CUET UG 2025 — Mathematics Calculus

The region is bounded by the curve $y^2 = 4x$y2=4x, the y-axis, and the line $y = 3$y=3.

The parabola $y^2 = 4x$y2=4x opens to the right and passes through the origin. When $y = 3$y=3, we get $9 = 4x$9=4x, so $x = \frac{9}{4}$x=49​.

The region has:

• Left boundary: y-axis ($x = 0$x=0)
• Right boundary: the parabola $y^2 = 4x$y2=4x
• y ranges from $0$0 to $3$3

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Integrating with respect to $y$y is more convenient since the boundaries are naturally expressed this way.

From the parabola $y^2 = 4x$y2=4x:

$$x = \frac{y^2}{4}$$x=4y2​

The area is given by:

$$\text{Area} = \int_0^3 \frac{y^2}{4} \, dy$$Area=∫03​4y2​dy

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$$\text{Area} = \frac{1}{4} \int_0^3 y^2 \, dy$$Area=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​

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