# Thread: Area between 3 curves?

1. ## Area between 3 curves?

So I'm finding this problem difficult:

What is the area of the region in the first quadrant (upper right quadrant) bounded by the curves y = sin(x)cos^2(x), y = 2xcos(x^2), and y = 4 – 4x?

I keep trying it, and graphing it on my calculator but I'm getting different answers every time. Sometimes I'll get around .38 other times I get 1.8 I'm not sure what I'm doing wrong and what I should be getting?

2. ## Re: Area between 3 curves?

solve

$$2x_c\cos(x_c^2)=4-4x_c$$

and solve

$$\sin(x_d)\cos^2(x_d)=4-x_d$$

then the area of the bounded region becomes

$$\int_0^{x_c}\left(2 x \cos(x^2)-\sin(x)\cos^2(x)\right) dx+\int_{x_c}^{x_d}\left((4-4x)-\sin(x)\cos^2(x)\right)dx$$

3. ## Re: Area between 3 curves?

Originally Posted by canyouhelp
So I'm finding this problem difficult:

What is the area of the region in the first quadrant (upper right quadrant) bounded by the curves y = sin(x)cos^2(x), y = 2xcos(x^2), and y = 4 – 4x?

I keep trying it, and graphing it on my calculator but I'm getting different answers every time. Sometimes I'll get around .38 other times I get 1.8 I'm not sure what I'm doing wrong and what I should be getting?
Have a look at this.

4. ## Re: Area between 3 curves?

I got 3.78718. Is this close?

5. ## Re: Area between 3 curves?

Originally Posted by Plato
I got 3.78718. Is this close?

6. ## Re: Area between 3 curves?

It doesn't look like it.

7. ## Re: Area between 3 curves?

Originally Posted by romsek
It doesn't look like it.

I don't really understand what that is saying (I'm not used to working problems out online) Doesn't it say once it's fully integrated it's about .378... Like what I ended up with? Or is it saying something else?

8. ## Re: Area between 3 curves?

Oh my gosh! Nevermind, I just realized I wrote it out wrong. I meant .378... not 3.78! Sorry. Now I see!