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.

]]>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.

]]>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...

]]>(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 TI-89 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.

]]>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?

]]>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 ???

]]>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.

]]>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

]]>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.

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