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# 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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Never argue with an idiot. They drag you down to their level then beat you with experience.

### #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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