CUET UG 2025 — Mathematics Calculus

The integral to evaluate is $\displaystyle \int_0^2 (|x| + |x - 2|) dx$∫02​(∣x∣+∣x−2∣)dx

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For $|x|$∣x∣ on the interval $[0, 2]$[0,2]:

Since $x \geq 0$x≥0 throughout $[0, 2]$[0,2], we have $|x| = x$∣x∣=x

For $|x - 2|$∣x−2∣ on the interval $[0, 2]$[0,2]:

Since $x \leq 2$x≤2 throughout $[0, 2]$[0,2], we have $x - 2 \leq 0$x−2≤0

Therefore $|x - 2| = -(x - 2) = 2 - x$∣x−2∣=−(x−2)=2−x

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Substituting into the expression:

$$|x| + |x - 2| = x + (2 - x)$$∣x∣+∣x−2∣=x+(2−x)

$$= x + 2 - x$$=x+2−x

$$= 2$$=2

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The integral becomes:

$$\int_0^2 2 \, dx$$∫02​2dx

$$= 2x \Big|_0^2$$=2x​02​

$$= 2(2) - 2(0)$$=2(2)−2(0)

$$= 4$$=4

Therefore, the value of the integral is $4$4.