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    Is this integrable? [(-y^2)-(2y)+3]^(1/2)

    How would i go about doing this?
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    \sqrt{-y^2 - 2y + 3}.

    Complete the square on -y^2 - 2y + 3, and then try making a suitable hyperbolic/trigonometric substitution. I've not tried it, but I think the following substitution will work:

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    y + 1 = 2 \sin u

    (after obtaining -y^2 - 2y + 3 = 4 - (y + 1)^2)
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    Hey thanks for the reply, i used the subsitution y+1=2sinu (as u said) and i got and answer of
    (1/2)(y+1)[(4-(y+1)^2)] + sin^-1 [(y+1)/2]

    do you reckon it looks right?

    sorry if ur having trouble with reading my answer (im not good with that math typing software thing)
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    (Original post by dopemann)
    Hey thanks for the reply, i used the subsitution y+1=2sinu (as u said) and i got and answer of
    (1/2)(y+1)[(4-(y+1)^2)] + sin^-1 [(y+1)/2]

    do you reckon it looks right?

    sorry if ur having trouble with reading my answer (im not good with that math typing software thing)
    I think it should be 2 sin^-1 etc.
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    Another question, if i was to put the limits of integration between y=1 and y=0 , would i get my answer in degrees? because sin^-1 [(y+1)/2] would give degrees if i put in y=1 or y=0?
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    No, you should get your answer in radians (which are dimensionless).
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    Yeh sorry mr m, i forgot to write that
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    Ok, no problem, as the general suggested, when working with differentiation and integration, you should ALWAYS work in radian measure.
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    are you sure its radians, because i get the answer 119.1339746 (seems like degrees to me)
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    (Original post by dopemann)
    are you sure its radians, because i get the answer 119.1339746 (seems like degrees to me)
    The value of that integral with those limits is 1.23 to 3 sf.

    Imagine the area represented by that integral. The value of the function when y = 0 is \sqrt3 and its value when y = 1 is . If you calculate the area of a rectangle of height \sqrt3 and width 1 you will only obtain \sqrt3 as the answer and the area of the integral must clearly be less than this.
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    how did you get that? that looks like the right answer but i dont understand how to get that...
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    (Original post by dopemann)
    are you sure its radians, because i get the answer 119.1339746 (seems like degrees to me)
    Is your calculator in degree mode?
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    o i understand now... i must use radian mode when plugging in y=1 and y=0 because after all these are in radians (obviously - silly me). so yeh i get the answer 1.23.

    Thankyou everyone, appreciate the help
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    You are welcome.
 
 
 
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