# Help requiredWatch

This discussion is closed.
#1
Hi a friend and myself were going over this p1 question and after a while it became apparent that we had either messed up, or that it was more complicated than we presumed for p1.... i mean the numbers involved.

can anyone have a go.....

thanks!
0
13 years ago
#2
(Original post by melbourne)
Hi a friend and myself were going over this p1 question and after a while it became apparent that we had either messed up, or that it was more complicated than we presumed for p1.... i mean the numbers involved.

can anyone have a go.....

thanks!
First of all lets find the coordinates of A and B. The way we can find this is by equating the two curves together and finding the points where they agree.

6x - x^2 - 3 = x + 1

=> x^2 - 5x + 4 = 0
=> (x-4).(x-1) = 0
=> x = 4, 1

So A is at the point x = 1, and B is the point x = 4.

Now to find the area we must integrate the curve y = 6x - x^2 - 3 between these two points, and take it away from the area underneath the curve y = x + 1

So first integral:

INT [4,1] 6x - x^2 - 3 dx
=> 3x^2 - (x^3)/3 - 3x [4,1]
=> 48 - 64/3 - 12 - 3 + 1/3 + 3
=> 15

Now for the second integral:

INT [4,1] x + 1 dx
=> (x^2)/2 + x [4,1]
=> 8 + 4 - 1/2 - 1
=> 21/2

Hence the area of the shaded region is

15 - (21/2) = 4.5

Galois.
0
13 years ago
#3
you dont even need to integrate the trapesium, just use 1/2 the sum of the parallel sides = 21/2

Or you could have integrated all in one go, which is what i suspect you did, by combining the two equation and integrating between 3+rt6 and 3-rt6 on the x axis? this is tricky yes, so just stick to doing them separately...
0
#4
(Original post by xXxKatiexXx)
you dont even need to integrate the trapesium, just use 1/2 the sum of the parallel sides = 21/2

Or you could have integrated all in one go, which is what i suspect you did, by combining the two equation and integrating between 3+rt6 and 3-rt6 on the x axis? this is tricky yes, so just stick to doing them separately...
yeh thats what we tried to do and we had to work out -(3+rt6)^3 and rubbish like that lol
0
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