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    Hi. Just doing Maths and come across this question with this mark scheme (OCR MEI C4 Jan 06)
    I am wondering how comes it is 1/2lny-1/2ln(2-y) instead of 1/2lny+1/2ln(2-y)?
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    (Original post by prepdream)
    Hi. Just doing Maths and come across this question with this mark scheme (OCR MEI C4 Jan 06)
    I am wondering how comes it is 1/2lny-1/2ln(2-y) instead of 1/2lny+1/2ln(2-y)?
    Integrate 1/[2(2-y)]. There's a minus in there.

    Differentiating 1/2 * ln (2 - y) creates a minus sign at the front.
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    (Original post by ombtom)
    Integrate 1/[2(2-y)]. There's a minus in there.

    Differentiating 1/2 * ln (2 - y) creates a minus sign at the front.
    Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
    Thanks for the reply btw
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    (Original post by prepdream)
    Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
    Thanks for the reply btw
    You're right that it's the same as 1/2 * integral of 1/(2-y).

    Integrating 1/(2-y) gives -ln(2-y).

    Differentiating ln(2-y) gives -1/(2-y), hence the need for a minus sign.

    Hope this helps.
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    (Original post by prepdream)
    Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
    Thanks for the reply btw
    The integral of \displaystyle \frac{1}{2-y} = -\frac{1}{y-2} is \displaystyle -\ln|y-2|.
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    (Original post by prepdream)
    Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
    Thanks for the reply btw
    We know that \int \frac{f'(x)}{f(x)} \, \mathrm{d}x = \ln f(x) + c, yeah? That is, if the numerator is the derivative of the denominator, then integrating it produces the logarithm of the denominator.

    In this case, we have: \displaystyle \frac{1}{2} \int \frac{1}{2-y} \, \mathrm{d}y - the numerator is almost the derivative of the denominator - the dervative of the denominator is \frac{\mathrm{d}}{\mathrm{d}y}(2-y) = -1.

    How do we fix this? Well, \displaystyle \frac{1}{2} \int \frac{1}{2-y} \, \mathrm{d}y = -\frac{1}{2} \int \frac{-1}{2-y} \, \mathrm{d}y = -\frac{1}{2} \ln |2-y| + c

    since we now have the numerator as the derivative of the numerator.

    Try this out on: \displaystyle \int \frac{x}{2-x^2} \, \mathrm{d}x - the numerator is almost the derivative of the numerator.

    If you can't spot what to do:
    Spoiler:
    Show
    \displaystyle \int \frac{x}{2-x^2} \, \mathrm{d}x = -\frac{1}{2} \int \frac{-2x}{2-x^2} \, \mathrm{d}x
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    (Original post by Zacken)
    x
    This doesn't look much like Physics.
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    (Original post by tinkerbella~)
    This doesn't look much like Physics.
    Physics is maths basically
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    (Original post by Zacken)
    We know that \int \frac{f'(x)}{f(x)} \, \mathrm{d}x = \ln f(x) + c, yeah? That is, if the numerator is the derivative of the denominator, then integrating it produces the logarithm of the denominator.

    In this case, we have: \displaystyle \frac{1}{2} \int \frac{1}{2-y} \, \mathrm{d}y - the numerator is almost the derivative of the denominator - the dervative of the denominator is \frac{\mathrm{d}}{\mathrm{d}y}(2-y) = -1.

    How do we fix this? Well, \displaystyle \frac{1}{2} \int \frac{1}{2-y} \, \mathrm{d}y = -\frac{1}{2} \int \frac{-1}{2-y} \, \mathrm{d}y = -\frac{1}{2} \ln |2-y| + c

    since we now have the numerator as the derivative of the numerator.

    Try this out on: \displaystyle \int \frac{x}{2-x^2} \, \mathrm{d}x - the numerator is almost the derivative of the numerator.

    If you can't spot what to do:
    Spoiler:
    Show
    \displaystyle \int \frac{x}{2-x^2} \, \mathrm{d}x = -\frac{1}{2} \int \frac{-2x}{2-x^2} \, \mathrm{d}x
    Ahah alright that makes so much sense. I completely forgot about the f'(x)/f(x) thing. Thank you!
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    (Original post by prepdream)
    Ahah alright that makes so much sense. I completely forgot about the f'(x)/f(x) thing. Thank you!
    No problem! Do you want to test your understanding on the problem I gave at the end?
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    (Original post by Zacken)
    No problem! Do you want to test your understanding on the problem I gave at the end?
    Yeah is that -1/2ln(2-x^2) if I am not wrong? :/
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    (Original post by prepdream)
    Yeah is that -1/2ln(2-x^2) if I am not wrong? :/
    Yes!
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    (Original post by prepdream)
    Yeah is that -1/2ln(2-x^2) if I am not wrong? :/
    Well technically it should be
     \displaystyle -\frac{1}{2} \ln |2-x^2| +c .
 
 
 
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