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    Hey, how would you work this out: The integral of 4 / 2x - 3 between the limits x is 6 and 2?
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    Do you know the integral of 1/x ?
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    (Original post by student92)
    Hey, how would you work this out: The integral of 4 / 2x - 3 between the limits x is 6 and 2?
    Do you mean \frac{4}{2x}-3 or \frac{4}{2x-3}
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    (Original post by steve2005)
    Do you mean \frac{4}{2x}-3 or \frac{4}{2x-3}
    the second oneee
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    (Original post by JB Johnstone)
    Do you know the integral of 1/x ?
    yeah lnx
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    (Original post by student92)
    Hey, how would you work this out: The integral of 4 / 2x - 3 between the limits x is 6 and 2?
    Rewrite that as 2(2/2x-3). Then use lnfx + c = int(f'(x)/f(x))
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    (Original post by Hippysnake)
    Then use lnx = f'(x)/f(x)
    That's wrong on many levels :p: I think you mean \dfrac{\mbox{d}}{\mbox{d}x}\bigg  ( \ln f(x) \bigg) = \dfrac{f'(x)}{f(x)}... or perhaps more relevantly, \displaystyle \int \dfrac{f'(x)}{f(x)}\, \mbox{d}x = \ln |f(x)| + C.
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    (Original post by nuodai)
    That's wrong on many levels :p: I think you mean \dfrac{\mbox{d}}{\mbox{d}x}\bigg  ( \ln f(x) \bigg) = \dfrac{f'(x)}{f(x)}... or perhaps more relevantly, \displaystyle \int \dfrac{f'(x)}{f(x)}\, \mbox{d}x = \ln |f(x)| + C.
    Ye! Wateva lol.

    My C3 career is officially over. It wasn't that wrong, I just got a little lazy and missed out some constants, which weren't needed tbf because the limits are defined, on and the f(x) too.

    But he gets what I'm saying...
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    (Original post by Hippysnake)
    Rewrite that as 2(2/2x-3). Then use lnfx + c = int(f'(x)/f(x))
    i get 2ln6 - 2ln1 but the books answer is 4ln3 :s
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    (Original post by Hippysnake)
    Ye! Wateva lol.

    My C3 career is officially over. It wasn't that wrong, I just got a little lazy and missed out some constants, which weren't needed tbf because the limits are defined, on and the f(x) too.

    But he gets what I'm saying...
    SHE!
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    (Original post by Hippysnake)
    Ye! Wateva lol.

    My C3 career is officially over. It wasn't that wrong, I just got a little lazy and missed out some constants, which weren't needed tbf because the limits are defined, on and the f(x) too.

    But he gets what I'm saying...
    You don't know that...
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    (Original post by student92)
    i get 2ln6 - 2ln1 but the books answer is 4ln3 :s
    Check the 2ln6 bit again.
    If stuck see spoiler for solution.
    Spoiler:
    Show

    should be 2ln9. Which can be written as 2ln3^2 which is 4ln3.
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    (Original post by Sakujo)
    You don't know that...
    Even if I get a U, I'm not doing it again.
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    (Original post by Hippysnake)
    Even if I get a U, I'm not doing it again.
    Lol, I'm sure you did fine.
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    i meaaaaaaaaaannn 2ln9 - 2ln1
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    (Original post by Hippysnake)
    My C3 career is officially over. It wasn't that wrong, I just got a little lazy and missed out some constants, which weren't needed tbf because the limits are defined, on and the f(x) too.
    You missed out constants, a function (inside the logarithm) and either a differential operator (before the logarithm) or integral operator (before the RHS), depending on which way round you're doing it. It wasn't "lazy but almost right", it was "wrong"... but it being wrong wasn't my quibble, it was that the OP doesn't seem to know what's going on that much so probably wouldn't see that what you said was wrong, and then would use it in her work.

    And you'll need it in C4 anyway.

    (Original post by student92)
    i meaaaaaaaaaannn 2ln9 - 2ln1
    Simplify it using the laws of logarithms. There are a few things to spot. First, since \ln a^b = b\ln a, we have that \ln 1 = \ln (e^0) = 0\ln e = 0, and also \ln 9 = \ln (3^2) = \cdots. You've got the right answer, you just haven't simplified it.
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    thanks
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    (Original post by nuodai)
    You missed out constants, a function (inside the logarithm) and either a differential operator (before the logarithm) or integral operator (before the RHS), depending on which way round you're doing it. It wasn't "lazy but almost right", it was "wrong"... but it being wrong wasn't my quibble, it was that the OP doesn't seem to know what's going on that much so probably wouldn't see that what you said was wrong, and then would use it in her work.

    And you'll need it in C4 anyway.


    Simplify it using the laws of logarithms. There are a few things to spot. First, since \ln a^b = b\ln a, we have that \ln 1 = \ln (e^0) = 0\ln e = 0, and also \ln 9 = \ln (3^2) = \cdots. You've got the right answer, you just haven't simplified it.
    Sorry dude....I got the answer right though...
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    (Original post by Hippysnake)
    Sorry dude....I got the answer right though...
    Hey, i got another question... if u could help me: find dy/dx of 3/x-2 - 2/x+3 when x is -1. The answer is 1/6 but i keep getting -1/6
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    (Original post by student92)
    Hey, i got another question... if u could help me: find dy/dx of 3/x-2 - 2/x+3 when x is -1. The answer is 1/6 but i keep getting -1/6
    Have you rewitten it as 3(x - 2)^{-1} - 2(x + 3)^{-1} and used the chain rule?
 
 
 
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