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    Just need quick help with two question.

    1) I=(integrate)
    Ixsquareroot(3x-1) dx using substitution u=3x-1

    Now I tried doing it by differentiating u with respect to x to give 3 but then it becomes jumbled up near the end. And when i checked the mark scheme i see they just rearranged u in terms of x and then differentiated x with respect to u. So my question is when do you know when to do this?

    2)a) describe the 2 geometrical that maps the graph of y=x^2 onto the graph of y=4x^2 -5
    I can do the translation part but when I did the enlargment part (stretch) I got it as stretch parallel to x-axis with scale factor of 1/4 but its wrong :confused: .

    b&c) solve the equation (modulus) of 4x^2 -5 so ([4x^2 -5] where []=modulus). Then hence solve the inequality [4x^2 -5]>(or equal to) 4.

    p.s. sory about it not being in latex form I tried to put in there but when i clicked preview it showed up as messed up.
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    for 1) let u^2 = 3x - 1
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    (Original post by Sambo2)
    for 1) let u^2 = 3x - 1
    They tell you to use u=3x-1
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    (Original post by thewait)
    They tell you to use u=3x-1
    oh ok.. it's been a long time :p:
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    (Original post by Sambo2)
    oh ok.. it's been a long time :p:
    A long day as well I take it? As i put that in my 1st post aswell :p:
    Thanks for trying to help tho.
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    (Original post by thewait)
    Just need quick help with two question.

    1) I=(integrate)
    Ixsquareroot(3x-1) dx using substitution u=3x-1

    Now I tried doing it by differentiating u with respect to x to give 3 but then it becomes jumbled up near the end. And when i checked the mark scheme i see they just rearranged u in terms of x and then differentiated x with respect to u. So my question is when do you know when to do this?

    2)a) describe the 2 geometrical that maps the graph of y=x^2 onto the graph of y=4x^2 -5
    I can do the translation part but when I did the enlargment part (stretch) I got it as stretch parallel to x-axis with scale factor of 1/4 but its wrong :confused: .

    b&c) solve the equation (modulus) of 4x^2 -5 so ([4x^2 -5] where []=modulus). Then hence solve the inequality [4x^2 -5]>(or equal to) 4.

    p.s. sory about it not being in latex form I tried to put in there but when i clicked preview it showed up as messed up.
    whats the answer?? - if it matches mine, ill post my solution, otherwise ill save my grace
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    ooh I was right.. ok here we go!!

    firstly differentiate to get du/dx = 3 so dx = du/3

    so u then have I of (u^1/2)/3 du which is 2/9u^3/2 - substitute back for you and that is the right answer!! I'm quite proud of that, haven't done any maths in ages....
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    (Original post by Sambo2)
    whats the answer?? - if it matches mine, ill post my solution, otherwise ill save my grace
    Actually my answer was right, they just factorised more than I did.

    If you're still interested answer was: 2/45 (3x-1)^5/2 + 2/27 (3x-1)^3/2 + c
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    couldnt quite get latex to work though... sorry not really being helpfull just amusing myself with a bit of maths..
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    (Original post by Sambo2)
    couldnt quite get latex to work though... sorry not really being helpfull just amusing myself with a bit of maths..
    Its cool at least you tried thanks. Any chance of help with 2)?
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    (Original post by thewait)
    Its cool at least you tried thanks. Any chance of help with 2)?
    your going to have to wait till the maths geeks wake up im afraid.. it's been too long for me to remember that I'm ashamed to say.. - if you ever need bio or chem help though I'm your man
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    let f(x) = x^2 so f(4x)=4x^2 and f(4x)-5 = 4x^2 -5 i believe thats a stretch in the y direction and a shift of the entire graph 5 spaces down in y direction
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    Aqa?
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    I=\int{\sqrt{(3x-1)}}dx
    Let u = 3x - 1
    Then \frac{du}{dx}=3
    Or du=3dx
    Take I to be:
    I=\frac{1}{3} \int3{ \sqrt{(3x-1)}}dx
    Substututing u=3x-1 we now have:
    I=\frac{1}{3} \int {u^{0.5}}du

    Now you can happily do the integration.

    As for the transformation:

    It is described by a stretch parallel to the y-axis by a scale factor of 4 and a shift of 5 in the negative y-direction.
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    Is the integral \displaystyle \int \sqrt{3x-1}\, \mbox{d}x or \displaystyle \int x\sqrt{3x-1}\, \mbox{d}x. Either way, you need to do more or less what tgodkin said above. If it's the latter, notice that if u = 3x-1 then x = \dfrac{1}{3}(u+1), and so after you make the substitution you'll get \displaystyle \dfrac{1}{3}\int (u+1)\sqrt{u} \times \dfrac{1}{3}\, \mbox{d}u -- just expand the brackets and integrate.

    For 2a the reason it's wrong is because a stretch of factor 1/4 in the x-direction would be in the form f(4x), however that would give 16x^2 rather than 4x^2. Instead you can either see it as a stretch of factor 1/2 in the x-direction (so you're squaring 2 to give 4), or a stretch of factor 4 in the y-direction. Either way it makes no difference to the geometry of the graph.

    For 2b I think you've copied out the question wrong. You can't "solve |4x²-5|", since there's nothing to solve... your answer to part (b) will, I imagine, help in part (c).
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    (Original post by tgodkin)
    Take I to be:
    I=\frac{1}{3} \int3{ \sqrt{(3x-1)}}dx
    Substututing u=3x-1 we now have:
    I=\frac{1}{3} \int {u^{0.5}}du
    What happened to the 3? shouldnt it be :


    I=3 \int {u^{0.5}}du ?


    Thanks!
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    (Original post by wizz_kid)
    What happened to the 3? shouldnt it be :


    I=3 \int {u^{0.5}}du ?


    Thanks!
    Because du = 3dx the 3 is swapped out with dx for du.
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    (Original post by tgodkin)
    Because du = 3dx the 3 is swapped out with dx for du.
    Agreed
 
 
 
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