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anonymous
 3 years ago
curve problems?
anonymous
 3 years ago
curve problems?

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Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Do you know what the graph looks like?

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Well it might be a good idea to sketch a graph

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would it possible to solve the problem without a graph?

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Well yes but it be a bit harder since we don't know which graph is higher

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i see... how do you draw a graph

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1wait, are you finding total area or net area?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it says find the area between the curves?

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1If I'm not mistaken then the graphs should look like this when graphed dw:1366855841011:dw

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Uhm no, the ends of the graph go off into infinity after they have intersected

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Now, do you notice anything about the two graphs?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0they get smaller and smaller? they intersects

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Well yes but also they are reflections of one another

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1So that means that whatever area on graph has the other has the opposite area. Such as x and x.

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1So that means the area would be zero

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i thought i need more calculations

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Well you could if you want but the answer would still be zero. These two graphs are going to cancel each others area out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i entered zero into the answer but it tells me that it was wrong

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Well let's solve it right now, do you know how to set it up as an integral?

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry I was trying to evaluate the problem

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1but anyways we would set up the problem like this: \[\int\limits_{0}^{9} (x^314x^2+45x)(x^3+14x^245x) dx\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0I'm betting that they want you to set up two different integrals...based on the two enclosed regions.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so between...= minus. i got it

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1Setting up two equations would take a bit longer

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=area+between+x^314x^2%2B45x+and+x^3%2B14x^245x

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1The answer I got was 253/2..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answer was wrong but thanks for ur help!

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.1I'm sorry I couldn't help you solve it..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No thanks for your time!
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