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# for those who want P2 practice watch

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1. i get 11.3 for a
b was easy
2. I don't get why between 5 and 1.
because its the area under the curve
3. (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?
4. 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
5. area underthe curve is (5*2) - R
6. 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
7. (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)
8. (Original post by Mathemagician)
.

Where do you get "20pi-pi" from???? (in your 1st line)
it says 20pi-pi * integral
9. (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?
I think that "R" was labelled wrong
10. guys 1 p2 question:

why is 4^x = (2^x)^2?
11. (Original post by lgs98jonee)
it says 20pi-pi * integral
yes, but where did you get it?
12. (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
13. (Original post by Mathemagician)
4^x = (2^x)^2
LHS = 2^2x = (2^2)^x = 4^x
ta
14. (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
15. (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 ... thanks

but why do you multiply the area by the function^2?? to get volume?
16. (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 ... 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
17. (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 . Now, I know its seperate! thank you!
18. 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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