The area bounded by the loop of the curve `y^(2) = x^(2)(1-x)` is

Explanation

Replacing y by - y in the equation ` y^2 = x ^2 ( 1 - x ) `, the equation does not change.
` therefore ` the curve is symmetric about X - axis.
To find the points of intersection of this curve with the X- axis,
Putting ` y = 0 `, we get
` x ^ 2 ( 1 - x ) = 0 " " therefore x = 0 , x = 1 `
Hence the curve intersects the X - axis at ` O( 0, 0 ) and A ( 1, 0 ) ` and it is symmetric about X - axis.
` therefore ` loop of the curve is forme between ` x = 0 ` and ` x = 1 `

Required area = area of region OCABO
= 2 ( area of region OABO )
= ` 2 int _ 0 ^ ( 1 ) y dx `, where ` y ^ ( 2 ) = x ^ ( 2 ) ( 1 - x ) `, i.e., ` y = x sqrt ( 1 - x ) `
` = 2 int _ 0 ^ ( 1 ) x sqrt ( 1 - x ) dx `
` = 2 int _ 0 ^ ( 1 ) ( 1 - x ) sqrt ( 1 - ( 1- x )) dx ... [ because int _ 0 ^ ( a) f ( x ) d x = int _ 0 ^ ( a) f ( a - x ) dx ] `
` = 2 int _ 0 ^ ( 1 ) sqrt x ( 1 - x ) dx " " = 2 int _ 0 ^ 1 ( x ^ (1//2 ) - x ^ ( 3//2) ) dx `
` = 2 [ ( x ^(3//2))/( 3//2) - ( x ^(5//2))/( 5//2) ] _ 0 ^ (1) " " = 2 [ ( 2 ) / ( 3) ( 1 - 0 ) - ( 2 ) / ( 5) ( 1 - 0 ) ] `
` = 2 ( (2 ) / ( 3) - ( 2) /( 5) ) = 2 ( ( 4 ) / ( 15) ) = ( 8 ) / ( 15) ` sq units.