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FP2 Rectangle Approximation

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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Zacken
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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.
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.
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.
Original post by ghostwalker
You may find it illuminating to plot the curve, and the points/rectangles.


I have sketched the curve as follows :
image.jpg
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..?
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 image.jpg
And got (3x1)+(3x1) = 6 which is right.
Thank you, SO MUCH more simple with a sketch :biggrin:
Original post by Mathematicus65
Oh I get it!!!! I drew the rectangles like this for the underestimate image.jpg
And got (3x1)+(3x1) = 6 which is right.


Good. I presume you're OK with the overestimate as well now.
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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