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    How do I find underestimates and overestimates under the curve for the interval 0<=x<=4 with rectangles of width 1 for y=4x-x^2. I thought the underestimate would be 1((4(0)-(0)^2)+...(4(3)-(3)^2) and the overestimate would be
    1((4(1)-(1)^2)+...(4(4)-(4)^2) like all the other rectangle approximation question I have come across in fp2 but this on this one I am incorrect and don't see why??
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    (Original post by Mathematicus65)
    How do I find underestimates and overestimates under the curve for the interval 0<=x<=4 with rectangles of width 1 for y=4x-x^2. I thought the underestimate would be 1((4(0)-(0)^2)+...(4(3)-(3)^2) and the overestimate would be
    1((4(1)-(1)^2)+...(4(4)-(4)^2) like all the other rectangle approximation question I have come across in fp2 but this on this one I am incorrect and don't see why??
    Post the question itself, either a picture or a link to it.
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    (Original post by Mathematicus65)
    on this one I am incorrect and don't see why??
    You may find it illuminating to plot the curve, and the points/rectangles.
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    (Original post by Mathematicus65)
    How do I find underestimates and overestimates under the curve for the interval 0<=x<=4 with rectangles of width 1 for y=4x-x^2. I thought the underestimate would be 1((4(0)-(0)^2)+...(4(3)-(3)^2) and the overestimate would be
    1((4(1)-(1)^2)+...(4(4)-(4)^2) like all the other rectangle approximation question I have come across in fp2 but this on this one I am incorrect and don't see why??
    As usual ghostwalker is right; you may find it helpful to note that

    4x - x^2 = 4 - (x-2)^2.
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    (Original post by ghostwalker)
    You may find it illuminating to plot the curve, and the points/rectangles.
    I have sketched the curve as follows :
    Name:  image.jpg
Views: 121
Size:  445.1 KB
    I am confused how to draw the rectangles for this? - (haven't done a question like this that has a negative parabola I have only done increasing curves such as the graph of 2^x)
    I also am not seeing anything from this..?
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    (Original post by ghostwalker)
    You may find it illuminating to plot the curve, and the points/rectangles.
    Oh I get it!!!! I drew the rectangles like this for the underestimate Name:  image.jpg
Views: 114
Size:  467.4 KB
    And got (3x1)+(3x1) = 6 which is right.
    Thank you, SO MUCH more simple with a sketch
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    (Original post by Mathematicus65)
    Oh I get it!!!! I drew the rectangles like this for the underestimate Name:  image.jpg
Views: 114
Size:  467.4 KB
    And got (3x1)+(3x1) = 6 which is right.
    Good. I presume you're OK with the overestimate as well now.
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    (Original post by ghostwalker)
    Good. I presume you're OK with the overestimate as well now.
    Yeah i am! Thank you!
 
 
 
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