I could use some help with the setup on this one: The line $\displaystyle y=c$ intersects the curve $\displaystyle y=2x-3x^3$ in the first quadrant (see image). Find c such that area a is equal to area b. Thanks.
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where a and b are the intersection points of these to functions then solve $\displaystyle \int_0^a~c-(2x-3x^3)~dx = \int_a^b~2x-3x^3-c~dx$ Not 100% sure where it will lead but I think the logic is something to follow.
Originally Posted by pickslides where a and b are the intersection points of these to functions then solve $\displaystyle \int_0^a~c-(2x-3x^3)~dx = \int_a^b~2x-3x^3-c~dx$ Not 100% sure where it will lead but I think the logic is something to follow. It seems to lead to $\displaystyle b^{2}-\frac{3}{4}b^{4}-cb =0 $ But you also know that $\displaystyle 2b-3b^{3}=c $ EDIT: I probably shouldn't be using a and b for limits.
Originally Posted by Random Variable It seems to lead to $\displaystyle b^{2}-\frac{3}{4}b^{4}-cb =0 $ But you also know that $\displaystyle 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.
I get $\displaystyle a = \frac{\text{-1}}{3} + \frac{\sqrt{3}}{3}, \ b = \frac{2}{3}, \ \text{and} \ c = \frac{12}{27} $
Originally Posted by Random Variable I get $\displaystyle a = \frac{\text{-1}}{3} + \frac{\sqrt{3}}{3}, \ b = \frac{2}{3}, \ \text{and} \ c = \frac{12}{27} $ I got the same thing, $\displaystyle c=\frac{4}{9}$. Thanks again.
Originally Posted by McScruffy I got the same thing, $\displaystyle c=\frac{4}{9}$. Thanks again. I need to retake third grade math and relearn how to reduce a fraction.
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