Question

favorite kind of calculus problem

Answer 1

I dont really need help :D

Answer 2

Cool.

Answer 3

Nin ain't pellet

Answer 4

laughing out loud

Answer 5

yeah. nin is NIN

Answer 6

Oh yeah, my kind of problem. Ok I'm gonna try to figure this one out now give me a few minutes...

Answer 7

I remember looking at the solution manual for this one

Answer 8

I need to study cubics, quartic, and higher order polynomials more haha. I'm making progress but it's fairly slow but fun.

Answer 9

I think I figured it out. Tricky problem!!

Answer 10

Nope, I'm stuck. Hmm. Ok someone tell me if I'm going in the right direction or what:

Basically I'm calling these two x values that have the height C on the curve as x_0 and x_1 So to be clearer, \[\LARGE 0<x_0<x_1\]
To make the areas equal I set their integrals equal:\[\Large \int\limits_0^{x_0}Cdx-\int\limits_0^{x_0}2x-3x^3dx=\int\limits_{x_0}^{x_1}2x-3x^3dx\] I combined the cubic integrals on the right \[\Large \int\limits_0^{x_0}Cdx=\int\limits_{0}^{x_1}2x-3x^3dx\] and got:\[\Large Cx_0=x_1^2-\frac{3}{4}x_1^4\]
Also, remember that since x_0 and x_1 fall on the curve and have the same height C, then I can plug them into the cubic from earlier before integrating to get: \[\Large 2x_0-3x_0^3=2x_1-3x_1^3=C\] So I do some clever substitution: \[\Large Cx_0=\frac{x_1}{4}(2x_1+C)\] But this really doesn't solve my problem after I solve for C I get:\[\Large C=\frac{2x_1^2}{4x_0-x_1}\] So I need to find some way to figure out more.

Answer 11

\[\int\limits_{a}^{b}[(-3x^3+2x)-c]dx=0\]

Answer 12

\[-\frac{ 3b^4 }{ 4 }+b^2-bc=0\]

Answer 13

What's a and b supposed to be?

Answer 14

pellet that's wrong

Answer 15

\[c = -3b^3 + 2b\]

Answer 16

a =0 @Kainui

Answer 17

And what's b? I am not seeing why this integral should be zero.

Answer 18

it has to be zero just the same when we setting up with the value for the y or straight line.

Answer 19

label this with a and b.

Answer 20

right? then after we set that one up, we equate