# Math Help - Equal Areas

1. ## Equal Areas

I could use some help with the setup on this one:

The line $y=c$ intersects the curve $y=2x-3x^3$ in the first quadrant (see image). Find c such that area a is equal to area b.

Thanks.

2. where a and b are the intersection points of these to functions

then solve $\int_0^a~c-(2x-3x^3)~dx = \int_a^b~2x-3x^3-c~dx$

Not 100% sure where it will lead but I think the logic is something to follow.

3. Originally Posted by pickslides
where a and b are the intersection points of these to functions

then solve $\int_0^a~c-(2x-3x^3)~dx = \int_a^b~2x-3x^3-c~dx$

Not 100% sure where it will lead but I think the logic is something to follow.
It seems to lead to $b^{2}-\frac{3}{4}b^{4}-cb =0$

But you also know that $2b-3b^{3}=c$

EDIT: I probably shouldn't be using a and b for limits.

4. Originally Posted by Random Variable
It seems to lead to $b^{2}-\frac{3}{4}b^{4}-cb =0$

But you also know that $2b-3b^{3}=c$

EDIT: I probably shouldn't be using a and b for limits.
Either way, I get the idea.
Thank you both.

5. I get $a = \frac{\text{-1}}{3} + \frac{\sqrt{3}}{3}, \ b = \frac{2}{3}, \ \text{and} \ c = \frac{12}{27}$

6. Originally Posted by Random Variable
I get $a = \frac{\text{-1}}{3} + \frac{\sqrt{3}}{3}, \ b = \frac{2}{3}, \ \text{and} \ c = \frac{12}{27}$
I got the same thing, $c=\frac{4}{9}$.
Thanks again.

7. Originally Posted by McScruffy
I got the same thing, $c=\frac{4}{9}$.
Thanks again.
I need to retake third grade math and relearn how to reduce a fraction.