IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: elemis on October 10, 2010, 02:37:09 pm

Title: P1 Help Requested
Post by: elemis 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 - $\frac{28}{3}$ = $10 \frac{2}{3}$
Title: Re: P1 Help Requested
Post by: Deadly_king on October 10, 2010, 02:45:47 pm: Quote from: Ari Ben Canaan 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 - $\frac{28}{3}$ = $10 \frac{2}{3}$

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

Can you post the whole question ???
Title: Re: P1 Help Requested
Post by: elemis on October 10, 2010, 02:52:57 pm: Quote from: Deadly_king on October 10, 2010, 02:45:47 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 ;)
Title: Re: P1 Help Requested
Post by: Deadly_king on October 10, 2010, 02:55:00 pm: Quote from: Ari Ben Canaan on October 10, 2010, 02:52:57 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 ;)
Title: Re: P1 Help Requested
Post by: elemis on October 10, 2010, 02:58:57 pm: Quote from: Deadly_king on October 10, 2010, 02:55:00 pm
Oopz.....sorry br0 ;)

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

A long one too ,eh ? ;D
Title: Re: P1 Help Requested
Post by: Deadly_king on October 10, 2010, 03:22:16 pm: Quote from: Ari Ben Canaan on October 10, 2010, 02:58:57 pm
A long one too ,eh ? ;D

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

But the great Ari can handle anything ;)
Title: Re: P1 Help Requested
Post by: thecandydoll on October 10, 2010, 03:30:02 pm: silly,but the area under the line same as area of trap?
Title: Re: P1 Help Requested
Post by: elemis on October 10, 2010, 03:40:07 pm: Quote from: thecandydoll on October 10, 2010, 03:30:02 pm
silly,but the area under the line same as area of trap?

Yes, in the set limits.