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curve problems?

Mathematics
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1 Attachment
Do you know what the graph looks like?
No, I have no idea!

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Other answers:

Well it might be a good idea to sketch a graph
would it possible to solve the problem without a graph?
would it be
Well yes but it be a bit harder since we don't know which graph is higher
oh i see... how do you draw a graph
wait, are you finding total area or net area?
it says find the area between the curves?
If I'm not mistaken then the graphs should look like this when graphed |dw:1366855841011:dw|
sin and cos graph?
Uhm no, the ends of the graph go off into infinity after they have intersected
Oh.
Now, do you notice anything about the two graphs?
-they get smaller and smaller? -they intersects
Well yes but also they are reflections of one another
Yes,
So that means that whatever area on graph has the other has the opposite area. Such as -x and x.
Yes,
So that means the area would be zero
the answer is zero?
i thought i need more calculations
Well you could if you want but the answer would still be zero. These two graphs are going to cancel each others area out
i entered zero into the answer but it tells me that it was wrong
Well let's solve it right now, do you know how to set it up as an integral?
No, I have no idea!!
Sorry I was trying to evaluate the problem
but anyways we would set up the problem like this: \[\int\limits_{0}^{9} (x^3-14x^2+45x)-(-x^3+14x^2-45x) dx\]
I'm betting that they want you to set up two different integrals...based on the two enclosed regions.
so between...= minus. i got it
Setting up two equations would take a bit longer
http://www.wolframalpha.com/input/?i=area+between+x^3-14x^2%2B45x+and+-x^3%2B14x^2-45x
whats next step?
The answer I got was 253/2..
*243/2
the answer was wrong but thanks for ur help!
I'm sorry I couldn't help you solve it..
No thanks for your time!

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