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Math Help - Integration.

  1. #1
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    Integration.

    Integration.-question.png

    I believe what's being shown is a method of finding an integral called "u substitution". So I understand that u=x^4+2. Also I understand that du=4x^3dx although I wonder at this point why is a derivative being taken? I thought we were finding the anti derivative? Following further I understand that [tex](x^4+2)^5=u^5. Then there is x^3dx=\frac{1}{4}du which I don't understand at all. I am just left wondering where the \frac{1}{4} came from.

    At this point in my online course they are talking about the anti derivative and we have yet only talked about the indefinite integral.

    Thanks for any responses...
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  2. #2
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    Re: Integration.

    you need to find the formula for du in terms of dx so that you can replace dx in the original integral by a function of du.

    Here we have

    u = x4 + 2

    du/dx = 4x3

    du = 4x3 dx

    1/4 du = x3 dx

    the right hand side is what appears in your original integral.

    So after substituting you end up with

    \int {u^5} \frac{1}{4}\,du = \frac{1}{4}\ \int u^5 \,du
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  3. #3
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    Re: Integration.

    I know that this opinion is in the vast minority but, I always discouraged students from using u-substitution.
    I think it is pathetically sad if a student sees \int {{x^3}{{\left( {{x^4} + 2} \right)}^5}dx} and does not see at once that the answer is \frac{1}{{24}}{\left( {{x^4} + 2} \right)^6}.

    For years I used such tests questions as: Prove that \int {{x^3}{{\left( {{x^4} + 2} \right)}^5}dx}=\frac{1}{{24}}{\left( {{x^4} + 2} \right)^6}+C.
    Students were often floored to see that only differentiation is needed as a proof.
    But it always got my point across: Learn the differentiation forms.
    Thanks from sepoto
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