CUET UG 2025 — Mathematics Calculus

The region is bounded by:

• $y^2 = 4x$y2=4x: A parabola opening to the right
• $y$y-axis: The line $x = 0$x=0
• $y = 3$y=3: A horizontal line at height 3

The region is a slice between the $y$y-axis (left edge) and the parabola (right edge), from $y = 0$y=0 to $y = 3$y=3.

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From the parabola equation $y^2 = 4x$y2=4x, we express $x$x in terms of $y$y:

$$x = \frac{y^2}{4}$$x=4y2​

The region extends from $x = 0$x=0 (the $y$y-axis) to $x = \frac{y^2}{4}$x=4y2​ (the parabola) for each value of $y$y from $0$0 to $3$3.

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The area is given by:

$$\text{Area} = \int_{0}^{3} \left(\frac{y^2}{4} - 0\right) dy$$Area=∫03​(4y2​−0)dy

$$= \int_{0}^{3} \frac{y^2}{4} dy$$=∫03​4y2​dy

$$= \frac{1}{4} \int_{0}^{3} y^2 \, dy$$=41​∫03​y2dy

$$= \frac{1}{4} \left[\frac{y^3}{3}\right]_{0}^{3}$$=41​[3y3​]03​

$$= \frac{1}{4} \left[\frac{27}{3} - 0\right]$$=41​[327​−0]

$$= \frac{1}{4} \times 9$$=41​×9

$$= \frac{9}{4}$$=49​

Therefore, the area of the region is $\frac{9}{4}$49​ square units.