Answer 1

Given equation of the curve is x² = 25y
∴ `5sqrt(y)`      ...[∵ In first quadrant, x > 0]
Required area = area of the region PRSVP
= 2(area of the region QRSTQ)

= `2int_1^4x*dy`

= `2int_1^4 5sqrt(y)*dy`

= `10 int_1^4 y^(1/2)*dy`

= `10[y^(3/2)/(3/2)]_1^4`

= `(20)/(3)[(4)(3/2) - (1)^(3/2)]`

= `(20)/(3)(8 - 1)`

= `(20)/(3)(7)`

= `(140)/(3)"sq. units"`.