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Math Help - Random Integration Problems

  1. #1
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    Exclamation Random Integration Problems

    2. For the q. attached, why can't a = -1 too?

    3. Calculate the area bounded by the graphs of f(x) = x^2, g(x) = 1/x^2, x>0 and the line x =3. Could someone please verify the answer for me? The txtbk says it's 8 units squared but for some reason I only get 1 ><

    Thanx a lot.
    Attached Thumbnails Attached Thumbnails Random Integration Problems-untitled.bmp  
    Last edited by xwrathbringerx; September 25th 2009 at 10:27 PM.
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    1. Show, by integration, that the area of a unit square is divided into four equal parts by the curves y = x^3, y = sqrt(x) and y = x. Could someone show me the correct way to do this proof?

    2. For the q. attached, why can't a = -1 too?

    Thanx a lot.
    2. is easy.

    \int_{-1}^a{x\,dx} = 0

    \left[\frac{1}{2}x^2\right]_{-1}^a = 0

    \frac{1}{2}[a - (-1)] = 0

    \frac{1}{2}(a + 1) = 0

    a + 1 = 0

    a = -1.
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  3. #3
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    Hmmm I get both 1 AND - 1. The txtbk answer says it's only 1 though...
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  4. #4
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    Corrections in red:

    Quote Originally Posted by Prove It View Post
    2. is easy.

    \int_{-1}^a{x\,dx} = 0

    \left[\frac{1}{2}x^2\right]_{-1}^a = 0

    \frac{1}{2} [a{\color{red}^2} - (-1){\color{red}^2}] = 0

    \frac{1}{2}(a{\color{red}^2 - } 1) = 0

    a{\color{red}^2 -} 1 = 0

    a = {\color{red}\pm} 1.
    @OP: a = -1 is a perfectly good (if trivial) solution. Perhaps the book only wanted NON-trivial solutions.
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    Quote Originally Posted by mr fantastic View Post
    Corrections in red:


    @OP: a = -1 is a perfectly good (if trivial) solution. Perhaps the book only wanted NON-trivial solutions.
    OMG my solutions have been shocking today... I must need sleep...

    Thanks Mr F.
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  6. #6
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    Quote Originally Posted by xwrathbringerx View Post
    [snip]
    3. Calculate the area bounded by the graphs of f(x) = x^2, g(x) = 1/x^2, x>0 and the line x =3. Could someone please verify the answer for me? The txtbk says it's 8 units squared but for some reason I only get 1 ><

    Thanx a lot.
    Area = \int_1^3 x^2 - \frac{1}{x^2} \, dx = 8.
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  7. #7
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    Ummm for q. 3, is this the diagram for the area or is it the shaded area onli ABOVE the x-axis?
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  8. #8
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    Quote Originally Posted by xwrathbringerx View Post
    Ummm for q. 3, is this the diagram for the area or is it the shaded area onli ABOVE the x-axis?
    No. Your graphs are wrong. Try plotting points. Note that:

    f(1) = g(1) = 1.

    f(3) = 9 and g(3) = 1/9.
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