CUET UG 2025 — Mathematics Calculus

The region is bounded by $y^2 = 9x$y2=9x, $x = 2$x=2, $x = 4$x=4, and the $x$x-axis in the first quadrant.

Given: $y^2 = 9x$y2=9x

In the first quadrant, $y$y is positive:

$$y = \sqrt{9x}$$y=9x​

$$y = 3\sqrt{x}$$y=3x​

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The area under the curve from $x = 2$x=2 to $x = 4$x=4 is:

$$\text{Area} = \int_2^4 3\sqrt{x} \, dx$$Area=∫24​3x​dx

$$= \int_2^4 3x^{1/2} \, dx$$=∫24​3x1/2dx

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$$\int 3x^{1/2} \, dx = 3 \times \dfrac{x^{3/2}}{3/2}$$∫3x1/2dx=3×3/2x3/2​

$$= 3 \times \dfrac{2}{3} \times x^{3/2}$$=3×32​×x3/2

$$= 2x^{3/2}$$=2x3/2

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$$\text{Area} = \left[2x^{3/2}\right]_2^4$$Area=[2x3/2]24​

$$= 2(4)^{3/2} - 2(2)^{3/2}$$=2(4)3/2−2(2)3/2

For $x = 4$x=4:

$$4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$$43/2=(4​)3=23=8

$$2(4)^{3/2} = 2 \times 8 = 16$$2(4)3/2=2×8=16

For $x = 2$x=2:

$$2^{3/2} = (\sqrt{2})^3 = \sqrt{2} \times \sqrt{2} \times \sqrt{2} = 2\sqrt{2}$$23/2=(2​)3=2​×2​×2​=22​

$$2(2)^{3/2} = 2 \times 2\sqrt{2} = 4\sqrt{2}$$2(2)3/2=2×22​=42​

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$\text{Area} = 16 - 4\sqrt{2}$Area=16−42​ square units

Therefore, the area of the region is $16 - 4\sqrt{2}$16−42​ square units.