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2002- Winter Diet - Paper II - Q11 - HSN forum

# 2002- Winter Diet - Paper II - Q11

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### #1herbeey

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Posted 19 May 2005 - 11:47 AM

I've been looking around threads on this site looking for an explanation of how to do this, but I couldn't find anything that explained it for me.

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!
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### #2Mr H

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Posted 19 May 2005 - 12:42 PM

This is quite a tough question but taken a step at a time is the same as the easy ones.

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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### #3herbeey

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Posted 19 May 2005 - 12:50 PM

Ah, I am seeing how to do it...
It shows the seperation of f(x) and g(x) very badly...
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### #4gary

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Posted 19 May 2005 - 02:09 PM

Can someone please write down how you would start this I am confused thanks.

### #5dfx

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Posted 19 May 2005 - 04:19 PM

A = Integration of (f (x) - 6) dx + Integration of (g(x) - (-6)) dx

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 ?

### #6Ally

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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.

### #7dfx

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Posted 19 May 2005 - 06:51 PM

Thats what I thought too since it logically made sense.

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?

### #8Dave

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Posted 19 May 2005 - 06:53 PM

if they dont meet then there is no enclosed area to measure or am i missing something

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### #9dfx

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Posted 19 May 2005 - 06:54 PM

That's what I thought too, there's no area enclosed. Lemme post a diagram..

### #10dfx

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Posted 19 May 2005 - 06:59 PM

Right, so I approached it as Ally did by integrating between 5 and 0, and then subtracting the area from the rectangle (6 x 5 = 30).

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) ...?

### #11Dave

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Posted 19 May 2005 - 07:02 PM

if you were asked to find the area between though two lines in the parameters shown then yes it would be fine

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### #12gary

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Posted 19 May 2005 - 07:03 PM

I think this question has gone over many peoples heads it is so difficult.

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