
Its just I don't recall ever having to do something you did and didn't know it was legal.
Typically if I made the substitution u = x^2 then du = 2x dx I would have figured u substitution wouldn't work, but you can replace x with sqrt u to get dx = 1/( 2sqrt u ) du ???
 irspow
 20060215 07:50:32
Yes, mikau, you can always replace the u with what you substituted it for after you are done integrating. I think it's just a matter of preference. You either do what I did or you change the limits of integration. I just think it is usually easier to do it the way I did because I already know what u equals. If you change the limits of integration then you have to solve for the values of u.
Oh, you won't get stuck much if you keep practicing. After a while, you will have a certain set of integrals that you remember best. What you find then is that you tend to manipulate equations to match the integrals that you do remember.
 mikau
 20060215 05:16:04
wow! Many thanks irspow! I was stuck on that one for weeks!
It's just another substitution...
 mikau
 20060214 15:24:58
hmm... are you sure its legal to replace x with expressions containing u after differentiating? I don't see why it wouldn't be but there are a lot of things that I think should work but don't. :/
 irspow
 20060214 15:13:05
Let u = x^(1/3)
du = 1/[3(x^(2/3))] dx
3x^(2/3)du = dx
3u^2 du = dx
∫√(1 + 4/[9(x^(2/3))] dx
3∫u^2√(1 + 4/(9u^2)) du
3∫(u^2)/3√((9u^2 + 4)/u^2) du
∫u^2√((9u^2 + 4)/u^2) du
∫u√(9u^2 + 4) du
∫u√(9(u^2 + 4/9)) du
3∫u√(u^2 + 4/9) du
3[1/3(u^2 + 4/9)^(3/2)]
[u^2 + 4/9]^(3/2)
u^2 = x^(2/3)
[x^(2/3) + 4/9]^(3/2) + C
 mikau
 20060214 15:04:16
lol. I do agree. I mean I use a calculator to add subtract multiply and divide when I can't do it in my head, rather then using long division or multiplication.
In the end I think it all comes down to this. A calculator is to avoid tedious time consuming tasks so your attention can be focused elsewhere when needed. A machine to do algorithms quickly and accurately. But never a replacement for understanding of a concept.
