Could 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=3xx^2 and the line with the equation y=x8. 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 ~

BeccAhhh
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 17042016 13:52

Notnek
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 17042016 14:09
(Original post by rebeccaslugocki)
Could 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=3xx^2 and the line with the equation y=x8. 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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 17042016 14:25
good point how can you find the area of the shaded bit?
doesn't doing get you that midgy bit above the xaxis??? 
Toasticide
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 17042016 14:35
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 xvalues. for each xvalue put into either equation to get the corresponding yvalue (y=x8 is easier). that'll get the coords for A and B
for part B integerate between the values (xvalue of A and xvalue of B) equation of curve minus equation of y=x8 (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 xaxis 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 mathsLast edited by Toasticide; 17042016 at 14:37. Reason: explanation as to why im not currently actually answering the question 
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 17042016 14:45
(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 xvalues. for each xvalue put into either equation to get the corresponding yvalue (y=x8 is easier). that'll get the coords for A and B
for part B integerate between the values (xvalue of A and xvalue of B) equation of curve minus equation of y=x8 (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 xaxis 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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 17042016 14:49
(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 xvalues. for each xvalue put into either equation to get the corresponding yvalue (y=x8 is easier). that'll get the coords for A and B
for part B integerate between the values (xvalue of A and xvalue of B) equation of curve minus equation of y=x8 (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 xaxis 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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 17042016 14:50
(Original post by thefatone)
good point how can you find the area of the shaded bit?
doesn't doing get you that midgy bit above the xaxis??? 
Notnek
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 17042016 14:54
(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 !
It doesn't matter if there are parts of the area above/below the xaxis.
Try this and post your working if you get stuck.Last edited by Notnek; 17042016 at 14:55. 
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 17042016 14:56
(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 above another curve then the area between the two curves from to is
It doesn't matter if there are parts of the area above/below the xaxis.
Try this and post your working if you get stuck. 
Toasticide
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 17042016 15:01
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 yaxis but with the xaxis 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 xaxis 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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 17042016 15:05
(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 yaxis but with the xaxis 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 xaxis 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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 17042016 15:08
(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 xaxis as long as one function is always the other function. The algebra works itself out. 
QuietEmbers
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 17042016 15:37
Integrating the curve y=3xx^{2} from the xcoordinate of A (2) to the origin (0) will give you the area bounded between the curve and the xaxis. The same applies when you integrate the line y=x8. When there are segments of area above and below the xaxis, you will not be able to just integrate the entire segment as the areas below the xaxis will be calculated as negative values which will cancel out the areas above the xaxis. This is a good example illustrating how you can get a 'zero area' answer after integration:
http://revisionmaths.com/advancedle...eaundercurve
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=x8 to find the area bounded between the line and the xaxis. You'll notice that it's a negative answer because the area is underneath the xaxis.
2. Integrate y=3xx^{2} to find the area bounded between the curve and the xaxis.
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 xaxis)
1. Integrate y=x8 to find the area bounded between the line and the xaxis.
2. Integrate y=3xx^{2} to find the area bounded between the curve and the xaxis.
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 xaxis to the xcoordinate of B)
I'll let you figure this one out by yourself. It's almost the same as part A.

HapaxOromenon2
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 17042016 16:12
(Original post by QuietEmbers)
Integrating the curve y=3xx^{2} from the xcoordinate of A (2) to the origin (0) will give you the area bounded between the curve and the xaxis. The same applies when you integrate the line y=x8. When there are segments of area above and below the xaxis, you will not be able to just integrate the entire segment as the areas below the xaxis will be calculated as negative values which will cancel out the areas above the xaxis. This is a good example illustrating how you can get a 'zero area' answer after integration:
http://revisionmaths.com/advancedle...eaundercurve
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=x8 to find the area bounded between the line and the xaxis. You'll notice that it's a negative answer because the area is underneath the xaxis.
2. Integrate y=3xx^{2} to find the area bounded between the curve and the xaxis.
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 xaxis)
1. Integrate y=x8 to find the area bounded between the line and the xaxis.
2. Integrate y=3xx^{2} to find the area bounded between the curve and the xaxis.
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 xaxis to the xcoordinate of B)
I'll let you figure this one out by yourself. It's almost the same as part A.

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 17042016 16:20
(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.
http://tutorial.math.lamar.edu/Class...eenCurves.aspx
That general rule doesn't apply to the shaded area here. 
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 17042016 16:22
(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. 
Notnek
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 17042016 16:53
(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.
Theorems are used all over the place in maths that can be quoted if they have been proven.
Example 3 doesn't work because isn't satisfied over the full interval.Last edited by Notnek; 17042016 at 16:55. 
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 17042016 20:56
(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 yaxis but with the xaxis 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 xaxis 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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 17042016 21:14
(Original post by QuietEmbers)
Integrating the curve y=3xx^{2} from the xcoordinate of A (2) to the origin (0) will give you the area bounded between the curve and the xaxis. The same applies when you integrate the line y=x8. When there are segments of area above and below the xaxis, you will not be able to just integrate the entire segment as the areas below the xaxis will be calculated as negative values which will cancel out the areas above the xaxis. This is a good example illustrating how you can get a 'zero area' answer after integration:
http://revisionmaths.com/advancedle...eaundercurve
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=x8 to find the area bounded between the line and the xaxis. You'll notice that it's a negative answer because the area is underneath the xaxis.
2. Integrate y=3xx^{2} to find the area bounded between the curve and the xaxis.
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 xaxis)
1. Integrate y=x8 to find the area bounded between the line and the xaxis.
2. Integrate y=3xx^{2} to find the area bounded between the curve and the xaxis.
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 xaxis to the xcoordinate of B)
I'll let you figure this one out by yourself. It's almost the same as part A

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 17042016 21:19
(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
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.Last edited by Notnek; 17042016 at 21:26.
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