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C4 integration using substitution help Watch

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    Hi, just wondering if someone could help me with an integration using substitution question (when you are given u=...). I know you work out du/dx and substitute this and u into the integration and cancel out dx, then work it out.. But I can't see how it works on some questions - like this one..

    Use the substitution u= 3x - 1 to express integral of x(3x-1)^4dx as an integral of x

    Thank you for any help
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    What is x in terms of u?

    Can you solve it now?
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    sometimes the x will not cancel as in this case... instead you rewrite x in terms of u and go from there...
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    (Original post by b4nanapancakes)
    Hi, just wondering if someone could help me with an integration using substitution question (when you are given u=...). I know you work out du/dx and substitute this and u into the integration and cancel out dx, then work it out.. But I can't see how it works on some questions - like this one..

    Use the substitution u= 3x - 1 to express integral of x(3x-1)^4dx as an integral of x

    Thank you for any help
    Make x the subject of u = 3x -1 then sub that in along with u4 into your integral.
    Then its just normal integration.

    EDIT: Beaten to it.
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    (Original post by aznkid66)
    What is x in terms of u?

    Can you solve it now?
    Still a bit stuck.. What about the dx? Do I replace the 3 in (u+1)/3 with du/dx ?


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    (Original post by b4nanapancakes)
    Still a bit stuck.. What about the dx? Do I replace the 3 in (u+1)/3 with du/dx ?


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    You've got du/dx = 3?
    Then dx = du/3
    Divide the function in the integrand by 3.
    Then integrate the function with respect to u.
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    (Original post by joostan)
    You've got du/dx = 3?
    Then dx = du/3
    Then integrate the function with respect to u.
    Ah ok! Great thank you for your help!


    This was posted from The Student Room's iPhone/iPad App
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    (Original post by b4nanapancakes)
    Ah ok! Great thank you for your help!


    This was posted from The Student Room's iPhone/iPad App
    you're welcome
 
 
 
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