

2002- Winter Diet - Paper II - Q11
#1
Posted 19 May 2005 - 11:47 AM
How do I get started in this question? I'm able to do the integration of a single equation, and beyond. But before that, I'm stumped.
There are sooooo many lines! What to do with them all!
#2
Posted 19 May 2005 - 12:42 PM
Basically you have two 'area between two curves' calculations added together with some fractions to make things harder for you!
Find the area between f(x) and y=-6 by integrating f(x) - (-6) repeat with g(x) and y=-6 and add your answers together should give the right answer.
Doing it twice with fractions is why it is 7 marks.
H tends 2 infinity
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#3
Posted 19 May 2005 - 12:50 PM
It shows the seperation of f(x) and g(x) very badly...
Cheers again!
#4
Posted 19 May 2005 - 02:09 PM
#5
Posted 19 May 2005 - 04:19 PM
You know the limits...
Out of curiosity, CAN you do the second bit integration (g(x) - (-6))dx EVEN IF g(x) and the line y = -6 never intersect? I actually did the area between 5 and 0 and then subtracted it from the total area of the whole rectangle which is 6 x 5 = 30 ?
#6
Posted 19 May 2005 - 06:46 PM

That's how I done it; make sure once you integrate it you plug the numbers correctly into your calculator.
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#7
Posted 19 May 2005 - 06:51 PM
But, if you look at the way they've done it in the marking scheme. Can you integrate for the area between a curve and line that dont meet? Like in the case of this question the line and curve never intersect - for the second bit that is, g(x) and y = -6. Or doesn't that matter?
#8
Posted 19 May 2005 - 06:53 PM
If i am not here i am somewhere else
#9
Posted 19 May 2005 - 06:54 PM
#10
Posted 19 May 2005 - 06:59 PM
However, technically, when you take this approach above, you're basically integrating between A = Integral of g(x) - 0 , right? cause y = 0 is the x-axis. That's perfectly fine cause the x-axis intersects the curve. But what about in this case where it doesn't, could you similarly do A = Integral of g(x) - (-6) ...?
#11
Posted 19 May 2005 - 07:02 PM
If i am not here i am somewhere else
#12
Posted 19 May 2005 - 07:03 PM
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