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C4 - Integration watch

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    Hi!

    Stuck on an easy question, need to clarify this. :o:

    Is the integration of

    \int \frac{1}{3t}

    \frac{1}{3} \ln |3t| + c

    In the answer booklet it says \frac{1}{3} \ln |t| + c
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    (Original post by uer23)
    Hi!

    Stuck on an easy question, need to clarify this. :o:

    Is the integration of

    \int \frac{1}{3t}

    \frac{1}{3} \ln |3t| + c

    In the answer booklet it says \frac{1}{3} \ln |t| + c
    The book is correct because \int \frac{1}{3t}dt = \frac{1}{3} \int \frac{1}{t}dt
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    The answer book is right. Differentiate your result (ignore the absolute for a second) and see what you get.
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    They are both right, it's just that the constant term will be different in each case.

    Differentiate the OPs answer and see what you get.
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    Take out the factor of 1/3 and you're left with 1/t.

    The integral of 1/t results in ln 'mod' t + c.

    Thus when you multiply by the factor of 1/3, you will obtain 1/3 ln 'mod' t + c.

    Don't forget the "+c"!

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    basically, you can move constants i.e. 1/3 in this case out of the integral so it becomes 1/3 (integral) 1/t dt

    1/ t integrates to ln [ t ] + c ( [] is modulus sign to make it clear)

    Therefore integral of 1/3t is...

    1/3 { ln [t] + c)

    which is 1/3 ln [t] + c

    C can = ln A

    therefore this can also integrate to

    1/3 ln A [t]. but 1/3 ln [t] + c is also fine
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    (Original post by ghostwalker)
    They are both right, it's just that the constant term will be different in each case.

    Differentiate the OPs answer and see what you get.
    Do you know if my answer is acceptable in A-Level Exams (Edexcel Board)
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    (Original post by uer23)
    Do you know if my answer is acceptable in A-Level Exams (Edexcel Board)
    To be honest, it's fairly unconventional as a result and in general I pull all constants to the front of the integral and then integrate the remaining variables. That way you're likely to be getting the result on the mark scheme.
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    (Original post by Farhan.Hanif93)
    To be honest, it's fairly unconventional as a result and in general I pull all constants to the front of the integral and then integrate the remaining variables. That way you're likely to be getting the result on the mark scheme.
    Right thanks.

    I'm going to do it the way you guys did it, just to be on the safe side.
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    dont mess things up when u get closer to exams! ... try hard and follow the syllabus!
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    (Original post by Farhan.Hanif93)
    The book is correct because
    If in doubt, use a substitution, e.g. u = 3t
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    (Original post by v-zero)
    If in doubt, use a substitution, e.g. u = 3t
    Cheers for the tip, I appreciate it
    ...But I'm not the one who's struggling with the concept :p:
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    (Original post by Farhan.Hanif93)
    Cheers for the tip, I appreciate it
    ...But I'm not the one who's struggling with the concept
    Oops. :p:
 
 
 
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