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    Cant do part b, anyone help me!?
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    just integrate in and put in the values in
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    (Original post by El Asesino)
    just integrate in and put in the values in
    Its x^2 = not x = ??
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    can't you just write the question PLEASE??

    My copy of review exercise is all the way upstairs - you honestly can't expect me to go that far can you???
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    (Original post by Silly Sally)
    can't you just write the question PLEASE??

    My copy of review exercise is all the way upstairs - you honestly can't expect me to go that far can you???
    Theres a diagram..
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    (Original post by imasillynarb)
    Theres a diagram..
    Ok the - you win - i will got get my copy
    :rolleyes:
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    (Original post by imasillynarb)
    Its x^2 = not x = ??
    you cant sqrt both sides?
    x= +- y^(1/2) * (4-y) ?
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    (Original post by kimoni)
    you cant sqrt both sides?
    x= +- y^(1/2) * (4-y) ?
    I tried, but Im not getting the right answer, hence I need help

    And what isnt the same is it, its the root of the WHOLE thing not root y x (4-y)
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    you have to integrate the positive part of the sqrt of the curve as its to the right side of the x-axis....
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    (Original post by toxi)
    you have to integrate the positive part of the sqrt of the curve as its to the right side of the x-axis....
    Yes, IVE TRIED, but Im not getting the right answer.
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    Tis a hard question - whats the answer you keep getting cos maybe the book is wrong?
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    (Original post by imasillynarb)
    Yes, IVE TRIED, but Im not getting the right answer.
    ok
    x^2 = blabla goes to..
    x= +- y^(1/2) * (4-y)

    integrate by parts:

    v=(4-y)
    dv/dy= -1

    du=y^1/2
    u= (2/3)y^(3/2)

    uv - Int:udv

    you do the rest
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    Ok heres what u need to do:

    take the square root so u have

    √y(4-y) = 4√y-y^3/2

    then you integrate this and u get

    2/3(4y^3/2)- 2/5(y^5/2)

    You substitute the values and u got it
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    (Original post by kimoni)
    ok
    x^2 = blabla goes to..
    x= +- y^(1/2) * (4-y)

    integrate by parts:

    v=(4-y)
    dv/dy= -1

    du=y^1/2
    u= (2/3)y^(3/2)

    uv - Int:udv

    you do the rest
    You don't really need to do integration by parts, you can do the multiplication and it'll be much easier that way
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    (Original post by Silly Sally)
    Tis a hard question - whats the answer you keep getting cos maybe the book is wrong?
    The book is right, but indeed it's a hard question.
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    (Original post by kimoni)
    ok
    x^2 = blabla goes to..
    x= +- y^(1/2) * (4-y)

    integrate by parts:

    v=(4-y)
    dv/dy= -1

    du=y^1/2
    u= (2/3)y^(3/2)

    uv - Int:udv

    you do the rest
    Cheers!
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    (Original post by toxi)
    You don't really need to do integration by parts, you can do the multiplication and it'll be much easier that way
    ... or you could do that!
    silly me
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    (Original post by kimoni)
    The book is right, but indeed it's a hard question.
    Solid question, didnt even think about using integration by parts ;(

    Help me with my other integration problem now! Its question 24, but Ive posted it..
 
 
 
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