Author Topic: P1 Help Requested  (Read 1433 times)

elemis

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P1 Help Requested
« on: October 10, 2010, 02:37:09 pm »
Okay looking at the diagram below, notice the red line. It is the shape of a trapezium.

You can find the area under the curve and from this value MINUS the area under the line.

First I found the two points at which the curve and line intersect :

x2 - 4x +4 = 7-2x

Hence, x2-2x -3 =0

Solving for x gives : 3    and  -1

Inputting the above x values into 7-2x = y  give corresponding y coordinates of : 1 and 9

Finding the area under the line :

\frac{1}{2}(1+9)*4=20

Next I integrate the curve to get :

\frac{(x-2)^3}{3}

Inputting 3 into above gives a value of 1/3 .

Inputting -1 into abov gives : -9

Hence 1/3 - - 9 = \frac{28}{3}

Hence, 20 - \frac{28}{3} = 10 \frac{2}{3}
« Last Edit: October 10, 2010, 02:52:21 pm by Ari Ben Canaan »

Offline Deadly_king

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Re: P1 Help Requested
« Reply #1 on: October 10, 2010, 02:45:47 pm »
Okay looking at the diagram below, notice the red line. It is the shape of a trapezium.

You can find the area under the curve and from this value MINUS the area under the line.

First I found the two points at which the curve and line intersect :

x2 - 4x +4 = 7-2x

Hence, x2-2x -3 =0

Solving for x gives : 3    and  -1

Inputting the above x values into 7-2x = y  give corresponding y coordinates of : 1 and 9

Finding the area under the line :

\frac{1}{2}(1+9)*4=20

Next I integrate the curve to get :

\frac{(x-2)^3}{3}

Inputting 3 into above gives a value of 1/3 .

Inputting -1 into abov gives : -9

Hence 1/3 - - 9 = \frac{28}{3}

Hence, 20 - \frac{28}{3} = 10 \frac{2}{3}

Sorry........but can't see any diagram below.

Can you post the whole question ???

elemis

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Re: P1 Help Requested
« Reply #2 on: October 10, 2010, 02:52:57 pm »
Sorry........but can't see any diagram below.

Can you post the whole question ???

A member requested some help regarding this question via PM.

This is my explanation to help that member ;)

Offline Deadly_king

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Re: P1 Help Requested
« Reply #3 on: October 10, 2010, 02:55:00 pm »
A member requested some help regarding this question via PM.

This is my explanation to help that member ;)

Oopz.....sorry br0 ;)

That's why I was wondering.....it looked more like an explanation ;)

elemis

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Re: P1 Help Requested
« Reply #4 on: October 10, 2010, 02:58:57 pm »
Oopz.....sorry br0 ;)

That's why I was wondering.....it looked more like an explanation ;)

A long one too ,eh ?  ;D

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Re: P1 Help Requested
« Reply #5 on: October 10, 2010, 03:22:16 pm »
A long one too ,eh ?  ;D

Yeah........a very long one.

But the great Ari can handle anything ;)

Offline thecandydoll

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Re: P1 Help Requested
« Reply #6 on: October 10, 2010, 03:30:02 pm »
silly,but the area under the line same as area of trap?

elemis

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Re: P1 Help Requested
« Reply #7 on: October 10, 2010, 03:40:07 pm »
silly,but the area under the line same as area of trap?

Yes, in the set limits.