I noticed there was an earlier thread on this question but i don't even no how to begin it!

Could someone possibly put up a solution for the full question please!

Thanks very much!

**0**

# 2004 p2 Q11

Started by eck231, May 08 2005 02:03 PM

2 replies to this topic

### #1

Posted 08 May 2005 - 02:03 PM

### #2

Posted 08 May 2005 - 03:26 PM

I was going to start typing but here's a link to someone who has done that already!

http://www.mathsrevision.com/index_files/M...2004_paper2.pdf

http://www.mathsrevision.com/index_files/M...2004_paper2.pdf

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 08 May 2005 - 03:37 PM

I already started to type

To get the limits, find where y=2x-1/2x^2 and y=1.5 collide.

2x-1/2x^2 = 1.5

4x-x^2 = 3

-x^2 + 4x - 3 = 0

(-x+3)(x-1)=0

Limits: x=3, x=1

then integrate y=2x-1/2x^2

and get this:

[x^2 - x^3/6] upper limit: 3, lower: 1

area = (9 - 27/6) - (1-1/6)

area = 3+2/3

then you need to subtract the area of the box underneath the segment:

area of box = 2*1.5 = 3

so area of segment = 3+2/3 - 3 = 2/3 metres^2.

To get the limits, find where y=2x-1/2x^2 and y=1.5 collide.

2x-1/2x^2 = 1.5

4x-x^2 = 3

-x^2 + 4x - 3 = 0

(-x+3)(x-1)=0

Limits: x=3, x=1

then integrate y=2x-1/2x^2

and get this:

[x^2 - x^3/6] upper limit: 3, lower: 1

area = (9 - 27/6) - (1-1/6)

area = 3+2/3

then you need to subtract the area of the box underneath the segment:

area of box = 2*1.5 = 3

so area of segment = 3+2/3 - 3 = 2/3 metres^2.

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