CUET UG 2025 — Mathematics Algebra

The region is bounded by:

• Parabola: $y = x^2 + 2$y=x2+2
• Line: $y = x$y=x
• Vertical boundaries: $x = 0$x=0 and $x = 2$x=2

To determine which curve is on top, test at $x = 1$x=1:

For $y = x^2 + 2$y=x2+2: $y = (1)^2 + 2 = 3$y=(1)2+2=3

For $y = x$y=x: $y = 1$y=1

Since $3 > 1$3>1, the parabola $y = x^2 + 2$y=x2+2 is above the line $y = x$y=x in this interval.

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The area between curves is given by:

$$\text{Area} = \int_0^2 [(x^2 + 2) - x] \, dx$$Area=∫02​[(x2+2)−x]dx

$$= \int_0^2 (x^2 - x + 2) \, dx$$=∫02​(x2−x+2)dx

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Integrating term by term:

$$= \left[\frac{x^3}{3} - \frac{x^2}{2} + 2x\right]_0^2$$=[3x3​−2x2​+2x]02​

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At $x = 2$x=2:

$$\frac{(2)^3}{3} - \frac{(2)^2}{2} + 2(2)$$3(2)3​−2(2)2​+2(2)

$$= \frac{8}{3} - \frac{4}{2} + 4$$=38​−24​+4

$$= \frac{8}{3} - 2 + 4$$=38​−2+4

At $x = 0$x=0:

$$\frac{0}{3} - \frac{0}{2} + 0 = 0$$30​−20​+0=0

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$$\text{Area} = \frac{8}{3} - 2 + 4$$Area=38​−2+4

$$= \frac{8}{3} - \frac{6}{3} + \frac{12}{3}$$=38​−36​+312​

$$= \frac{8 - 6 + 12}{3}$$=38−6+12​

$$= \frac{14}{3}$$=314​

Therefore, the area of the region is $\frac{14}{3}$314​ square units.