MHTCET 20 Apr 2026 Shift 1 Paper

Concept:The region is bounded between $y = 2$ and $y = 4$, so we use horizontal strips and integrate with respect to $y$.For each strip, the length is the $x$-coordinate of the curve in the first quadrant.Explanation:The given curve is $y = 4x^2$.Since the region lies in the first quadrant, we rewrite the curve as:$$x = \frac{\sqrt{y}}{2}$$The region is bounded on the left by the Y-axis, so the strip length is:$$\frac{\sqrt{y}}{2} - 0 = \frac{\sqrt{y}}{2}$$These strips are taken from $y = 2$ to $y = 4$.Hence, the required area is:$$\int_{2}^{4} \frac{\sqrt{y}}{2} \, dy = \frac{1}{2} \int_{2}^{4} y^{1/2} \, dy$$Applying the power rule of integration:$$= \frac{1}{2} \left[ \frac{2}{3} y^{3/2} \right]_{2}^{4} = \frac{1}{3} \left[ y^{3/2} \right]_{2}^{4}$$Substituting the limits:$$= \frac{1}{3} \left( 4^{3/2} - 2^{3/2} \right) = \frac{1}{3} (8 - 2\sqrt{2})$$Answer:The area of the region is $\frac{1}{3}[8 - 2\sqrt{2}]$ sq. units.Therefore, the correct option is Option B.