How do you find the ariea between these curves? y=4x^2 y=7x^2 4x+y=3 x>=0?
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 . . .
thanks
the exact answer is 29/294
f(X)3x^2 + 7x-20 G(x)= x+4 FIND f/g
To find the area between two curves, you need to determine the points of intersection between the curves and integrate the difference in their y-values.
First, let's find the points of intersection between the curves y = 4x^2 and y = 7x^2. To find these points, set the two equations equal to each other:
4x^2 = 7x^2
Subtracting 7x^2 from both sides gives:
3x^2 = 0
Dividing both sides by 3, we get:
x^2 = 0
Taking the square root of both sides, we find:
x = 0
So, the curves intersect at x = 0.
Next, we need to determine the bounds of integration. Since the constraint x >= 0 is given, the lower bound of integration will be 0. We need to find the upper bound of integration, which is the x-value where the two curves intersect.
Now we can calculate the area between the curves by integrating the difference in their y-values. The integral will be:
∫[0, upper bound] (7x^2 - 4x^2) dx
Simplifying the integral gives:
∫[0, upper bound] (3x^2) dx
Integrating with respect to x, we get:
(x^3) | [0, upper bound]
Substituting the limits of integration into the integral, we obtain:
(upper bound^3) - (0^3) = upper bound^3
Finally, the area between the curves is given by the upper bound cubed. To find the upper bound, we need to substitute the x-value where the curves intersect into the expression for the upper bound cubed. In this case, since the curves intersect at x = 0, the upper bound is 0.
Therefore, the area between the curves y = 4x^2 and y = 7x^2, with the constraint x >= 0, is 0.