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    Name:  image.jpg
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Size:  393.1 KBCould anyone help me with this question , i just can't seem to answer it.

    "The diagram shows the graphs of the curve with equation y=3x-x^2 and the line with the equation y=x-8. The graph intersect in the points A and B. Find
    A) the co ordinates of A and B
    B) the area contained between the curve and the line

    Thank you ~
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    (Original post by rebeccaslugocki)
    Name:  image.jpg
Views: 411
Size:  393.1 KBCould anyone help me with this question , i just can't seem to answer it.

    "The diagram shows the graphs of the curve with equation y=3x-x^2 and the line with the equation y=x-8. The graph intersect in the points A and B. Find
    A) the co ordinates of A and B
    B) the area contained between the curve and the line

    Thank you ~
    Can you please post your thoughts / working?
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    good point how can you find the area of the shaded bit?

    doesn't doing \dfrac {\mathrm{d}x}{\mathrm{d}y} get you that midgy bit above the x-axis???
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    righto, so for part A you must let the equations equal each other (it's handy that they both begin with y=)
    after some subtractions you should get a quadratic, solve it to get two x-values. for each x-value put into either equation to get the corresponding y-value (y=x-8 is easier). that'll get the co-ords for A and B
    for part B integerate between the values (x-value of A and x-value of B) equation of curve minus equation of y=x-8 (be careful of signs, i keep making that mistake .-.) and that should give you the answer.

    so be careful though i may be wrong as what i have done involves areas being above the x-axis so this MAY by wrong, is this by any chance a C2 question, or 3/4? i WOULD do this question on paper but currently i'm busy revising some chemistry stuff, if this question isn't answered in about 2 hours though, ill happily give it a try, i do enjoy a good bit of maths
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    (Original post by Toasticide)
    righto, so for part A you must let the equations equal each other (it's handy that they both begin with y=)
    after some subtractions you should get a quadratic, solve it to get two x-values. for each x-value put into either equation to get the corresponding y-value (y=x-8 is easier). that'll get the co-ords for A and B
    for part B integerate between the values (x-value of A and x-value of B) equation of curve minus equation of y=x-8 (be careful of signs, i keep making that mistake .-.) and that should give you the answer.

    so be careful though i may be wrong as what i have done involves areas being above the x-axis so this MAY by wrong, is this by any chance a C2 question, or 3/4? i WOULD do this question on paper but currently i'm busy revising some chemistry stuff, if this question isn't answered in about 2 hours though, ill happily give it a try, i do enjoy a good bit of maths
    It's C2 thank you ! I'm currently revising chemistry too because u need a break from this question XD but I will try it again thank you
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    (Original post by Toasticide)
    righto, so for part A you must let the equations equal each other (it's handy that they both begin with y=)
    after some subtractions you should get a quadratic, solve it to get two x-values. for each x-value put into either equation to get the corresponding y-value (y=x-8 is easier). that'll get the co-ords for A and B
    for part B integerate between the values (x-value of A and x-value of B) equation of curve minus equation of y=x-8 (be careful of signs, i keep making that mistake .-.) and that should give you the answer.

    so be careful though i may be wrong as what i have done involves areas being above the x-axis so this MAY by wrong, is this by any chance a C2 question, or 3/4? i WOULD do this question on paper but currently i'm busy revising some chemistry stuff, if this question isn't answered in about 2 hours though, ill happily give it a try, i do enjoy a good bit of maths
    It's C2 thank you ! I'm currently revising chemistry too because I need a break from this question XD but I will try it again thank you
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    (Original post by thefatone)
    good point how can you find the area of the shaded bit?

    doesn't doing \dfrac {\mathrm{d}x}{\mathrm{d}y} get you that midgy bit above the x-axis???
    DY DX find the gradient of a curve at any point , to find the area you have to use integration but I just can't work it out !
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    (Original post by rebeccaslugocki)
    DY DX find the gradient of a curve at any point , to find the area you have to use integration but I just can't work it out !
    If you have a curve f(x) above another curve g(x) then the area between the two curves from x=a to x=b is

    \displaystyle \int^a_b f(x) - g(x) \ dx

    It doesn't matter if there are parts of the area above/below the x-axis.

    Try this and post your working if you get stuck.
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    (Original post by rebeccaslugocki)
    DY DX find the gradient of a curve at any point , to find the area you have to use integration but I just can't work it out !

    oops supposed to be dy/dx lol

    (Original post by notnek)
    If you have a curve f(x) above another curve g(x) then the area between the two curves from x=a to x=b is

    \displaystyle \int^a_b f(x) - g(x) \ dx

    It doesn't matter if there are parts of the area above/below the x-axis.

