Integration by substitution

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  1. TheJ0ker's Avatar
    1 18-04-2012  14:47
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    Integration by substitution
    I=\displaystyle\int_0^a \frac{f(x)}{f(x)+f(a-x)} dx

    Use a substitution to show I =\displaystyle\int_0^a \frac{f(a-x)}{f(x)+f(a-x)} dx

    I used u = a - x and got the required result then

    Hence evaluate I in terms of a

    I have no idea of how to start I was thinking that \displaystyle\int_0^a f(x) dx = F(a) - F(0) and \displaystyle\int_0^a f(a-x) dx = F(0) - F(a) but I don't know how to proceed.

    Please don't post the full solution I want to do it by myself as much as possible.

    Cheers
  2. notnek's Avatar
    2 18-04-2012  15:03
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: Integration by substitution
    Hint: Can you write the numerator (in the second form of I) another way so that the fraction can be split up?

    Spoiler:
    Show
    f(a-x)=f(x)+f(a-x)-f(x)
    Last edited by notnek; 18-04-2012 at 15:18.
  3. TheJ0ker's Avatar
    3 18-04-2012  15:20
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    Re: Integration by substitution
    (Original post by notnek)
    Hint: f(a-x)=f(x)+f(a-x)-f(x)
    Ah ok so I =\displaystyle\int_0^a \frac{f(x) + f(a-x)}{f(x)+f(a-x)} -\frac{f(x)}{f(x)+f(a-x)} dx

    So I=\displaystyle\int_0^a 1 dx - I

    So I=\frac{1}{2}\displaystyle\int_0 ^a 1 dx?
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  4. notnek's Avatar
    4 18-04-2012  15:25
    • TSR Demigod
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    Re: Integration by substitution
    (Original post by TheJ0ker)
    Ah ok so I =\displaystyle\int_0^a \frac{f(x) + f(a-x)}{f(x)+f(a-x)} -\frac{f(x)}{f(x)+f(a-x)} dx

    So I=\displaystyle\int_0^a 1 dx - I

    So I=\frac{1}{2}\displaystyle\int_0 ^a 1 dx?
    Yes, that's right. Can you finish it off or did you stop because you were stuck?
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  5. TheJ0ker's Avatar
    5 18-04-2012  15:28
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    Re: Integration by substitution
    (Original post by notnek)
    Yes, that's right. Can you finish it off or did you stop because you were stuck?
    No I can finish off from there. It's amazing how something simple like that can completely change a fairly nasty looking integral, I need to be sharper next time. Thanks for the help.
  6. notnek's Avatar
    6 18-04-2012  15:40
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: Integration by substitution
    (Original post by TheJ0ker)
    No I can finish off from there. It's amazing how something simple like that can completely change a fairly nasty looking integral
    I'm with you on this - I used to like these kinds of questions. Is this a STEP/AEA question or has A level maths become harder?

    And the more tricky integrals you do, the sharper you'll get. Next time you see a question with two forms of the same integral I, there's a good chance that you'll know what to do.
  7. elldeegee's Avatar
    7 18-04-2012  15:46
    • Exalted and Worshipped Member
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    Re: Integration by substitution
    (Original post by TheJ0ker)
    Ah ok so I =\displaystyle\int_0^a \frac{f(x) + f(a-x)}{f(x)+f(a-x)} -\frac{f(x)}{f(x)+f(a-x)} dx

    So I=\displaystyle\int_0^a 1 dx - I

    So I=\frac{1}{2}\displaystyle\int_0 ^a 1 dx?

    (Original post by notnek)
    Yes, that's right. Can you finish it off or did you stop because you were stuck?
    I tried helping, but i got stuck

    however, i have a question now,

    How did ytou get from So I=\displaystyle\int_0^a 1 dx - I
    to
    I=\frac{1}{2}\displaystyle\int_0 ^a 1 dx

    I'm probably being a bit dim, have i forgotten some special rule?

    DOESN'T MATTER
    I thought the - I was inside the integral, didnt take into account the fact that dx came before it! duh!

    so yes, i was being dim
    Last edited by elldeegee; 18-04-2012 at 15:47.
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  8. TheJ0ker's Avatar
    8 18-04-2012  16:23
    • Peer Of The TSR Realm
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    Re: Integration by substitution
    (Original post by notnek)
    I'm with you on this - I used to like these kinds of questions. Is this a STEP/AEA question or has A level maths become harder?

    And the more tricky integrals you do, the sharper you'll get. Next time you see a question with two forms of the same integral I, there's a good chance that you'll know what to do.
    Yeah this is a STEP q. These are the kind of questions I go straight for when I start a new paper, I really like them. I am a bit slow at the moment with them but hopefully in the 9 weeks I have left I will get much better.
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