Posted by johnathon on Monday, October 24, 2011 at 11:42am.
How do you find the ariea between these curves? y=4x^2 y=7x^2 4x+y=3 x>=0?

calculus  Steve, Monday, October 24, 2011 at 12:11pm
Find where the graphs intersect. That will give you the limits of integration.
4x^2 = 7x^2 at x=0
4x^2 = 34x at x = 0.5
7x^2 = 34x at .43
4x^2 < 7x^2, so we need to integrate
7x^2  4x^2 from 0 to 0.43
34x  4x^2 from 0.43 to 0.5
Int(3x^2) = x^3 [0,0.43] = 0.08
Int(34x4x^2) = 3x  2x^2  4/3 x^3 [0.43,0.5) = 0.02
So, the total area = 0.10
If my math is right . . . 
calculus  johnathon, Monday, October 24, 2011 at 1:52pm
thanks
the exact answer is 29/294 
calculus  Anonymous, Sunday, September 9, 2012 at 8:41pm
f(X)3x^2 + 7x20 G(x)= x+4 FIND f/g