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    My question;

    For the curve and straight lines, find their points of intersection, draw a sketch on the same axes and find the area of segment cut off from each curve by the corresponding straight line.

    a)  y=(x+1(x-2), x-y+1=0

    sOLUTION
    points of coordinates (-1,0) (3,4)

    Area required = Area of trianagle -  \int_{3}^{2} x^{2}-x-2
    =  (\frac{1}{2}\times 4 \times4 )- \int_{3}^{2} x^{2}-x-2
    =  8 - \int_{3}^{2} x^{2}-x-2
    break into two
     \int_{2}^{3} x^{2}-x-2 = [\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x]_{2}^{3}

     =(\frac{3^{3}}{3}-\frac{3^{2}}{2}-2(3))-(\frac{2^{3}}{3}-\frac{2^{2}}{2}-2(2))
      = -\frac{3}{2} - (-\frac{10}{3}) = \frac{11}{6}

    bring it together
    area required =
     8-\frac{11}{6} = \frac{37}{6}

    please help me and point out what i did wrong
    the book says the answer is  \frac{32}{3}
    please help me understand where i went wrong
    i have attached my sketch Name:  q2.png
Views: 92
Size:  28.9 KB
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    (Original post by bigmansouf)
    My question;

    For the curve and straight lines, find their points of intersection, draw a sketch on the same axes and find the area of segment cut off from each curve by the corresponding straight line.

    a)  y=(x+1(x-2), x-y+1=0

    sOLUTION
    points of coordinates (-1,0) (3,4)

    Area required = Area of trianagle -  \int_{3}^{2} x^{2}-x-2
    =  (\frac{1}{2}\times 4 \times4 )- \int_{3}^{2} x^{2}-x-2
    =  8 - \int_{3}^{2} x^{2}-x-2
    break into two
     \int_{2}^{3} x^{2}-x-2 = [\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x]_{2}^{3}

     =(\frac{3^{3}}{3}-\frac{3^{2}}{2}-2(3))-(\frac{2^{3}}{3}-\frac{2^{2}}{2}-2(2))
      = -\frac{3}{2} - (-\frac{10}{3}) = \frac{11}{6}

    bring it together
    area required =
     8-\frac{11}{6} = \frac{37}{6}

    please help me and point out what i did wrong
    the book says the answer is  \frac{32}{3}
    please help me understand where i went wrong
    i have attached my sketch Name:  q2.png
Views: 92
Size:  28.9 KB
    I also got 32/3 by integrating x+1 -(x+1)(x-2) between -1 and 3
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    (Original post by bigmansouf)
    My question;

    For the curve and straight lines, find their points of intersection, draw a sketch on the same axes and find the area of segment cut off from each curve by the corresponding straight line.

    a)  y=(x+1(x-2), x-y+1=0

    sOLUTION
    points of coordinates (-1,0) (3,4)

    Area required = Area of trianagle -  \int_{3}^{2} x^{2}-x-2
    =  (\frac{1}{2}\times 4 \times4 )- \int_{3}^{2} x^{2}-x-2
    =  8 - \int_{3}^{2} x^{2}-x-2
    break into two
     \int_{2}^{3} x^{2}-x-2 = [\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x]_{2}^{3}

     =(\frac{3^{3}}{3}-\frac{3^{2}}{2}-2(3))-(\frac{2^{3}}{3}-\frac{2^{2}}{2}-2(2))
      = -\frac{3}{2} - (-\frac{10}{3}) = \frac{11}{6}

    bring it together
    area required =
     8-\frac{11}{6} = \frac{37}{6}

    please help me and point out what i did wrong
    the book says the answer is  \frac{32}{3}
    please help me understand where i went wrong
    i have attached my sketch Name:  q2.png
Views: 92
Size:  28.9 KB
    what area are you finding?
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    (Original post by TeeEm)
    what area are you finding?
    i assume that the area i am asked to find is the one coloured in the picName:  q3.png
Views: 69
Size:  33.9 KB

    thanks for the help
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    (Original post by bigmansouf)
    i assume that the area i am asked to find is the one coloured in the picName:  q3.png
Views: 69
Size:  33.9 KB

    thanks for the help
    I think it is that plus the bit under the x axis, bounded by the curve.
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    (Original post by TeeEm)
    I think it is that plus the bit under the x axis, bounded by the curve.
    in that sense would you say that my method was wrong
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    (Original post by TeeEm)
    I also got 32/3 by integrating x+1 -(x+1)(x-2) between -1 and 3
    thanks found this answer it seems that i was wrong
    i was not looking for the right area

    i think if it was suppose to be the area that i thought it would have said the curve, straight line and x-axis

    thank i understand where i was going wrong
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    (Original post by bigmansouf)
    in that sense would you say that my method was wrong
    no
    you just need to add the bit under

    integrate the curve between -1 and 2
    turn the answer positive and add it to 37/6 and see what you get
    (I did it differently)
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    (Original post by TeeEm)
    no
    you just need to add the bit under

    integrate the curve between -1 and 2
    turn the answer positive and add it to 37/6 and see what you get
    (I did it differently)
    your way would be the best and simplest method but i decide to apply the alternative of using areas of shapes and then subtracting it i thought it would be easier
    in most cases it is but not in this one
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    (Original post by bigmansouf)
    your way would be the best and simplest method but i decide to apply the alternative of using areas of shapes and then subtracting it i thought it would be easier
    in most cases it is but not in this one
    I am advising you to stick to your method
 
 
 
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