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x² + y² - 4x + y - 1 = 0
( x² - 4x ) + ( y² + y ) = 1
( x² - 4x + 4 ) + [ y² + y + (1/4) ] = 1 + 4 + (1/4)
( x - 2 )² + [ y + (1/2) ]² = [ √(21) / 2 ]²
This equation represents a Circle with
Centre ≡ ( 2, -1/2 ) and Radius = √(21) / 2.
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What does y24x-15 equa?
If you mean: y = 24x-15 then it is a straight line equation
How do you find the area between the curves y24x and y2x using integration?
You need to be more accurate and precise in submitting problems of this nature. Regardless, I think I have it figured out. The functionsY2 = 4XY = 2XTo find area here you need bounds. The graph admits of only these bounds; 0 to 1. Y2 = 4X is the top function.Rewrite.Y2 = 4XY = sqrt(4X)Y = 4X 1/2------------------integrateint( 4X 1/2 - 2X) dx= (8/3)X 3/2 - X2----------------------------------insert bounds(8/3)(1) 3/2 - (0)2= 8/3======the area between these functions
