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    Question:

    Find the areas enclosed by the curve and the straight lines
    a)  Y=\frac{1}{x^{2}}-1
     y=-1  x=\frac{1}{2}  x=2



    My answer
    Are required =  \int_{\frac{1}{2}}^{1} (x^{-2}-1)dx + + Area of rectangle
    First find  \int_{\frac{1}{2}}^{1} (x^{-2}-1)dx

    = \int_{\frac{1}{2}}^{1} (x^{-2}-1)dx
    =  [-x^{-1}-x]_{\frac{1}{2}}^{1}
    = (-1-1)-(-\frac{1}{\frac{1}{2}}-0.5) = 0.5

    Area of rectangle

    = width= 2-0.5 = \frac{3}{2}
    Area =  \frac{3}{2} \times 1 = \frac{3}{2}

    therefore area required =  \frac{3}{2} +0.5 =2
    However I was wrong the book gave an answer of  \frac{3}{2}

    Please help me understand where I went wrong

    Thank you very much
    Attached Images
      
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    Very Important Poster
    (Original post by bigmansouf)
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    Think carefully about the area of the rectangle and what the question requires - you're almost there though
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    (Original post by SeanFM)
    Think carefully about the area of the rectangle and what the question requires - you're almost there though
    pleasecan i ask for a hiny
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    (Original post by bigmansouf)
    pleasecan i ask for a hiny
    I'll say the same as my last hint but a bit more. Look at the green shaded area in one of your attachments - you have more of the rectangle than you need.
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    as SEAN pointed out you have included a non-enclosed bit of area...
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    (Original post by the bear)
    as SEAN pointed out you have included a non-enclosed bit of area...
    I have tried again to find it but this time I calculated the area from coordinates x=0.5 to x=1 and then add the intergal vaule for   \int_{\frac{1}{2}}^{1} (x^{-2}-1)dx  and add   \int_{1}^{2} (x^{-2}-1)dx
    I got a result of 7/4
    I have attached the area I calculated as a picture
    Please point in the right of the direction this question is quite confusing
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    to find the area between two curves ( one of which may be a straight line such as y = -1 ) you just find ∫ ( top curve - bottom curve ) dx

    here ∫ ( 1/x2 - 1 - ( -1 )) dx

    with limits 1/2 and 2

    ANOTHER WAY

    if you imagine the curves translated upwards by 1 you would find the area by integrating 1/x2 - 1 + 1 or just 1/x2
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    (Original post by the bear)
    to find the area between two curves ( one of which may be a straight line such as y = -1 ) you just find ∫ ( top curve - bottom curve ) dx

    here ∫ ( 1/x2 - 1 - ( -1 )) dx

    with limits 1/2 and 2

    ANOTHER WAY

    if you imagine the curves translated upwards by 1 you would find the area by integrating 1/x2 - 1 + 1 or just 1/x2
    thank yuo very much i mistakenly got x=1/2 and x=-2 to be the straight lines
 
 
 
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