CUET UG 2025 — Mathematics Calculus

The area bounded by the curve, the x-axis, and $x \in [0,1]$x∈[0,1] is:

$$A = \int_0^1 3\sqrt{1-x^2}\;dx$$A=∫01​31−x2​dx

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Let $x = \sin\theta \implies dx = \cos\theta\;d\theta$x=sinθ⟹dx=cosθdθ

When $x = 0$x=0, $\theta = 0$θ=0
When $x = 1$x=1, $\theta = \frac{\pi}{2}$θ=2π​

$$A = 3\int_0^{\pi/2} \sqrt{1 - \sin^2\theta}\;\cos\theta\;d\theta$$A=3∫0π/2​1−sin2θ​cosθdθ

Since $1 - \sin^2\theta = \cos^2\theta$1−sin2θ=cos2θ:

$$A = 3\int_0^{\pi/2} \cos^2\theta\;d\theta$$A=3∫0π/2​cos2θdθ

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Using $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$cos2θ=21+cos2θ​:

$$A = \dfrac{3}{2}\int_0^{\pi/2} (1 + \cos 2\theta)\;d\theta$$A=23​∫0π/2​(1+cos2θ)dθ

$$= \dfrac{3}{2}\left[\theta + \dfrac{\sin 2\theta}{2}\right]_0^{\pi/2}$$=23​[θ+2sin2θ​]0π/2​

$$= \dfrac{3}{2}\left[\left(\dfrac{\pi}{2} + \dfrac{\sin \pi}{2}\right) - \left(0 + 0\right)\right]$$=23​[(2π​+2sinπ​)−(0+0)]

$$= \dfrac{3}{2} \times \dfrac{\pi}{2}$$=23​×2π​

$$= \dfrac{3\pi}{4}$$=43π​

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The curve $y = 3\sqrt{1-x^2}$y=31−x2​ represents the upper half of the ellipse $\dfrac{x^2}{1} + \dfrac{y^2}{9} = 1$1x2​+9y2​=1 with $a=1$a=1, $b=3$b=3.

Full ellipse area $= \pi ab = 3\pi$=πab=3π

First quadrant portion $= \dfrac{1}{4} \times 3\pi = \dfrac{3\pi}{4}$=41​×3π=43π​ sq. units