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for those who want P2 practice watch

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    i get 11.3 for a
    b was easy
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    I don't get why between 5 and 1.
    because its the area under the curve
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    (Original post by RussianDude)
    because ts the area under the curve
    I don't understand. What is the area under the curve? What is ts and isn't it meant to be the volume? :confused:
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    I don't understand. What is the area under the curve? What is ts and isn't it meant to be the volume?
    the curve crosses x axis at y=0, so if you substitute 0 in the equaton u get

    0=x-1
    so x = 1

    It is the area, but if this area will be rotated it will form a volume
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    area underthe curve is (5*2) - R
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    yeah i understand now....
    cos R+area under curve (between 5 and 1)=10
    so R=10-area undercurve
    =20pi-(x^2/2-x)
    =20pi-(12.5-.5-5+1)pi
    =12pi
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    (Original post by lgs98jonee)
    a. volume= 20pi-pi*int(x-1)[5 and 1]
    =20pi-pi(x^2/2-x)
    =20pi-pi(12.5-5-0.5+1)
    =20pi-8pi
    =12pi
    .


    Where do you get "20pi-pi" from???? (in your 1st line):confused:
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    (Original post by Mathemagician)
    .


    Where do you get "20pi-pi" from???? (in your 1st line):confused:
    it says 20pi-pi * integral
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    (Original post by mik1a)
    I don't understand. What is the area under the curve? What is ts and isn't it meant to be the volume? :confused:
    I think that "R" was labelled wrong
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    guys 1 p2 question:

    why is 4^x = (2^x)^2?
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    (Original post by lgs98jonee)
    it says 20pi-pi * integral
    yes, but where did you get it?
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    (Original post by TheWolf)
    guys 1 p2 question:

    why is 4^x = (2^x)^2?
    4^x = (2^x)^2
    LHS = 2^2x = (2^2)^x = 4^x
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    (Original post by Mathemagician)
    4^x = (2^x)^2
    LHS = 2^2x = (2^2)^x = 4^x
    ta
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    (Original post by Mathemagician)
    yes, but where did you get it?
    well because R=the rectangle from y=2 and x=5 to the origin - area under curve between 1 and 5
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    (Original post by lgs98jonee)
    well because R=the rectangle from y=2 and x=5 to the origin - area under curve between 1 and 5
    oh! i see now. lol, i thought that it was supposed to be under the curve...but that would lead to an infinite area :confused: ... thanks

    but why do you multiply the area by the function^2?? to get volume?
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    (Original post by Mathemagician)
    oh! i see now. lol, i thought that it was supposed to be under the curve...but that would lead to an infinite area :confused: ... thanks

    but why do you multiply the area by the function^2?? to get volume?
    well volume = pi * ∫ y² dx
    the line is y=2
    so y²=4

    4*5=20pi
    this is the volume of the rectangular bit to which you subtract the volume: 20pi-8pi=12pi
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    (Original post by hihihihi)
    well volume = pi * ∫ y² dx
    the line is y=2
    so y²=4

    4*5=20pi
    this is the volume of the rectangular bit to which you subtract the volume: 20pi-8pi=12pi
    oh I see! Thank you very much... before, i was trying to solve them together somehow :confused: . Now, I know its seperate! thank you!
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    man, i'm confused. ok, so y=2 and when y=2, x=5 BUT this cant be a limit can it? cus i mean, x=5 isnt where area R starts/finishes, whatever
 
 
 
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