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Integration

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    Hi, the brackets are not in the original equation, I've just added them to make it simpler.

    The original equation is (x/1)-(1/x1/2)dx

    I got up to

    (x2/2)-(x1/2/1/2)

    I'm pretty sure it is correct up until then, I'm unsure were to go from there... The numbers we sub in are 9 & 4 respectively.
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    Sub 9 in first then sub 4 in
    Then take the first answer where you subbed in 9 and take away the second answer that you got from subbing in 4
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    (Original post by 0range)
    Sub 9 in first then sub 4 in
    Then take the first answer where you subbed in 9 and take away the second answer that you got from subbing in 4
    No, thats no finished, it has to be simplified further... You can't have a fraction at the bottom of a fraction.... When I simplify the last fraction, I still dont get the right answer which is (10.2/3)
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     \frac{x^2}{2} - \frac{x^\frac{1}{2}}{(\frac{1}{2  })} = \frac{x^2}{2} - 2x^\frac{1}{2}

    substituting in 9 and 4:

    [\frac{(9)^2}{2} - 2(9)^\frac{1}{2}] - [\frac{(4)^2}{2} - 2(4)^\frac{1}{2}]

    = [40.5-6] - [8-4] = 30.5

    So i guess must have gone wrong before this point if thats not the right answer :L

    Are you sure the original equation is correct?
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    (Original post by just george)
     \frac{x^2}{2} - \frac{x^\frac{1}{2}}{(\frac{1}{2  })} = \frac{x^2}{2} - 2x^\frac{1}{2}

    substituting in 9 and 4:

    [\frac{(9)^2}{2} - 2(9)^\frac{1}{2}] - [\frac{(4)^2}{2} - 2(4)^\frac{1}{2}]

    = [40.5-6] - [8-4] = 30.5

    So i guess must have gone wrong before this point if thats not the right answer :L

    Are you sure the original equation is correct?
    Yes, except in the original equation the second denominator is root x, which x1/2, thus (x/1)-(1/x1/2)dx, also its (x minus 1 over root x), which is the same as (x/1)-(1/x1/2)dx... I get the same answer as you, have you tried using the original equation?
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    Anyone able to do it?
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    (Original post by King-Panther)
    No, thats no finished, it has to be simplified further... You can't have a fraction at the bottom of a fraction.... When I simplify the last fraction, I still dont get the right answer which is (10.2/3)
    You don't have to simplify it, you won't get anymore marks unless it tells you to simplify your answer. Where I told you to sub it in is right. You may have copied the question down wrong or are looking at a different answer.
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    (Original post by 0range)
    You don't have to simplify it, you won't get anymore marks unless it tells you to simplify your answer. Where I told you to sub it in is right. You may have copied the question down wrong or are looking at a different answer.
    No, the answer is (10.2/3)

    (x/1)-(1/x1/2)dx is the original equation, although its just (x-1/root x) which is the same as (x/1)-(1/x1/2)
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    (Original post by King-Panther)
    No, the answer is (10.2/3)

    (x/1)-(1/x1/2)dx is the original equation, although its just (x-1/root x) which is the same as (x/1)-(1/x1/2)
    Surely if the original equation was ( x-1/root x ), this becomes ( x-1/x^1/2 ), and that can be written as; x^-1/2 (x-1) which equals [ x^1/2 - x^-1/2 ], which when integrated becomes;
    ( x^3/2 / 3/2 ) - ( x^1/2 / 1/2 )

    However if what you say is the correct answer is right, this is also wrong (although i cant see how, sorry if there's any stupid mistakes) as i got the answer 32/3
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    (Original post by J.G.M)
    Surely if the original equation was ( x-1/root x ), this becomes ( x-1/x^1/2 ), and that can be written as; x^-1/2 (x-1) which equals [ x^1/2 - x^-1/2 ], which when integrated becomes;
    ( x^3/2 / 3/2 ) - ( x^1/2 / 1/2 )

    However if what you say is the correct answer is right, this is also wrong (although i cant see how, sorry if there's any stupid mistakes) as i got the answer 32/3
    I don't understand how you have written it, x-(1/rootx) is the same as (x/1)-(1/x1/2)

    We both agree with that. I 'm lost with the rest, could you write how I have done please because 32/3 is the same as (10.2/3), thus the correct answer
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     \frac{x-1}{x^\frac{1}{2}} = x^\frac{1}{2}-x^-^\frac{1}{2}

    integrates to  \frac{2x^\frac{3}{2}}{3} - 2x^\frac{1}{2}

    substituting in 9 and 4:

