CUET UG 2025 — Mathematics Calculus

The region is bounded by the line $x + y = 2$x+y=2, the y-axis ($x = 0$x=0), the vertical line $x = 3$x=3, and the x-axis.

From $x + y = 2$x+y=2:

$$y = 2 - x$$y=2−x

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To find where the line crosses the x-axis, set $y = 0$y=0:

$$0 = 2 - x$$0=2−x

$$x = 2$$x=2

The line crosses the x-axis at $(2, 0)$(2,0).

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The region splits into two parts at $x = 2$x=2:

From $x = 0$x=0 to $x = 2$x=2: The line is above the x-axis

• At $x = 0$x=0: $y = 2$y=2
• At $x = 2$x=2: $y = 0$y=0

From $x = 2$x=2 to $x = 3$x=3: The line is below the x-axis

• At $x = 3$x=3: $y = -1$y=−1

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For the region from $x = 0$x=0 to $x = 2$x=2 (triangle above x-axis):

Base $= 2$=2 units

Height $= 2$=2 units

Area $= \dfrac{1}{2} \times 2 \times 2$=21​×2×2

Area $= 2$=2

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For the region from $x = 2$x=2 to $x = 3$x=3 (triangle below x-axis):

Base $= 1$=1 unit

Height $= 1$=1 unit

Area $= \dfrac{1}{2} \times 1 \times 1$=21​×1×1

Area $= \dfrac{1}{2}$=21​

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Total Area $= 2 + \dfrac{1}{2}$=2+21​

Total Area $= \dfrac{4}{2} + \dfrac{1}{2}$=24​+21​

Total Area $= \dfrac{5}{2}$=25​

Therefore, the area of the region is $\dfrac{5}{2}$25​ square units.