The correct option is D π/8
∫10√x(1−x)dx=∫10√x−x2dx
∫10√(12)−(x−12)2dx
∫√(12)2−(x−12)2dx=x2√(12)2−(x−12)2+(12)22sin−1((x−12)12)+c
=x2√x−x2+18sin−1(2x−1)+c
∫60√x(1−x)dx=(x2)√x(1−x)+18sin−1(2x−1)+c)∫10
=18[sin−1(1)−sin−1(−1)]
=18[π2−(−π2)]
=18(π)
∫10√x(1−x)dx=π8