    Try this and post your working if you get stuck.
    i see so you can just integrate normally and you'd still get the right answer?
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    oh yeh quick update for the second part i think my method is ever so slightly wrong because some of the area is in the positive axis and some in the negative, i think it doesn't matter about the y-axis but with the x-axis in the negative section you'll get a negative area so at when integrating you have to split it up into two parts, one of the integral of the positive x-axis and one of the negative and since the negative gives a negative area, you need to literally just remove the negative sign and add the two areas (or if you want to be really specific, make the negative area modular, but that word doesnt matter currently)

    basically, what im trying to say is because it goes wierd with the axis and goes through all four quadrants, look at the bottom of page 177, that's what im trying to explain. (i think you're doing edexcel, if not don't look at that page in the textbook, it's probs completly different XD)
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    (Original post by Toasticide)
    oh yeh quick update for the second part i think my method is ever so slightly wrong because some of the area is in the positive axis and some in the negative, i think it doesn't matter about the y-axis but with the x-axis in the negative section you'll get a negative area so at when integrating you have to split it up into two parts, one of the integral of the positive x-axis and one of the negative and since the negative gives a negative area, you need to literally just remove the negative sign and add the two areas (or if you want to be really specific, make the negative area modular, but that word doesnt matter currently)

    basically, what im trying to say is because it goes wierd with the axis and goes through all four quadrants, look at the bottom of page 177, that's what im trying to explain. (i think you're doing edexcel, if not don't look at that page in the textbook, it's probs completly different XD)
    If you follow the method I explained in my last post (the same method you gave) then it doesn't matter if some of the area is above or below the x-axis as long as one function is always \geq the other function. The algebra works itself out.
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    (Original post by notnek)
    If you follow the method I explained in my last post (the same method you gave) then it doesn't matter if some of the area is above or below the x-axis as long as one function is always \geq the other function. The algebra works itself out.
    ah thanks, couldn't revise my chemistry properly as that kept bugging me
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    Integrating the curve y=3x-x2 from the x-coordinate of A (-2) to the origin (0) will give you the area bounded between the curve and the x-axis. The same applies when you integrate the line y=x-8. When there are segments of area above and below the x-axis, you will not be able to just integrate the entire segment as the areas below the x-axis will be calculated as negative values which will cancel out the areas above the x-axis. This is a good example illustrating how you can get a 'zero area' answer after integration:
    http://revisionmaths.com/advanced-le...ea-under-curve

    I'll assume you know how to find the coordinates of A and B and that the problem lies in the second part of the question. To get the shaded area, you'll need to break the integration process into three regions:

    A) From x=-2 to x=0

    1. Integrate y=x-8 to find the area bounded between the line and the x-axis. You'll notice that it's a negative answer because the area is underneath the x-axis.
    2. Integrate y=3x-x2 to find the area bounded between the curve and the x-axis.
    3. The shaded area will be the difference between the absolute value of the answer in 1. and the absolute value of the answer in 2. (Absolute value basically means ignore the negative sign and just look at the 'magnitude' of the area)

    B) From x=0 to x=3 (from the origin to the second intersection between the curve and the x-axis)

    1. Integrate y=x-8 to find the area bounded between the line and the x-axis.
    2. Integrate y=3x-x2 to find the area bounded between the curve and the x-axis.
    3. The shaded area will be the sum of the absolute value of the answer in 1. and the value of the answer in 2.

    C) From x=3 to x=4 (from the second intersection between the curve and the x-axis to the x-coordinate of B)
    I'll let you figure this one out by yourself. It's almost the same as part A.


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    (Original post by QuietEmbers)
    Integrating the curve y=3x-x2 from the x-coordinate of A (-2) to the origin (0) will give you the area bounded between the curve and the x-axis. The same applies when you integrate the line y=x-8. When there are segments of area above and below the x-axis, you will not be able to just integrate the entire segment as the areas below the x-axis will be calculated as negative values which will cancel out the areas above the x-axis. This is a good example illustrating how you can get a 'zero area' answer after integration:
    http://revisionmaths.com/advanced-le...ea-under-curve

    I'll assume you know how to find the coordinates of A and B and that the problem lies in the second part of the question. To get the shaded area, you'll need to break the integration process into three regions:

    A) From x=-2 to x=0

    1. Integrate y=x-8 to find the area bounded between the line and the x-axis. You'll notice that it's a negative answer because the area is underneath the x-axis.
    2. Integrate y=3x-x2 to find the area bounded between the curve and the x-axis.
    3. The shaded area will be the difference between the absolute value of the answer in 1. and the absolute value of the answer in 2. (Absolute value basically means ignore the negative sign and just look at the 'magnitude' of the area)

    B) From x=0 to x=3 (from the origin to the second intersection between the curve and the x-axis)

    1. Integrate y=x-8 to find the area bounded between the line and the x-axis.
    2. Integrate y=3x-x2 to find the area bounded between the curve and the x-axis.
    3. The shaded area will be the sum of the absolute value of the answer in 1. and the value of the answer in 2.

    C) From x=3 to x=4 (from the second intersection between the curve and the x-axis to the x-coordinate of B)
    I'll let you figure this one out by yourself. It's almost the same as part A.

