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What's the point of integration by inspection? Watch

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    (Original post by htn)
    because integration by substitution (sometimes) is an invalid argument for a lot functions that are to be integrated

    for example if you try to integrate x^2 with the substitution u=x^2, it is not a valid argument
    Of course it's a valid argument.

    doing it by inspection and understanding the reverse-differentiation is a valid argument, faster, and is what separates you from a robot
    Conversely, I don't think I've ever seen someone talk about "reverse-differentiation" and not think invalid stuff like \int e^{x^2}\,dx = \frac{e^{x^2}}{2x} + C is true.

    The big problem with integration by inspection/recognition is that it only really works if you can more or less see the answer before you start.

    In contrast, integration by substitution lets you solve integrals where you don't have a clue what the answer will look like before you start.

    Out of the two, substitution is far more useful. (Lots of 'recognized' results rely on someone having found them first by substitution).
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    (Original post by DFranklin)
    Of course it's a valid argument.

    Conversely, I don't think I've ever seen someone talk about "reverse-differentiation" and not think invalid stuff like \int e^{x^2}\,dx = \frac{e^{x^2}}{2x} + C is true.

    The big problem with integration by inspection/recognition is that it only really works if you can more or less see the answer before you start.

    In contrast, integration by substitution lets you solve integrals where you don't have a clue what the answer will look like before you start.

    Out of the two, substitution is far more useful. (Lots of 'recognized' results rely on someone having found them first by substitution).
    for the example i put in, it is not a valid argument with the substitution that ive put in.

    the smarter you are, the smarter you get, the more you start to inspect by just sight - saying it only works "if you can see the answer before you start" is very constrained, it's not creative or even logical to think like that, it's the pure idea of trying to solve it with your mind, rather than always applying a framework or a formula/substitution to it, that separates good mathematicians from average ones. integration by substitution has down-falls and when it doesn't work it can absolutely boggle students and those who use it - that a formula or a framework cannot solve a problem, that they have to use their mind, that they have to inspect it themselves.

    maybe that's why the OP doesnt like integration by sight.
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    htn you're a tool
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    (Original post by htn)
    for the example i put in, it is not a valid argument with the substitution that ive put in.
    Yes it is.

    \int x^2\,dx. Sub u = x^2, then du/dx = 2x, and so our integral becomes

    \int x^2 \dfrac{1}{2x} \, du = \dfrac{1}{2} \int x\, du = \dfrac{1}{2} \int \sqrt{u} \, du = \dfrac{1}{2} \dfrac{2}{3} u^{3/2} + C = \dfrac{1}{3} u^{3/2} + C = \dfrac{1}{3}x^3 + C

    the smarter you are, the smarter you get, the more you start to inspect by just sight - saying it only works "if you can see the answer before you start" is very constrained, it's not creative or even logical to think like that
    Somewhat more logical than claiming integration by substitution is an invalid method, however.

    it's the pure idea of trying to solve it with your mind, rather than always applying a framework or a formula/substitution to it, that separates good mathematicians from average ones
    If you ever hang out on AoPS, you'll see about half the integral problems get solved by substitution. (And pretty much none of them by recognition).

    integration by substitution has down-falls and when it doesn't work it can absolutely boggle students and those who use it - that a formula or a framework cannot solve a problem, that they have to use their mind, that they have to inspect it themselves.

    maybe that's why the OP doesnt like integration by sight.
    I think you're projecting your own issues with integration by substitution here, to be honest.
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    (Original post by DFranklin)
    Yes it is.

    \int x^2\,dx. Sub u = x^2, then du/dx = 2x, and so our integral becomes

    \int x^2 \dfrac{1}{2x} \, du = \dfrac{1}{2} \int x\, du = \dfrac{1}{2} \int \sqrt{u} \, du = \dfrac{1}{2} \dfrac{2}{3} u^{3/2} + C = \dfrac{1}{3} u^{3/2} + C = \dfrac{1}{3}x^3 + C
    But you could solve that question in less than a second by inspection; isn't that faster, rather than just using a formula?
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    (Original post by htn)
    But you could solve that question in less than a second by inspection; isn't that faster, rather than just using a formula?
    That has nothing to do with whether it's a valid argument.
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    "Integration by inspection" isn't an algorithm for finding integrals, unlike integration by substitution or integration by parts. Integration by inspection is just a name we give the method of recognising patterns which we've derived using other methods (e.g. by integration by substitution or by parts, or by a back-to-basics approach using the definition of an integral). As such, integration by inspection is just a short-cut. The same goes for "differentiation by inspection". When we write \dfrac{d}{dx} x^n = nx^{n-1} we're differentiating by inspection, as opposed to going to the trouble of finding the limit of \dfrac{(x+h)^n-x^n}{h} as h \to 0 each time. Similarly, when you write \displaystyle \int x^n\, dx = \dfrac{x^{n+1}}{n+1} + C, it's a short-cut either for using the fundamental theorem of calculus (if you know the analogous result for differentiation), or for finding the supremum over all partitions of an arbitrary interval of the lower sums and the infimum over all partitions of the same interval of the upper sums and checking if they're equal... which, believe me, isn't something you want to do every time.

    If you look at \displaystyle \int \dfrac{\sin x - \cos x}{\sin x + \cos x}\, dx and say "ah, well the derivative of the denominator is the numerator" and then write that it's equal to \ln |\sin x + \cos x| + C, that's inspection. The result is derived from integration by substitution using the substitution u=\sin x + \cos x. [In fact, the rule \displaystyle \int \dfrac{f'(x)}{f(x)}\, dx = \ln |f(x)| + C is derived using integration by substitution with u=f(x).]

    So that's the answer to the OP's question really: integration by inspection is just a short-cut to other methods. As such, an integration by inspection method can never be "more valid" than a method from which it was derived.
 
 
 
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