Recent content by RUStudent

Problem Statment: Int (32/((x^2)(sqrt(16-(x^2)))))dx Problem Attempt: a = 4 x = 4sin(w) w = theta dx = 4cos(w) Worked the problem down to: 2*int(1/(sin^2(w)))dw

Wow, thank you so much, I don't know why I couldn't see that. a=0.41421. Thanks again.

Common denominator is (1+a) so I have ((3(1+a)-2-2a)/(1+a))(1/(1+a))^2. This would give me (3(1+a)-2-2a)(1/(1+a))^3=1/2. I am so frustrated. I don't get it!

Yes I did that but I don't get it, there is still a cubic.

I'm confused. Now I have (1/1+a)^2(3-2(1/1+a)-2a(1/1+a))=1/2. I still have the cubic terms.

I'm not sure I understand. After I integrate substitute x for (1/1+a). If I do this I get "((1/1+a)^2/2)-((1/1+a)^3/3)-a((1/1+a)^3/3) and they still do not cancel.

Yes sorry that was a typo. After integration I get (x^2/x)-(x^3/3)-a(x^3/3). I set this equal to 1/12 but as you see the cubic terms do not cancel.

That is what I thought I was doing wrong but I have an addition of the cubics. My integral is x-x^2 from 0 to (1/1+a) - ax^2 from x to (1/1+a).

That is what I have done. But I come up with an equation too hard to solve for a. I believe my problem lies in my algebra.

Homework Statement Find the parabola y=a(x^2) that divides the area under the curve y=x(1-x) over [0,1] into two regions of equal area. Homework Equations I set the two equations equal to each other to solve that the intersection point is x=1/(1+a). I solved for the entire area "definite...