Asked by johnathon
How do you find the ariea between these curves? y=4x^2 y=7x^2 4x+y=3 x>=0?
Answers
Answered by
Steve
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 . . .
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 . . .
Answered by
johnathon
thanks
the exact answer is 29/294
the exact answer is 29/294
Answered by
Anonymous
f(X)3x^2 + 7x-20 G(x)= x+4 FIND f/g
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