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C1 Integration watch

1. Okay so I did the 2014 C1 paper about an hour ago and I realised I have absolutely no clue on how to do integration and finding the area under a curve would anyone be able to give me the run down? I've attached the question I struggled on, I think its all just knowing where to start, it really got me stumped thank you very much
2. (Original post by bethanygriffin)
Okay so I did the 2014 C1 paper about an hour ago and I realised I have absolutely no clue on how to do integration and finding the area under a curve would anyone be able to give me the run down? I've attached the question I struggled on, I think its all just knowing where to start, it really got me stumped thank you very much
What did you try?
If you explain what you did, we might be able to point out where you went wrong - or was the whole question a bit of a mess?
3. to intergrate you just raise the power by one and then divide by the new power e.g. the intergrel of x^2 is x^3 divided by 3

and to find the area you just intergrate the equation and then sub in the limits, so for the question you posted the limits would be -2 and 3 (they're the point A and C). you sub in 3 into the intergtaed equation then sub -2 into the intergrated equation and subtract the two
4. (Original post by chloemelissa1996)
to intergrate you just raise the power by one and then divide by the new power e.g. the intergrel of x^2 is x^3 divided by 3

and to find the area you just intergrate the equation and then sub in the limits, so for the question you posted the limits would be -2 and 3 (they're the point A and C). you sub in 3 into the intergtaed equation then sub -2 into the intergrated equation and subtract the two
You might want to go back and re-read the question.

OP, I was actually doing that paper before - where did you go wrong/ which bits do you need help with?
5. (Original post by aoxa)
You might want to go back and re-read the question.

OP, I was actually doing that paper before - where did you go wrong/ which bits do you need help with?
why might i want to re-read the question, I know what to do and I know what its asking me.

I was using x^2 as an example not taking it from the question.

Trust me what I'm saying is correct
6. that is c2
7. (Original post by chloemelissa1996)
why might i want to re-read the question, I know what to do and I know what its asking me.I was using x^2 as an example not taking it from the question.Trust me what I'm saying is correct
The bit in bold was why you may have wanted to re-read the question. The limits you put were wrong for the question, which is why I told you to re-read the question.

(Original post by indignation)
that is c2
No it's not. That is a C1 paper.
8. (Original post by aoxa)
The bit in bold was why you may have wanted to re-read the question. The limits you put were wrong for the question, which is why I told you to re-read the question.

No it's not. That is a C1 paper.
I'm pretty sure the limits are x=-2 and x=3 because when you factorise x^2-x-6 you get (x+2)(x-3)=0, hence x=-2 and x=3, which are the x co-ords of A and C and hence the limits
9. (Original post by Neuth)
What did you try?
If you explain what you did, we might be able to point out where you went wrong - or was the whole question a bit of a mess?
The whole question was a bit of a mess , I think if I knew how to get the coordinates of A and C I'd be able to do it. I looked on the mark scheme to try and work out what to do but I have no clue where they got their numbers from
10. (Original post by bethanygriffin)
The whole question was a bit of a mess , I think if I knew how to get the coordinates of A and C I'd be able to do it. I looked on the mark scheme to try and work out what to do but I have no clue where they got their numbers from

what do they say are the x co-ordinates for A and C?
11. (Original post by indignation)
that is c2
It's definitely C1 on AQA, maybe we're on different exam boards?
12. (Original post by chloemelissa1996)
I'm pretty sure the limits are x=-2 and x=3 because when you factorise x^2-x-6 you get (x+2)(x-3)=0, hence x=-2 and x=3, which are the x co-ords of A and C and hence the limits
Your factorisation is correct. If you look at the question however, you're finding the shaded area R. This gives your limits to be 0 and -2 (pt c of the question does specify this 'find the area R bounded by the curve and the line segment AB' - where A is (-2,5) and B is (0,7)) Once you've found that area, you need to subtract the area below R (area of a trapezium) from the area, to leave you with area R.

The only thing you've done wrong is misread the question - make sure to double check every question before you answer them tomorrow!
13. (Original post by chloemelissa1996)
what do they say are the x co-ordinates for A and C?
-2 and 3
14. (Original post by bethanygriffin)
The whole question was a bit of a mess , I think if I knew how to get the coordinates of A and C I'd be able to do it. I looked on the mark scheme to try and work out what to do but I have no clue where they got their numbers from
Did you manage to get pt ai? Showing that the x-coordinates satisfy x^2 -x-6=0?
15. (Original post by aoxa)
Your factorisation is correct. If you look at the question however, you're finding the shaded area R. This gives your limits to be 0 and -2 (pt c of the question does specify this 'find the area R bounded by the curve and the line segment AB' - where A is (-2,5) and B is (0,7)) Once you've found that area, you need to subtract the area below R (area of a trapezium) from the area, to leave you with area R.

The only thing you've done wrong is misread the question - make sure to double check every question before you answer them tomorrow!
Oh my gosh thank you
16. (Original post by aoxa)
Your factorisation is correct. If you look at the question however, you're finding the shaded area R. This gives your limits to be 0 and -2 (pt c of the question does specify this 'find the area R bounded by the curve and the line segment AB' - where A is (-2,5) and B is (0,7)) Once you've found that area, you need to subtract the area below R (area of a trapezium) from the area, to leave you with area R.

The only thing you've done wrong is misread the question - make sure to double check every question before you answer them tomorrow!
oh yes of course! thank you
17. (Original post by bethanygriffin)
Oh my gosh thank you
You're welcome? I'm not sure what I've done, but if the proverbial penny has dropped I'm glad
18. (Original post by chloemelissa1996)
I'm pretty sure the limits are x=-2 and x=3 because when you factorise x^2-x-6 you get (x+2)(x-3)=0, hence x=-2 and x=3, which are the x co-ords of A and C and hence the limits
You are correct (not sure what the issue is here ).

(Original post by bethanygriffin)
The whole question was a bit of a mess , I think if I knew how to get the coordinates of A and C I'd be able to do it. I looked on the mark scheme to try and work out what to do but I have no clue where they got their numbers from
To start with (and find the x co-ords) you just need to factorise the equation, which gives you (x-3)(x+2)
19. (Original post by Neuth)
You are correct (not sure what the issue is here ).

I cannot go through what limits are the correct limits discussion again... just go back, and read the question!
20. (Original post by aoxa)

I cannot go through what limits are the correct limits discussion again... just go back, and read the question!
I never disputed that

EDIT: I've just realised I quoted the wrong post...

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