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    Hi, just something how do you know what u can be? I know most of the time, u is what is inside the bracket or square root. if you get a question in the pic, can you just choose what u is ? Also in this question why isn't u =e^x^2 but instead just x^2?


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    (Original post by coconut64)
    Hi, just something how do you know what u can be? I know most of the time, u is what is inside the bracket or square root. if you get a question in the pic, can you just choose what u is ? Also in this question why isn't u =e^x^2 but instead just x^2?


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    Substitution is for a function of a function. So the function of a function here is e^{x^2}

    Generally, for fg(x), you set u=g(x)
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    (Original post by RDKGames)
    Substitution is for a function of a function. So the function of a function here is e^{x^2}
    But u is set as x^2 only tho
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    (Original post by RDKGames)
    Substitution is for a function of a function. So the function of a function here is e^{x^2}

    Generally, for fg(x), you set u=g(x)
    But with fg(x) f can be either one function so in this example, I could say that g(x) can be 6x or e^x^2 right ?
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    that function is probably easier to integrate by parts if you've done that??
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    (Original post by coconut64)
    But with fg(x) f can be either one function so in this example, I could say that g(x) can be 6x or e^x^2 right ?
    What your example involves is essentially 3 functions in the form h(x) \cdot fg(x) where h(x)=6x, f(x)=e^x and g(x)=x^2 so you cannot apply that logic here, where only two of which are used to make a composite function - so your 6x is not involved within the substitution whatsoever since it is not the one used for composition.
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    (Original post by azizadil1998)
    that function is probably easier to integrate by parts if you've done that??
    Haven't learnt that yet but thanks anyway
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    (Original post by RDKGames)
    What your example involves is essentially 3 functions in the form h(x) \cdot fg(x) where h(x)=6x, f(x)=e^x and g(x)=x^2 so you cannot apply that logic here, where only two of which are used to make a composite function - so your 6x is not involved within the substitution whatsoever since it is not the one used for composition.
    Thanks. How about when you just have two functions for example x(x-1)^0.5 how do you decide which part is g(x) and which is f(x). I know most of the time u just set whatever is inside the bracket to u but why is that?
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    (Original post by coconut64)
    Thanks. How about when you just have two functions for example x(x-1)^0.5 how do you decide which part is g(x) and which is f(x). I know most of the time u just set whatever is inside the bracket to u but why is that?
    For x\sqrt{x-1} you would set u=x-1 because the square root is one function, and x-1 is another which is inside it.

    The reason for this is covered when you first learn the chain rule, because this is essentially what it is just going in the opposite way.
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    (Original post by RDKGames)
    For x\sqrt{x-1} you would set u=x-1 because the square root is one function, and x-1 is another which is inside it.

    The reason for this is covered when you first learn the chain rule, because this is essentially what it is just going in the opposite way.
    Oh that makes sense ! Thanks so much.
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    (Original post by coconut64)
    Thanks. How about when you just have two functions for example x(x-1)^0.5 how do you decide which part is g(x) and which is f(x). I know most of the time u just set whatever is inside the bracket to u but why is that?
    Don't think of substitution in terms of rules. Mathematics, and integration especially, is fundamentally an art, and these things come with experience. In this case you can see that setting u=x-1 gives an integrand of (u+1)*u^0.5, which is easier to integrate as you can expand the brackets and then integrate each term as they're each powers of u. However, with integration at university level, you often will choose a substitution that doesn't necessarily appear completely in the integrand. The idea is just to choose something that will reduce it to a function you know how to integrate, and so you can think of various substitutions in your head and see which of them will work, before putting pen to paper.
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    For this integral you don't even need to make a substitution, if you consider the derivative of  e^{x^2} then the antiderivative you're after falls out right from this result.
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    (Original post by coconut64)
    Haven't learnt that yet but thanks anyway
    Should have a look at it. its uv - integral of v du/dx.

    You use it in a case where two functions are being multiplied e.g. 6xe^x. You let one of them equal u and the other dv/dx; so you differentiate u and integrate dv/dx. Usually you would let an x or a lnx as the u just because it makes integration easier (x gets priority).
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    Thanks everyone for helping
 
 
 
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