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is it just me or is this question incomplete??

Am I being a ding dong here?

I am asked to find the area between the bound of x in the second part of the question.

I can find part a of the question, however I cannot determine c without being given more information.

Therefore, without c I can’t draw the graph needed in order to know whether i’m dealing with one area or two or three areas….

I know there must be some error because if I consider the values given, the area is negative meaning there must be two separate areas and one must be at least negative.

So, is it possible to find c with the relevant information given in the question? Am I missing something?


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(edited 6 months ago)
Reply 1
Original post by KingRich
Am I being a ding dong here?

I am asked to find the area between the bound of x in the second part of the question.

I can find part a of the question, however I cannot determine c without being given more information.

Therefore, without c I can’t draw the graph needed in order to know whether i’m dealing with one area or two or three areas….

I know there must be some error because if I consider the values given, the area is negative meaning there must be two separate areas and one must be at least negative.

So, is it possible to find c with the relevant information given in the question? Am I missing something?

Whre is the question?
Reply 2
Original post by Muttley79
Whre is the question?


Sorry, it must have failed to upload. I’ve attached it now
Reply 3
Original post by KingRich
Sorry, it must have failed to upload. I’ve attached it now


You need to sketch y = x^3 - 3x^2 -4x

That's the curve you are dealing with - when you know what the curve looks like you can put values in and c becomes irrelevant
Reply 4
Original post by Muttley79
You need to sketch y = x^3 - 3x^2 -4x

That's the curve you are dealing with - when you know what the curve looks like you can put values in and c becomes irrelevant


I know I can ignore c when integrating it to find the area but I was unaware I could draw the graph without c to find the necessary bounds too ..

Makes a life a lot more simpler I knew that lol

Thank you
Reply 5
Original post by KingRich
I know I can ignore c when integrating it to find the area but I was unaware I could draw the graph without c to find the necessary bounds too ..

Makes a life a lot more simpler I knew that lol

Thank you

c isn't in the curve - you are sketching the cubic!
Reply 6
Original post by Muttley79
c isn't in the curve - you are sketching the cubic!


Is it possible to answer the question without drawing the graph by any chance?
Reply 7
Original post by KingRich
Is it possible to answer the question without drawing the graph by any chance?


y = x^3 - 3x^2 -4x = x(x^2 - 3x -4) = x (x - 4)(x + 1)

So you can find where it cuts the x-axis
Reply 8
Original post by KingRich
Is it possible to answer the question without drawing the graph by any chance?

Yes. Find the roots of the cubic; you'll need to split the range of integration at each root and then add the absolute value of the integral in each subinterval.

E.g. if your cubic f(x) has roots 2, 4 and 6 and you want the area between x=1 and x=5 you'd need to subdivide into intervals [1,2], [2,4], [4,5]. Find the area of each interval and add them together (taking the absolute value for each sub-area).
Reply 9
Original post by Muttley79
y = x^3 - 3x^2 -4x = x(x^2 - 3x -4) = x (x - 4)(x + 1)

So you can find where it cuts the x-axis


Okay, great thanks for confirming.

I see my confusion now. I would usually go ahead and integrate without the c and then input the two seperate x bounds to find the area. Without the graph given, it went right over my head… this has certainly cleared things up for me. Appreciated
Reply 10
Original post by DFranklin
Yes. Find the roots of the cubic; you'll need to split the range of integration at each root and then add the absolute value of the integral in each subinterval.

E.g. if your cubic f(x) has roots 2, 4 and 6 and you want the area between x=1 and x=5 you'd need to subdivide into intervals [1,2], [2,4], [4,5]. Find the area of each interval and add them together (taking the absolute value for each sub-area).


Thanks for the extra information. I can do that part. I was just being a little dumb in regards to the cubic crossing the x axis at zero.

I think it was just a case of misunderstanding the question and the fact I had to consider drawing the graph or deducing the generic shape to figure out which was negative and which was positive and then adding their absolute values

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