Thread: areas bounded by curves

I have to find the area of the area bound by y^2=4x and y=2x-4

I have to find the points of intersection. would I do this by solving 2x-4 = 4x or by 2x-4 = sqrt(4x) or both?

No, y is not equal to y^2! You can write y= sqrt(4x)= 2sqrt(x) and set that equal to 2x- 4. But it might be simpler to write x= y^2/4 so that y= 2(y^2/4)- 4= y^2/2- 4 and solve that quadratic equation.

im slightly confused by the response. what would a, b ,and c be then for the quadratic equation. well c = -4, but what would a and b be?

I keep getting x=4, but there should be another one that is x=-2 according to the back of the book. because the interval is -2<x<4

Originally Posted by colorado

I keep getting x=4, but there should be another one that is x=-2 according to the back of the book. because the interval is -2<x<4

No. This is the integral .
Because the two curves meet at .

now I am confused more. one of those days. how are you getting the intersection points, and I still don't see how i can get a quadratic equation out of this

Originally Posted by colorado

now I am confused more. one of those days. how are you getting the intersection points, and I still don't see how i can get a quadratic equation out of this

Just solve .

and there it is, thank you