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    edit: I am fine with this now. thanks!
    Hello there, I am having problems with part d) of this Integration question:

    'a)Show that the normal to the curve y=x2+6x+5 at (-2,-3) have the equation 2y+x+8=0

    b) Find the coordinates of the point where the normal cuts the curve again.

    c) draw a sketch of the curve and line

    d) show that the area enclosed between the curve and line it (125)/(48)

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    What exactly do I need to integrate and what limits do i need to take? Thanks in advance.
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    Draw two verticle lines from the points of intersection to the x-axis, is it clear how to do it now?
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    (Original post by raiden95)
    Draw two verticle lines from the points of intersection to the x-axis, is it clear how to do it now?
    I was doing that but then if we take area that area and the larger area below the axis are we not left with more than just the area below the curve? This is where I am getting confused
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    (Original post by upthegunners)
    Hello there, I am having problems with part d) of this Integration question:

    'a)Show that the normal to the curve y=x2+6x+5 at (-2,-3) have the equation 2y+x+8=0

    b) Find the coordinates of the point where the normal cuts the curve again.

    c) draw a sketch of the curve and line

    d) show that the area enclosed between the curve and line it (125)/(48)

    Name:  sdqdqdqwdwqd.png
Views: 78
Size:  7.5 KB

    What exactly do I need to integrate and what limits do i need to take? Thanks in advance.
    I think you need to use the formula for finding the area between a curve and a line = y1 - y2

    Basically you do the equation of the line minus the equation of the curve : simplify and get a single equation , intergrate it

    then the limits you use are the two x-values !
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    (Original post by upthegunners)
    I was doing that but then if we take area that area and the larger area below the axis are we not left with more than just the area below the curve? This is where I am getting confused
    The area that is found is enclosed by the x axis and curve in integration. If your graph was upside down it would work the same way.
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    (Original post by upthegunners)
    I was doing that but then if we take area that area and the larger area below the axis are we not left with more than just the area below the curve? This is where I am getting confused
    Hm?

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    (Original post by Fas)
    I think you need to use the formula for finding the area between a curve and a line = y1 - y2

    Basically you do the equation of the line minus the equation of the curve : simplify and get a single equation , intergrate it

    then the limits you use are the two x-values !
    I was integrating the curve from -5 to -1 and integrating the line from -9/2 to -2

    then I was subtracting the values that I got by integrating. But some how I am not getting the answer they are wanting me to show
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    Find the area between the curve and the x-axis using integration, find the area between the line and the x-axis (forms a trapezium) and then minus. Upside down they work exactly the same way
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    (Original post by aznkid66)
    Hm?

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    I think i was taking the wrong limits when integrating the curve

    do we take the limits from -4.5 to -2 ?
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    Yup, you are to integrate both from -4.5 to 2, as that will ensure you don't have the stuff "more than just the area below" that you previously mentioned :P

    (Also, you can combine the two integrals I(u)-I(v) into I(u-v) since the terminals are the same.)
 
 
 
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