CUET UG 2025 — Mathematics Calculus

The curve $y^2 = x$y2=x is a sideways parabola opening to the right with vertex at the origin $(0,0)$(0,0).

The line $x = 4$x=4 is a vertical line.

Since $y^2 = x$y2=x gives $y = +\sqrt{x}$y=+x​ (upper half) and $y = -\sqrt{x}$y=−x​ (lower half), the parabola is symmetric about the $x$x-axis.

The region runs from $x = 0$x=0 (vertex) to $x = 4$x=4.

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$$\text{Total Area} = \int_0^4 \left[\sqrt{x} - (-\sqrt{x})\right] dx$$Total Area=∫04​[x​−(−x​)]dx

$$= \int_0^4 2\sqrt{x} \, dx$$=∫04​2x​dx

$$= 2\int_0^4 x^{1/2} \, dx$$=2∫04​x1/2dx

$$= 2 \left[\dfrac{x^{3/2}}{3/2}\right]_0^4$$=2[3/2x3/2​]04​

$$= 2 \times \dfrac{2}{3}\left[x^{3/2}\right]_0^4$$=2×32​[x3/2]04​

$$= \dfrac{4}{3}\left[x^{3/2}\right]_0^4$$=34​[x3/2]04​

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

$$= \dfrac{4}{3}\left[8 - 0\right]$$=34​[8−0]

$$= \dfrac{4}{3} \times 8$$=34​×8

$= \dfrac{32}{3}$=332​ sq. units