Solving for C: Calculating Area Between Two Functions in Calculus

  • Thread starter Thread starter CACain
  • Start date Start date
  • Tags Tags
    Calculus
CACain
Messages
6
Reaction score
0
Okay, you have 2 equations (and no points on the graph given) -- You have f(x) = 2x-3x^3 and f(x) = c. You are to find the area under the curve between those 2 functions and are to make it equal with the area between the y-axis and the 2 curves.

So, I started this problem initially by creating a graph, and putting all my points on it. I integrated both the areas with respect to x.. and I used (for the area bounded by the y-axis and the curves -- which I marked A1) the bounds 0 to 2x-3x^3.

For the second one, I don't know what bounds to use.. and I think this is where I messed up initially... I used my original bounds which are now clearly wrong...

I'm so stumped, and it's most likely something silly...

Edit: in my sleepy stuper, I forgot to include (although it may be obvious) that you are to be solving for C :)
 
Last edited:
Physics news on Phys.org
I can't make heads or tails out of what you are saying. To start with you are given y= 2x- 3x3 and y= c and you are asked to find the area between them. Okay, that would be the integral of 2x- 3x3- c where the limits of integration are two points at which the curves intersect (NOT "0 to 2x- 3x3"). Of course for many values of c, those two curves might not intersect twice so there would be no area. I have no idea what you mean by "the area between the y-axis and the 2 curves". There is no one region "between the y-axis and the 2 curves" so there is no one area.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top