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    (Original post by creativebuzz)
    Oh okay! But when I plug the limits into my calculator I get 0 for 1 and an error for -1
    Remember that when you integrate you will get 0.5ln|2x-1| so remember the modulus signs
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    (Original post by TenOfThem)
    The answer I would expect would be 0.5(ln1 - ln3) = -0.5ln3
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    (Original post by TeeEm)
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    It wouldn't be asked in an exam (hopefully)
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    (Original post by TenOfThem)
    It wouldn't be asked in an exam (hopefully)
    I hope not.

    (an answer can be obtained of course but definitely wrong)
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    (Original post by TenOfThem)
    Remember that when you integrate you will get 0.5ln|2x-1| so remember the modulus signs
    Does that mean that instead of getting 0.5ln|-3| I would get 0.5ln|3|?
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    (Original post by TeeEm)
    I hope not.

    (an answer can be obtained of course but definitely wrong)
    Sure

    But the point of the exercise is to test a limited understanding ... That is how text books are produced

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    (Original post by creativebuzz)
    Does that mean that instead of getting 0.5ln|-3| I would get 0.5ln|3|?
    Yes
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    (Original post by TenOfThem)
    Yes
    Thanks!

    As for the last one, I did the fourth root got
    I = 2sinxcos
    = sin2x

    therefore it's =1/2cos2x + c

    Where did I go wrong? :/
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    (Original post by creativebuzz)
    Thanks!

    As for the last one, I did the fourth root got
    I = 2sinxcos
    = sin2x

    therefore it's =1/2cos2x + c

    Where did I go wrong? :/
    Taking the 4th root was only so you could find the sin2x

    So you need to integrate (sin2x)^4
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    [QUOTE=TenOfThem;53124173]Taking the 4th root was only so you could find the sin2x

    Woo, I got the right answers! Thanks

    As for K, I think I've got in a muddle but this is my working out so far:

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    [QUOTE=creativebuzz;53126211]
    (Original post by TenOfThem)
    Taking the 4th root was only so you could find the sin2x

    Woo, I got the right answers! Thanks

    As for K, I think I've got in a muddle but this is my working out so far:

    You do not want a mix of x and 2x

    You can always integrate cosx sin^nx by inverse chain rule

    So change your cos^3x into cocs^2xcosx and then change the cos^2x into 1-sin^2x
 
 
 
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