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    Jus giving you lot some more practice :p: hehe!

    Well ive dun this q twice and keep getting different answers.. its from a c4 mock paper, so cant find a mark scheme either!

    Integrate: (boundaries are 1/3 and 1/6)

    (5/1-x) - (4/2x+1)

    Give the answer in the form lnp, where p is a rational number

    thank you sooo much, also if any1 finds a mark scheme or has one for the mock paper could the pm me it or post it up here!!!

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    i assume that is:
     \int \frac{5}{1-x} - \frac {4}{2x+1}  dx
    = -5 ln(1-x) - 2 ln(2x+1) and plug the boundaries

    that would be:
    [-5 ln(5/6) - 2 ln(8/6)] - [ -5 ln(2/3) - 2ln(5/3) ] =
    ln[(2/3)5] + ln[(5/3)²] + ln[(6/5)5 + ln[(3/4)²] =
    make it as one ln... and then u will get the answer
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    I got p=125/64
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    yep thats wat i got the latest time ive dun it!! thank youu!!
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    yazan, how come you got -5ln(1-x)

    i got +5ln(1-x)
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    (Original post by __WiZaRD__)
    yazan, how come you got -5ln(1-x)

    i got +5ln(1-x)

    well.. it is 5/(1-x) ... there is a MINUS before the x... i.e. d[1-x]/dx = -1 not 1
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    yup i see it, sorry.
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    [-5 ln(5/6) - 2 ln(8/6)] - [ -5 ln(2/3) - 2ln(5/3) ] =

    I keep getting it as:

    [-5ln(2/3) - 2ln(5/3)] - [-5ln(5/6)-2ln(8/6)]

    surely its the (1/3) boundary take away the 1/6 boundary? i havent done any integration for over a month so please point out if ive made a trivial error....
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    ^^ yep uve dun it rite :, but you cant just put all that on your calculator coz they want the answer in terms of ln, so you need to simplify, thats where all the mistakes happen tho!! arghh!!
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    yeah the 2nd line is my own working, i got ln (125/64) the top line is from yazan's working, but i think he's done the boundaries the wrong way round....
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    (Original post by sebbie)
    [-5 ln(5/6) - 2 ln(8/6)] - [ -5 ln(2/3) - 2ln(5/3) ] =

    I keep getting it as:

    [-5ln(2/3) - 2ln(5/3)] - [-5ln(5/6)-2ln(8/6)]

    surely its the (1/3) boundary take away the 1/6 boundary? i havent done any integration for over a month so please point out if ive made a trivial error....


    well ya... i should've put the 1/3 first... but it doesnt matter after all.. the only difference is that my answer would be -ve...


    as  \int^a _b f(x) dx \,= \, - \int ^b _a f(x) dx
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    actually no...
    (boundaries are 1/3 and 1/6) means that the integral is FROM 1/3 TO 1/6... i know it should've been (boundaries are 1/6 and 1/3)... but it's tweety's mistake... hehehe
    but according to what the question she gave: my answer is the correct answer!
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    sorry i rote it rong, but you no wat i relli meant :p: hehe

    does any1 have the markscheme for c4 mock paper?
 
 
 
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