    No need, since regardless of if the area is above or below the axis, or both, when finding the area between two curves, there is no need to split the integral.
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    (Original post by HapaxOromenon2)
    No need, since regardless of if the area is above or below the axis, or both, when finding the area between two curves, there is no need to split the integral.
    I think it would be better if she understood the concept of the problem instead of relying on a general rule. Check out example 4 on this page:
    http://tutorial.math.lamar.edu/Class...eenCurves.aspx
    That general rule doesn't apply to the shaded area here.
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    (Original post by QuietEmbers)
    I think it would be better if she understood the concept of the problem instead of relying on a general rule. Check out example 4 on this page:
    http://tutorial.math.lamar.edu/Class...eenCurves.aspx
    That general rule doesn't apply to the shaded area here.
    Yes, because it changes from curve above line to line above curve back to curve above line. But in cases where the entire region has either the curve above the line, or the line above the curve, there's no need to overcomplicate it: a good mathematician never uses a bigger tool than he needs to.
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    (Original post by QuietEmbers)
    I think it would be better if she understood the concept of the problem instead of relying on a general rule. Check out example 4 on this page:
    http://tutorial.math.lamar.edu/Class...eenCurves.aspx
    That general rule doesn't apply to the shaded area here.
    For most C2 area questions I would agree with you. But here using the rule is much faster and will lead to fewer mistakes.

    Theorems are used all over the place in maths that can be quoted if they have been proven.

    Example 3 doesn't work because f\geq g isn't satisfied over the full interval.
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    (Original post by Toasticide)
    oh yeh quick update for the second part i think my method is ever so slightly wrong because some of the area is in the positive axis and some in the negative, i think it doesn't matter about the y-axis but with the x-axis in the negative section you'll get a negative area so at when integrating you have to split it up into two parts, one of the integral of the positive x-axis and one of the negative and since the negative gives a negative area, you need to literally just remove the negative sign and add the two areas (or if you want to be really specific, make the negative area modular, but that word doesnt matter currently)

    basically, what im trying to say is because it goes wierd with the axis and goes through all four quadrants, look at the bottom of page 177, that's what im trying to explain. (i think you're doing edexcel, if not don't look at that page in the textbook, it's probs completly different XD)
    This is what I'm doing now thank you , I'm doing WJEC haha
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    (Original post by QuietEmbers)
    Integrating the curve y=3x-x2 from the x-coordinate of A (-2) to the origin (0) will give you the area bounded between the curve and the x-axis. The same applies when you integrate the line y=x-8. When there are segments of area above and below the x-axis, you will not be able to just integrate the entire segment as the areas below the x-axis will be calculated as negative values which will cancel out the areas above the x-axis. This is a good example illustrating how you can get a 'zero area' answer after integration:
    http://revisionmaths.com/advanced-le...ea-under-curve

    I'll assume you know how to find the coordinates of A and B and that the problem lies in the second part of the question. To get the shaded area, you'll need to break the integration process into three regions:

    A) From x=-2 to x=0

    1. Integrate y=x-8 to find the area bounded between the line and the x-axis. You'll notice that it's a negative answer because the area is underneath the x-axis.
    2. Integrate y=3x-x2 to find the area bounded between the curve and the x-axis.
    3. The shaded area will be the difference between the absolute value of the answer in 1. and the absolute value of the answer in 2. (Absolute value basically means ignore the negative sign and just look at the 'magnitude' of the area)

    B) From x=0 to x=3 (from the origin to the second intersection between the curve and the x-axis)

    1. Integrate y=x-8 to find the area bounded between the line and the x-axis.
    2. Integrate y=3x-x2 to find the area bounded between the curve and the x-axis.
    3. The shaded area will be the sum of the absolute value of the answer in 1. and the value of the answer in 2.

    C) From x=3 to x=4 (from the second intersection between the curve and the x-axis to the x-coordinate of B)
    I'll let you figure this one out by yourself. It's almost the same as part A
    Our Teacher told us to split it up and do it that way , he gave us answers for what each section should be and the total area. I'm doing this method but just not getting the answers? I'm beginning to wonder if he gave us the right ones or that i'm just being stupid D; He wrote 5 answers and didn't say which answer is for which section, just that we should end up with these numbers for them. 48 , 64/3 , 12 , 8/3 , 36 are the answers he wrote down ( one of them is the total answer) the others are just the areas of the separate sections. I'm not getting these numbers
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    (Original post by rebeccaslugocki)
    Our Teacher told us to split it up and do it that way , he gave us answers for what each section should be and the total area. I'm doing this method but just not getting the answers? I'm beginning to wonder if he gave us the right ones or that i'm just being stupid D; He wrote 5 answers and didn't say which answer is for which section, just that we should end up with these numbers for them. 48 , 64/3 , 12 , 8/3 , 36 are the answers he wrote down ( one of them is the total answer) the others are just the areas of the separate sections. I'm not getting these numbers
    What did you get as your final answer for the area?

    EDIT: The answer should be 36. If you didn't get this, please post your working. You can attach a photo of your working if you like.
 
 
 
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