     [\frac{2(9)^\frac{3}{2}}{3} - 2(9)^\frac{1}{2}] - [\frac{2(4)^\frac{3}{2}}{3} - 2(4)^\frac{1}{2}]

     = [18 - 6] - [\frac{16}{3} - 4] = 12 - \frac{4}{3} = \frac{32}{3}

    if thats the right answer, then it is because you integrated x-\frac{1}{x^\frac{1}{2}} instead of  \frac{x-1}{x^\frac{1}{2}}


    edit: and writing 10.2/3 confused me.. that looks like  \frac{10.2}{3} whereas i think your meaning to say  10\frac{2}{3} ?
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    (Original post by King-Panther)
    I don't understand how you have written it, x-(1/rootx) is the same as (x/1)-(1/x1/2)

    We both agree with that. I 'm lost with the rest, could you write how I have done please because 32/3 is the same as (10.2/3), thus the correct answer
    ohhhh, sorry the 10.2/3 confused me, as it did with george, i thought you were writing it as a fraction, not a mixed number, but if you were meaning to say the answer's 102/3 then yeah, i guess my answer of 32/3 was right, woo! hah, okay i know george re-wrote it pretty clearly, but ill do so aswell just incase

    okay, so X-1/root X is the same as X-1/X1/2, and the X1/2 can be bought up and made into X-1/2.

    Therfore, X-1/root X = X-1/2 (X-1), now you work out this multiplication to get; X1/2 - X-1/2

    Now you integrate the equation; X1/2 - X-1/2 dx

    Which is ( X3/2 / 3/2 ) - ( X1/2 / 1/2 )

    Now simply substitute 9 into the now integrated equation and work it out, then do the same with 4. Then minus the value you found while substituting 4 from the value you found from substituting 9. And voila! you get 32/3, which can also be writen as 102/3
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    (Original post by just george)
     \frac{1}{x^\frac{1}{2}} = x^\frac{1}{2}-x^-^\frac{1}{2}

    integrates to  \frac{2x^\frac{3}{2}}{3} - 2x^\frac{1}{2}

    substituting in 9 and 4:

     [\frac{2(9)^\frac{3}{2}}{3} - 2(9)^\frac{1}{2}] - [\frac{2(4)^\frac{3}{2}}{3} - 2(4)^\frac{1}{2}]

     = [18 - 6] - [\frac{16}{3} - 4] = 12 - \frac{4}{3} = \frac{32}{3}

    if thats the right answer, then it is because you integrated x-\frac{1}{x^\frac{1}{2}} instead of  \frac{x-1}{x^\frac{1}{2}}


    edit: and writing 10.2/3 confused me.. that looks like  \frac{10.2}{3} whereas i think your meaning to say  10\frac{2}{3} ?
    (Original post by J.G.M)
    ..


    x - \frac{1}{x^\frac{1}{2}} Is what is written in the book....
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    (Original post by King-Panther)
    x - \frac{1}{x^\frac{1}{2}} Is what is written in the book....
    Sounds like the book has had a mare and got it wrong then
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    (Original post by just george)
    Sounds like the book has had a mare and got it wrong then
    Good, so i'm not wrong ..
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    What would be the nth term for this sequence 3, 6, 11, 18, 27
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    What would be the nth term for this sequence 3, 6, 11, 18, 27

    (Original post by J.G.M)
    ....
    (Original post by just george)
    ...
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    n^2 +2
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    (Original post by J.G.M)
    ....
    (Original post by just george)
    ...
    (Original post by 0range)
    ...
    (Original post by f1mad)
    ...
    (Original post by raheem94)
    ...
    (Original post by LeeM1)
    n^2 +2

    Thanks, I don't recall being taught finding the nth term in class... Maybe you can help me with the next problem, how would I find the second derivative of something? (x-1/x)^2 for example...... Also, how would I know what to sub into 4x^3-6x^2+2x when finding the turning points?
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    (Original post by King-Panther)
    Thanks, I don't recall being taught finding the nth term in class... Maybe you can help me with the next problem, how would I find the second derivative of something? (x-1/x)^2 for example...... Also, how would I know what to sub into 4x^3-6x^2+2x when finding the turning points?
    Is this your question  \displaystyle \left(x -\frac1{x}\right)^2 ?

    For your second question,  \displaystyle f(x) = 4x^3-6x^2+2x
    Differentiate f(x), then make the dy/dx =0, because dy/dx is zero at turning points. So by solving it equal to zero you will get the x-coordinates. Now substitute the values you will find for 'x' in the f(x) expression to find the y-coordinates.

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Updated: April 5, 2012
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