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    That moment when u have no idea what these 2 ppl are arguing about because u yourself left that question because you were unsure about the equation 😧

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    (Original post by I <3 WORK)
    Yes but this is my point lol. When you subtract area under the curve from triangle area you get the shaded area (R) AND the bit under the x axis (above the curve). So in order to get the area that's shaded ONLY you will further have to subtract area under line (under x axis) from area under curve. And this value is subtracted from 11/64.
    No. I used the limits 0.5 to 1. This only includes the area above the x-axis and below the curve.
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    (Original post by kprime2)
    No. I used the limits 0.5 to 1. This only includes the area above the x-axis and below the curve.
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    That is true, you're right about having the limits between 1/2 to 1 - and I would think that is actually a quicker method. However even with this method, you have only subtracted the area under the curve from the triangle, and therefore the resultant area would be R aswell as the area (between1/2 and 1) under the x axis. So you will need to find the area under the x axis (between 1/2 and 1) and further subtract this from 11/64.

    I hope I'm making sense here.
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    (Original post by I <3 WORK)
    That is true, you're right about having the limits between 1/2 to 1 - and I would think that is actually a quicker method. However even with this method, you have only subtracted the area under the curve from the triangle, and therefore the resultant area would be R aswell as the area (between1/2 and 1) under the x axis. So you will need to find the area under the x axis (between 1/2 and 1) and further subtract this from 11/64.

    I hope I'm making sense here.
    When you integrate from 1/2 to 1 the ONLY area you found is the one I've shaded in the previous post, nothing else. Therefore the triangle minus the area I've shaded results in area R.
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    (Original post by kprime2)
    When you integrate from 1/2 to 1 the ONLY area you found is the one I've shaded in the previous post, nothing else. Therefore the triangle minus the area I've shaded results in area R.
    Okay well to be honest I've just realised that the area under the x axis will always be zero! I think that's where the confusion arised and that's why you're saying the area will be the same.The way I calculated it was a much lengthier way by integrating between 1 and 2 the area under the line minus area under the curve. This would leave R + the area underneath x axis and area above curve between 1 and 2. So I integrated again to find this area and then subtracted it from previous answer. I'm pretty sure using this method (although a lot lengthier) should give me the same answer so I think I must have made errors in the calculation. How many marks do you think I will lose for using the correct method but wrong calculations?
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    (Original post by I <3 WORK)
    Okay well to be honest I've just realised that the area under the x axis will always be zero! I think that's where the confusion arised and that's why you're saying the area will be the same.The way I calculated it was a much lengthier way by integrating between 1 and 2 the area under the line minus area under the curve. This would leave R + the area underneath x axis and area above curve between 1 and 2. So I integrated again to find this area and then subtracted it from previous answer. I'm pretty sure using this method (although a lot lengthier) should give me the same answer so I think I must have made errors in the calculation. How many marks do you think I will lose for using the correct method but wrong calculations?
    I don't even understand what you did. Maybe you can post what you did on a piece of paper?
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    (Original post by kprime2)
    I don't even understand what you did. Maybe you can post what you did on a piece of paper?
    Right so basically when I used the same method as I did in the exam I truly did get 11/64 so I'm very sorry about all of that confusion! However I just would like to know please how many marks I would lose for using this method but doing some sort of calculation mistake.

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    (Original post by I <3 WORK)
    Right so basically when I used the same method as I did in the exam I truly did get 11/64 so I'm very sorry about all of that confusion! However I just would like to know please how many marks I would lose for using this method but doing some sort of calculation mistake.

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    That's such a lengthy method
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    (Original post by Imperion)
    That's such a lengthy method
    Lol yes I know right! I wasn't intelligent enough to think of such a simple and short method even though I literally had 5 mins to do that question. :argh:
    :banghead:
 
 
 
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