curve problems?

- anonymous

curve problems?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

##### 1 Attachment

- Luigi0210

Do you know what the graph looks like?

- anonymous

No, I have no idea!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Luigi0210

Well it might be a good idea to sketch a graph

- anonymous

would it possible to solve the problemã€€without a graph?

- anonymous

would it be

- Luigi0210

Well yes but it be a bit harder since we don't know which graph is higher

- anonymous

oh i see...
how do you draw a graph

- Luigi0210

wait, are you finding total area or net area?

- anonymous

it says find the area between the curves?

- Luigi0210

If I'm not mistaken then the graphs should look like this when graphed
|dw:1366855841011:dw|

- anonymous

sin and cos graph?

- Luigi0210

Uhm no, the ends of the graph go off into infinity after they have intersected

- anonymous

Oh.

- Luigi0210

Now, do you notice anything about the two graphs?

- anonymous

-they get smaller and smaller?
-they intersects

- Luigi0210

Well yes but also they are reflections of one another

- anonymous

Yes,

- Luigi0210

So that means that whatever area on graph has the other has the opposite area.
Such as -x and x.

- anonymous

Yes,

- Luigi0210

So that means the area would be zero

- anonymous

the answer is zero?

- anonymous

i thought i need more calculations

- Luigi0210

Well you could if you want but the answer would still be zero. These two graphs are going to cancel each others area out

- anonymous

i entered zero into the answer but it tells me that it was wrong

- Luigi0210

Well let's solve it right now, do you know how to set it up as an integral?

- anonymous

No, I have no idea!!

- anonymous

@Luigi0210

- Luigi0210

Sorry I was trying to evaluate the problem

- Luigi0210

but anyways we would set up the problem like this:
\[\int\limits_{0}^{9} (x^3-14x^2+45x)-(-x^3+14x^2-45x) dx\]

- Zarkon

I'm betting that they want you to set up two different integrals...based on the two enclosed regions.

- anonymous

so between...= minus.
i got it

- Luigi0210

Setting up two equations would take a bit longer

- Zarkon

http://www.wolframalpha.com/input/?i=area+between+x^3-14x^2%2B45x+and+-x^3%2B14x^2-45x

- anonymous

whats next step?

- Luigi0210

The answer I got was 253/2..

- Luigi0210

*243/2

- anonymous

the answer was wrong but thanks for ur help!

- Luigi0210

I'm sorry I couldn't help you solve it..

- anonymous

No thanks for your time!

Looking for something else?

Not the answer you are looking for? Search for more explanations.