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    Attachment 616856Could domeone help me with this question and show me what to do step by step?
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    (Original post by Mf1999)
    Could domeone help me with this question and show me what to do step by step?
    9 a) drop a vertical line at the intersection point. Work out the area of the triangle and add on the area under the curve to the left.

    b) Drop 2 vertical lines at the intersection points and work out the area of the rectangle, then add on the areas of the left and right under the curve.

    Also another way to do these questions is to substitute y into the curves and make it equal 0. Integrate the remaining expression between the area bounds. Though this is only safe to do if there is no negative area -- which there isn't on these.
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    (Original post by RDKGames)
    9 a) drop a vertical line at the intersection point. Work out the area of the triangle and add on the area under the curve to the left.

    b) Drop 2 vertical lines at the intersection points and work out the area of the rectangle, then add on the areas of the left and right under the curve.

    Also another way to do these questions is to substitute y into the curves and make it equal 0. Integrate the remaining expression between the area bounds. Though this is only safe to do if there is no negative area -- which there isn't on these.
    I tried to do the 2nd method however I ended up with x^3+x-2 and I didn't kno where to go from there
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    (Original post by Mf1999)
    I tried to do the 2nd method however I ended up with x^3+x-2 and I didn't kno where to go from there
    Integrate between x=0 and x=a where a is where the line crosses the x-axis.
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    (Original post by RDKGames)
    Integrate between x=0 and x=a where a is where the line crosses the x-axis.
    I dont really understand what you mean
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    (Original post by Mf1999)
    I dont really understand what you mean
    The line crosses at x=2 so that will be your other bound. The reason why it's between x=0 and x=2 is because that's where the area you want to find is when you consider the x-axis.

    So you have \displaystyle \int_0^2 x^3+x-2 .dx and this will give the area.

    EDIT: Scratch that, I overlooked something. Anyway, always stick with the first method. Find the point of intersection between the line and curve. Drop a vertical line from the point of intersection the x-axis, this should split your area into a right-angled triangle an and area exclusively under the cubic. You can find the area exclusively under the cubic using integration of x^3, and you know how to find the area of a right-angled triangle, then just add them together.
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    (Original post by RDKGames)
    The line crosses at x=2 so that will be your other bound. The reason why it's between x=0 and x=2 is because that's where the area you want to find is when you consider the x-axis.

    So you have \displaystyle \int_0^2 x^3+x-2 .dx and this will give the area.

    EDIT: Scratch that, I overlooked something. Anyway, stick with the first method. Find the point of intersection between the line and curve. Drop a vertical line from the point of intersection the x-axis, this should split your area into a right-angled triangle and area exclusively under the cubic. You can find the area exclusively under the cubic using integration of x^3, and you know how to find the area of a right-angled triangle, then just add them together.
    Oh okay thank you
 
 
 
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