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C2 intergration between a curve and a line

SophieL1996

Here is the question from C2 edexcel maths:Find the area of the finite region bounded by the curve with equation y=(1-x)(x+3) and the line (x+3)

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Reply 1

m4ths/maths247

So you have a parabola opening up downwards and a line....
You have a couple of choices, mine would be to take the line from the quadratic, integrate and evaluate.

Reply 2

Indeterminate

Original post by SophieL1996

Here is the question from C2 edexcel maths:Find the area of the finite region bounded by the curve with equation y=(1-x)(x+3) and the line (x+3)

If you draw both on the same axes, it should be obvious what you need to do to find the area between them.

Reply 3

Rainingshame

Original post by SophieL1996

Here is the question from C2 edexcel maths:Find the area of the finite region bounded by the curve with equation y=(1-x)(x+3) and the line (x+3)

A) Draw a sketch and shade the area you're trying to find.
B) Find the two points of intersection. This will give you the two points to integrate between. Integrate the quadratic between those two points.
C) Find the area of the rectangle/triangle/trapezium formed under the line y=x+3 .
D) This step is dependant on your sketch but it should become obvious which area to take away from which.

Reply 4

SophieL1996

How would you draw y=x+3 on the graph ? As I am not too sure.

Reply 5

SophieL1996

I have the points of intersection as (0.3) and (-3,0). Surely you would take the area under the curve- line??

Reply 6

SophieL1996

Here is the question from C2 edexcel maths:Find the area of the finite region bounded by the curve with equation y=(1-x)(x+3) and the line (x+3)

Reply 7

Mr M

Study Forum Helper

So what have you tried?

Reply 8

Indeterminate

Original post by SophieL1996

How would you draw y=x+3 on the graph ? As I am not too sure.

Notice that y=x goes through points such as (1,1), (2,2), etc, but y=x+3 goes through points such as (1,4), (2,5) etc. So it's shifted 3 up.

(1-x)(x+3) is an n shaped curve going through the x axis at the points where x=1 and x=-3 (the x axis is the line y=0).

You should know all this from C1/GCSE.

(edited 11 years ago)

Reply 9

SophieL1996

Original post by Indeterminate

Notice that y=x goes through points such as (1,1), (2,2), etc, but y=x+3 goes through points such as (1,4), (2,2) etc. So it's shifted 3 up.

(1-x)(x+3) is an n shaped curve going through the x axis at the points where x=1 and x=-3 (the x axis is the line y=0).

You should know all this from C1/GCSE.

Thanks I now remember :smile:

But I have drawn the graph and the points of intersection do not match up so I am confused.

Reply 10

SophieL1996

Original post by Mr M

So what have you tried?

Well firstly I found the points of intersection. I have a rough drawing and I thought that to fine the area you would take away the curve-line.
So I intergrated (x^2-2x+3)-(x+3) to 1/2 x^2-1/3x^3+x^2 in which I put the limits 0 and -3. I took the answer with -3 plugged in away from the 0 and got -45/2... The answer should be 4.5 ... :smile:

Reply 11

SophieL1996

Any ideas?? I am really stuck :l

Reply 12

davros

Original post by SophieL1996

Any ideas?? I am really stuck :l

check your integration carefully!

I did it slightly differently by working out the area between the curve and the x-axis and then subtracting the area between the line y=x+3 and the x-axis.

The latter part is just a triangle with area 0.5 x 3 x 3, and the area under the curve is given by the integral from -3 to 0 of:

3 - 2x - x^2

Reply 13

SophieL1996

Ok I will check that, thanks!

Reply 14

SophieL1996

Your method does not give the correct answer of 4.5 ...

Reply 15

davros

Original post by SophieL1996

Your method does not give the correct answer of 4.5 ...

Was that a reply to me - you really need to quote me so I see it :biggrin:

well it gave the right answer this morning when I did it!!

You should have found from a quick sketch that the curve and line intersect at (-3, 0) and (0, 3). The curve lies above the line in that region, so the area you want is (the area between the curve and the axis) - (the area between the line and the axis).

You can either subtract the equation of the line from that of the curve and integrate the whole thing between limits of -3 and 0; or do what I did which is just integrate the equation of the curve and recognise that the area under the line is just a right-angled triangle whose area is easy to find.

Post the result(s) of your integral(s) if you're still not getting the right answer!

Reply 16

SophieL1996

Original post by davros

Was that a reply to me - you really need to quote me so I see it :biggrin:

Thanks! I finally got the correct answer doing it both ways :biggrin:

When I intergrated by subtracting the line from the curve my overall answer was -4.5 When I used your method I got +4.5 .... does the negative matter as we can see from the sketch it is in a positive region?

Reply 17

davros

Original post by SophieL1996

Thanks! I finally got the correct answer doing it both ways :biggrin:

You should get a positive answer either way since as you say it's a positive region. A rough sketch shows you that the curve is above the line in the region in question, so your method should work with the equation of the line subtracted from the equation of the curve.

Remember that you're integrating from left to right, so your bottom limit is -3 and your upper limit is 0; if the result of your integration is F(x) then you want to evaluate F(0) - F(-3) as the final step.

You've probably just dropped or gained a sign somewhere in your working :smile:

Reply 18

SophieL1996