Non-Fractional - If Du is Smaller than Dx

Let me attempt to give you some constructive criticism and a little advice. There is no "here are the steps" answer, and in many cases there is more than one way to do a problem. You will need to be prepared for many kinds of algebraic manipulations to put the integral into a particular form which is recognizable, and it gets far more complicated than in your example integral. I think you will benefit from reading your textbook where it should discuss particular strategies prior to the exercises. Again, there is no set algorithm for evaluating integrals in general, and most functions (though probably none of those in your book) do not even have elementary anti-derivatives (so you cannot "integrate" them).

It also does not make sense to say "du is smaller than dx". If you choose to try the u-substitution method to evaluate an integral, you will, 99% of the time, get a du different from dx. And not just a constant multiple of it as in this case. Your inability to follow the book's solution in your other post leads me to believe you lack some basic understanding of what is going on (I hope I'm not coming across as mean here). That is why I suggest you read your textbook and maybe consult with a face-to-face tutor.
 
daon2 said:
Let me attempt to give you some constructive criticism and a little advice. There is no "here are the steps" answer, and in many cases there is more than one way to do a problem. You will need to be prepared for many kinds of algebraic manipulations to put the integral into a particular form which is recognizable, and it gets far more complicated than in your example integral. I think you will benefit from reading your textbook where it should discuss particular strategies prior to the exercises. Again, there is no set algorithm for evaluating integrals in general, and most functions (though probably none of those in your book) do not even have elementary anti-derivatives (so you cannot "integrate" them).

It also does not make sense to say "du is smaller than dx". If you choose to try the u-substitution method to evaluate an integral, you will, 99% of the time, get a du different from dx. And not just a constant multiple of it as in this case. Your inability to follow the book's solution in your other post leads me to believe you lack some basic understanding of what is going on (I hope I'm not coming across as mean here). That is why I suggest you read your textbook and maybe consult with a face-to-face tutor.
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Actually the reason I didn't catch on to Jeff B at first was I couldn't understand that DX was being integrated and the rest was being multiplied to DX. For instance with intergal of 5 is actually 5 TIMES the intergal of DX. But you know, if you study it's easy to skip over fundementals. You see patterns, you follow them, but you don't get the whole picture, which doesn't hurt you at first, but it does later on.

Of course the thread with Jeff B involved(at some point) the distributive property of Dx with 1 and a fraction (because the long division yielded that), a situation that doesn't come up often. After that you had to integrate two terms which both had DX, so that's why the answer was unusual (different from usual integration answers). Most integration answers involve some reciprocal fraction times the integration formula.
 
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Fair enough. But as I said there is no straight-forward approach. For example would you ever guess that \(\displaystyle \dfrac{x}{e^x}\) is easily integrated, but that \(\displaystyle \dfrac{1}{xe^x}\) cannot be? (Though the latter was such an important kind of integral they gave a name to it, the exponential integral function.)

Knowledge of these little things comes from experience. Just don't expect there to be step by step approaches for all problems.

Jason76 said:
For instance with intergal of 5 is actually 5 TIMES the intergal of DX.
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Yes, but it is important to understand that it is not a definition that those two are the same. It is more or less a coincidence, a result of a derivative being a special kind of operator. Indefinite integrals output families of antiderivatives - that is how they are introduced in pretty much any calculus book I have looked at. So \(\displaystyle \int 5 dx\) means "any function whose derivative is 5." You know one of them: 5x (feel free to add any constant you like). Look at the other one you mention: \(\displaystyle 5\int 1 dx\). This means "5 times any function whose derivative is 1." The derivative of the function y=x is 1, and 5 times it is 5x. It turns out these two are the same, as they should be.

While \(\displaystyle \int dx\) is fine notation, I would avoid it for now, as it takes away from the above reasoning. Instead, I would write it as \(\displaystyle \int 1 dx\).
 
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