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# C2 intergration between a curve and a line watch

1. 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)
2. 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.
3. (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.
4. (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 .
D) This step is dependant on your sketch but it should become obvious which area to take away from which.
5. How would you draw y=x+3 on the graph ? As I am not too sure.
6. I have the points of intersection as (0.3) and (-3,0). Surely you would take the area under the curve- line??
7. 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)
8. So what have you tried?
9. (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.
10. (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 But I have drawn the graph and the points of intersection do not match up so I am confused.
11. (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 ...
12. Any ideas?? I am really stuck :l
13. (Original post by SophieL1996)
Any ideas?? I am really stuck :l

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:

14. Ok I will check that, thanks!
15. Your method does not give the correct answer of 4.5 ...
16. (Original post by SophieL1996)
Was that a reply to me - you really need to quote me so I see it

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!
17. (Original post by davros)
Was that a reply to me - you really need to quote me so I see it

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!
Thanks! I finally got the correct answer doing it both ways 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?
18. (Original post by SophieL1996)
Thanks! I finally got the correct answer doing it both ways 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?
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
19. (Original post by davros)
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
Ok thanks I shall have a look
20. Yes I misplaced a sign.. thanks for your help

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