CUET UG 2025 — Mathematics Calculus

The equation of the curve is $y^2 = 4x$y2=4x, which represents a rightward-opening parabola.

Since the region is restricted to the first quadrant, we take the positive root for $y$y:
$$y = 2\sqrt{x}$$y=2x​

Step-by-Step Solution:

The area bounded by the curve between the lines $x = 1$x=1 and $x = 2$x=2 is given by the definite integral:
$$\text{Area} = \int_{1}^{2} y \, dx$$Area=∫12​ydx
$$\text{Area} = \int_{1}^{2} 2\sqrt{x} \, dx$$Area=∫12​2x​dx

Integrating with respect to $x$x:
$$\text{Area} = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{2}$$Area=2[3/2x3/2​]12​
$$\text{Area} = \frac{4}{3} \left[ x^{3/2} \right]_{1}^{2}$$Area=34​[x3/2]12​

Now, substitute the upper and lower limits:
$$\text{Area} = \frac{4}{3} \left( 2^{3/2} - 1^{3/2} \right)$$Area=34​(23/2−13/2)
$$\text{Area} = \frac{4}{3} (2\sqrt{2} - 1) \text{ sq. units}$$Area=34​(22​−1) sq. units