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 = 3-4x at x = 0.5

7x^2 = 3-4x at .43

4x^2 < 7x^2, so we need to integrate

7x^2 - 4x^2 from 0 to 0.43

3-4x - 4x^2 from 0.43 to 0.5

Int(3x^2) = x^3 [0,0.43] = 0.08

Int(3-4x-4x^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 + 7x-20 G(x)= x+4 FIND f/g

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