CUET UG 2025 — Mathematics Calculus

The absolute value $|x - 2|$∣x−2∣ changes behavior at $x = 2$x=2.

When $x < 2$x<2: $(x - 2)$(x−2) is negative, so $|x - 2| = -(x - 2) = 2 - x$∣x−2∣=−(x−2)=2−x

When $x \geq 2$x≥2: $(x - 2)$(x−2) is positive, so $|x - 2| = x - 2$∣x−2∣=x−2

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Since the limits go from 1 to 4, and the behavior changes at $x = 2$x=2:

$$\int_1^4 |x - 2|dx = \int_1^2 |x - 2|dx + \int_2^4 |x - 2|dx$$∫14​∣x−2∣dx=∫12​∣x−2∣dx+∫24​∣x−2∣dx

$$= \int_1^2 (2 - x)dx + \int_2^4 (x - 2)dx$$=∫12​(2−x)dx+∫24​(x−2)dx

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For the first integral:

$$\int_1^2 (2 - x)dx = \left[2x - \frac{x^2}{2}\right]_1^2$$∫12​(2−x)dx=[2x−2x2​]12​

$$= \left(2(2) - \frac{4}{2}\right) - \left(2(1) - \frac{1}{2}\right)$$=(2(2)−24​)−(2(1)−21​)

$$= (4 - 2) - (2 - 0.5)$$=(4−2)−(2−0.5)

$$= 2 - 1.5$$=2−1.5

$$= \frac{1}{2}$$=21​

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For the second integral:

$$\int_2^4 (x - 2)dx = \left[\frac{x^2}{2} - 2x\right]_2^4$$∫24​(x−2)dx=[2x2​−2x]24​

$$= \left(\frac{16}{2} - 8\right) - \left(\frac{4}{2} - 4\right)$$=(216​−8)−(24​−4)

$$= (8 - 8) - (2 - 4)$$=(8−8)−(2−4)

$$= 0 - (-2)$$=0−(−2)

$$= 2$$=2

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Adding both parts:

$$\int_1^4 |x - 2|dx = \frac{1}{2} + 2$$∫14​∣x−2∣dx=21​+2

$$= \frac{1}{2} + \frac{4}{2}$$=21​+24​

$$= \frac{5}{2}$$=25​