The curve for y = √x and the currve for y = x³ meet at a point P (x, y) where,
y = √x = x³
=> x = x⁶
=> x⁵ = 1 Or, x = 0
=> x = 1 and 0 are real solutions. perhaps the others are imaginary solutions.
The curve y = √x is above the curve for y= x³ in the interval [ 0, 1 ].
The integral for
The area enclosed by the curves in the first quadrant :
.Not very difficult.. right.