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    This is from an AS textbook but its pretty hard...
    I just need help integrating the equations.
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    To integrate each expression just expand all the brackets out.
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    How would you integrate the circle?


    (Original post by B_9710)
    To integrate each expression just expand all the brackets out.
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    (Original post by TheAlphaParticle)
    How would you integrate the circle?
    You don't need to integrate it. You can find the radius and the area is r^2\pi
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    (Original post by TheAlphaParticle)
    This is from an AS textbook but its pretty hard...
    I just need help integrating the equations.
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    area of a circle is  \displaystyle \pi r^2 , you dont need to integrate for it
    Spoiler:
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    This question is actually amazing, much more creative than most other integration questions ive ever seen
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    (Original post by DylanJ42)
    area of a circle is  \displaystyle \pi r^2 , you dont need to integrate for it
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    This question is actually amazing, much more creative than most other integration questions ive ever seen

    (Original post by RDKGames)
    You don't need to integrate it. You can find the radius and the area is r^2\pi
    (Original post by B_9710)
    To integrate each expression just expand all the brackets out.
    I need help integrating the black curve, I don't know the limits of the curve so how am i meant to integrate it.
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    (Original post by TheAlphaParticle)
    I need help integrating the black curve, I don't know the limits of the curve so how am i meant to integrate it.
    Just count the squares. Each one is 1 in a given direction of either up/down or left/right. So the limits for the black curve are 9 and 14. But that will give area of everything below it. I'm sure you know how to get the actual black area from there.
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    (Original post by TheAlphaParticle)
    I need help integrating the black curve, I don't know the limits of the curve so how am i meant to integrate it.
    just count the squares on the x axis until you reach the point directly under the furthest left point of the black shape, thats your lower limit

    do the same but for the furthest right point of the black shape, this is your upper limit
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    (Original post by RDKGames)
    Just count the squares. Each one is 1. So the limits for the black curve are 9 and 14. But that will give area of everything below it. I'm sure you know how to get the actual black area from there.
    I've tried that and it gives 2.2222m^2 which means it needs one can of paint but the answer is 3 cans.
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    (Original post by TheAlphaParticle)
    I've tried that and it gives 2.2222m^2 which means it needs one can of paint but the answer is 3 cans.
    The area should be (8x5) - integral between 9 and 14.
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    (Original post by RDKGames)
    The area should be (8x5) - integral between 9 and 14.
    Name:  IMG_20160902_150514.jpg
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    (Original post by TheAlphaParticle)
    Where did i go wrong?
    \frac{(x-9)(x-15)}{3}+8 \not= x^2-24x+143

    Not sure why you completely ignored the fact that the quadratic is being divided by 3.
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    (Original post by RDKGames)
    \frac{(x-9)(x-15)}{3}+8 \not= x^2-24x+143

    Not sure why you completely ignored the fact that the quadratic is being divided by 3.
    It's being multiplied by 1/3 later on it makes it easier to integrate
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    (Original post by TheAlphaParticle)
    It's being multiplied by 1/3 later on it makes it easier to integrate
    You did it wrong then. If you want to be taking that route then you need to get the 8 under the same denominator and add them before factoring out the third altogether and place it outside the integral.
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    (Original post by RDKGames)
    You did it wrong then. If you want to be taking that route then you need to get the 8 under the same denominator and add them before factoring out the third altogether and place it outside the integral.
    Ahh thanks so much, I understand now, what is another route (out of curiosity?)
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    (Original post by TheAlphaParticle)
    Ahh thanks so much, I understand now, what is another route (out of curiosity?)
    Not a very different one. You just divide each term of the quadratic by 3 then integrate each one between the limits. It's neater the way you want to do it.
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    (Original post by RDKGames)
    Not a very different one. You just divide each term of the quadratic by 3 then integrate each one between the limits. It's neater the way you want to do it.
    It still doesn't work it gives 28.888 now which is still wrong
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    (Original post by TheAlphaParticle)
    It still doesn't work it gives 28.888 now which is still wrong
    Well of course it doesn't work when you consider it that way because when you integrate you find the area below the curve...
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    (Original post by RDKGames)
    Well of course it doesn't work when you consider it that way because when you integrate you find the area below the curve...
    Oh!! the area of Black and Red are equal, so are Green and Blue.
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    I'm starting to get frustrated i have been doing this for an hour!
 
 
 
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