CUET UG 2025 — Mathematics Calculus

The region is bounded by the parabola $y = x^2$y=x2 and the horizontal line $y = 16$y=16.

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The curves intersect where:

$$x^2 = 16$$x2=16

$$x = \pm 4$$x=±4

The intersection points are at $x = -4$x=−4 and $x = 4$x=4.

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The area between the curves is:

$$\text{Area} = \int_{-4}^{4} (16 - x^2) \, dx$$Area=∫−44​(16−x2)dx

The upper curve is $y = 16$y=16 and the lower curve is $y = x^2$y=x2.

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The region is symmetric about the y-axis:

$$\text{Area} = 2 \int_{0}^{4} (16 - x^2) \, dx$$Area=2∫04​(16−x2)dx

$$= 2 \left[ 16x - \frac{x^3}{3} \right]_{0}^{4}$$=2[16x−3x3​]04​

$$= 2 \left[ \left(16(4) - \frac{4^3}{3}\right) - \left(16(0) - \frac{0^3}{3}\right) \right]$$=2[(16(4)−343​)−(16(0)−303​)]

$$= 2 \left[ 64 - \frac{64}{3} - 0 \right]$$=2[64−364​−0]

$$= 2 \left[ \frac{192}{3} - \frac{64}{3} \right]$$=2[3192​−364​]

$$= 2 \left[ \frac{128}{3} \right]$$=2[3128​]

$$= \frac{256}{3}$$=3256​

Therefore, the area of the region is $\dfrac{256}{3}$3256​ square units.