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    Find the shaded area .
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    equation of the curve =  y=\frac{1}{x^{2}}
    therefore  x=y^{-\frac{1}{2}}

    books's answer = 4
    my solution

     A = \int_{1}^{4}y^{-\frac{1}{2}}
     = [\frac{y^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}] lower limit = 1, upper limit= 4
     = [\frac{y^{\frac{1}{2}}}{\frac{1}{  2}}] lower limit = 1, upper limit= 4
     = [2y\frac{1}{2}] lower limit = 1, upper limit= 4

     = (2  \sqrt{4}) - (2\sqrt{1})= 4-2 = 2

    The book says that the answer is 4 however i got 2
    I dont understand if the book is wrong
    i only got 4 if the lower limit is changed to 0
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    (Original post by bigmansouf)
    Question

    Find the shaded area .
    Name:  q1.png
Views: 47
Size:  25.9 KB


    books's answer = 4
    my solution

     A = \int_{1}^{4}y^{-\frac{1}{2}}
     = [\frac{y^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}] lower limit = 1, upper limit= 4
     = [\frac{y^{\frac{1}{2}}}{\frac{1}{  2}}] lower limit = 1, upper limit= 4
     = [2y\frac{1}{2}] lower limit = 1, upper limit= 4

     = (2  \sqrt{4}) - (2\sqrt{1})= 4-2 = 2

    The book says that the answer is 4 however i got 2
    I dont understand if the book is wrong
    i only got 4 if the lower limit is changed to 0
    What is the equation of the curve?

    Edit: I assume  x = +/-  y^{\frac{-1}{2}}  or y = \frac{1}{x^2}?

    If so then your integral is only accounting for the area on one side of the y-axis so you would have to double it.
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    jeez that looks boring.

    Hope i helped
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    (Original post by 16Characters....)
    What is the equation of the curve?

    Edit: I assume  x = +/-  y^{\frac{-1}{2}} ?

    If so then your integral is only accounting for the area on one side of the y-axis so you would have to double it.
    sorry i edited it thank you

    equation  x=y^{-\frac{1}{2}}
     x =  y^{\frac{-1}{2}}
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    (Original post by bigmansouf)
    sorry i edited it thank you

    equation  x=y^{-\frac{1}{2}}
     x =  y^{\frac{-1}{2}}
    The issue is that if  y = \frac{1}{x^2} then  x = +/- y^{\frac{-1}{2}} so when you find

     \displaystyle \int_1^4 y^{\frac{-1}{2}} dy you are actually finding only the area between 1 of the two parts of the curve and the y-axis as opposed to the two different parts of the curve. Not to worry however as symmetry comes to your rescue; the area between the two parts of the curve is double the area between one part of the curve and the y-axis.
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    (Original post by 16Characters....)
    The issue is that if  y = \frac{1}{x^2} then  x = +/- y^{\frac{-1}{2}} so when you find

     \displaystyle \int_1^4 y^{\frac{-1}{2}} dy you are actually finding only the area between 1 of the two parts of the curve and the y-axis as opposed to the two different parts of the curve. Not to worry however as symmetry comes to your rescue; the area between the two parts of the curve is double the area between one part of the curve and the y-axis.

    thanks since it is two curves thanks
 
 
 
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