CUET Preparation Today
The area in the first quadrant bounded by y = 4x2, x = 0 y = 1, and y = 4 is |
2 sq. units $2\frac{1}{2}$ sq. units $2\frac{1}{3}$ sq. units 3 sq. units |
$2\frac{1}{3}$ sq. units |
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. |
