The area in the first quadrant bounded by y = 4x2, x = 0 y = 1, and y…

Required area =$\int\limits_1^4x\,dy$

$ =\int\limits_1^4\frac{\sqrt{y}}{2}dy =\frac{1}{2}[\frac{2}{3}y^{3/2}]_1^4$

$=\frac{1}{3}[4^{3/2}-1] = 2\frac{1}{3}$ sq. units.

Hence (C) is the correct answer.