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C2-integration question.

I can't find the ms for this paper, could you tell me of its correct and if not where I've gone wrong?

Thanks:tongue:

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This topic is a nuisance :colonhash:
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Reply 1
Avatar for GMQ
GMQ
13
I think you were supposed to add kx^2 rather than minus it
Also I'm not sure what you've done when using your limits. To integrate between 2 and 0, sub in 0 to your expression and take it away from you result when subbing in 2.
Reply 2
Avatar for abitpissy
abitpissy
OP
14
Original post by GMQ
I think you were supposed to add kx^2 rather than minus it
Also I'm not sure what you've done when using your limits. To integrate between 2 and 0, sub in 0 to your expression and take it away from you result when subbing in 2.

You've confused me, is it not 100/3? That's what I get when I sub in 2.
Reply 3
Avatar for GMQ
GMQ
13
Original post by Questioness
Right I don't know how i got that either, looking back at it I got -152/3


yeah try again with +14x^2 you should get 100/3
Reply 4
Avatar for abitpissy
abitpissy
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14
Original post by GMQ
yeah try again with +14x^2 you should get 100/3


So where did I go wrong?
Reply 5
Avatar for GMQ
GMQ
13
Original post by Questioness
So where did I go wrong?


Ahh sorry I misread the question :biggrin:

You have to find the area under the line PN and then subtract the area under the curve (which is 100/3)

area under PN is 2x24 = 48
area under curve is 100/3
which gives 44/3

Sorry for confusing you :redface:
Reply 6
Avatar for abitpissy
abitpissy
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14
Original post by GMQ
Ahh sorry I misread the question :biggrin:

You have to find the area under the line PN and then subtract the area under the curve (which is 100/3)

area under PN is 2x24 = 48
area under curve is 100/3
which gives 44/3

Sorry for confusing you :redface:

Right okay thanks, I don't know where you got 24. Isn't the line graph y=2. Where X=2
Reply 7
Avatar for GMQ
GMQ
13
Original post by Questioness
Right okay thanks, I don't know where you got 24. Isn't the line graph y=2. Where X=2


I got y=24 for x=2, using the original equation in the question
Reply 8
Avatar for abitpissy
abitpissy
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Original post by GMQ
I got y=24 for x=2, using the original equation in the question


Ahh I think o see were your getting at, shouldn't y=28 since it's the maximum is P and the line touches p?
So it must have the same y value.
Reply 9
Avatar for GMQ
GMQ
13
Okay I don't think I can make it any clearer than this:

The equation for the curve is "y = x^3 - 10x^2 + 28x" (as k = 28)

The x value of point P is 2. Putting in x as 2 in the equation you get: (2)^3 - 10(2)^2 + 28(2)

This is equal to the y value at P, equal to "8 - 40 + 56 = 24"

Original post by Questioness
Ahh I think o see were your getting at, shouldn't y=28 since it's the maximum is P and the line touches p?
So it must have the same y value.
Reply 10
Avatar for abitpissy
abitpissy
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14
Original post by GMQ
Okay I don't think I can make it any clearer than this:

The equation for the curve is "y = x^3 - 10x^2 + 28x" (as k = 28)

The x value of point P is 2. Putting in x as 2 in the equation you get: (2)^3 - 10(2)^2 + 28(2)

This is equal to the y value at P, equal to "8 - 40 + 56 = 24"

Ahh right. Thanks :smile:
Original post by Questioness
Ahh right. Thanks :smile:


If you are still stuck/don't understand, there is no shame in saying so :tongue:
Original post by Questioness
I can't find the ms for this paper, could you tell me of its correct and if not where I've gone wrong?

Thanks:tongue:

image.jpg

Attachment not found


This topic is a nuisance
Which year? And which exam board?
Original post by Questioness
I can't find the ms for this paper, could you tell me of its correct and if not where I've gone wrong?

Thanks:tongue:

image.jpg

Attachment not found


This topic is a nuisance :colonhash:


you can do this as a single integral:

∫ 24 - ( x3 - 10x2 + 28x ) dx with limits 0 & 2
Reply 14
Avatar for abitpissy
abitpissy
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14
Original post by SeanFM
If you are still stuck/don't understand, there is no shame in saying so :tongue:


I understand now : ). Probs need to practice this more :bawling:.
Reply 15
Avatar for abitpissy
abitpissy
OP
14
Original post by 04MR17
Which year? And which exam board?


Edexcel. I don't know what year :s-smilie:
Original post by Questioness
I understand now : ). Probs need to practice this more :bawling:.


You'll be okay, miss flower :tongue:
Reply 17
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abitpissy
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14
Original post by SeanFM
You'll be okay, miss flower :tongue:


'tis my name :wink:

Spoiler

Original post by Questioness
'tis my name :wink:

Spoiler



Oh, as in H...? Nice :tongue:
Reply 19
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abitpissy
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Original post by SeanFM
Oh, as in H...? Nice :tongue:


That's the one. You can read Japanese?

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