CUET UG 2025 — Mathematics Calculus

The absolute value function $|x|$∣x∣ is defined as:

$$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$∣x∣={x−x​if x≥0if x<0​

Since $|x|$∣x∣ changes its definition at $x = 0$x=0, the integral must be split at this point.

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$$\int_{-1}^{1}|x|dx = \int_{-1}^{0}|x|dx + \int_{0}^{1}|x|dx$$∫−11​∣x∣dx=∫−10​∣x∣dx+∫01​∣x∣dx

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For $x \in [-1, 0]$x∈[−1,0], we have $x \leq 0$x≤0, so $|x| = -x$∣x∣=−x

For $x \in [0, 1]$x∈[0,1], we have $x \geq 0$x≥0, so $|x| = x$∣x∣=x

$$\int_{-1}^{1}|x|dx = \int_{-1}^{0}(-x)dx + \int_{0}^{1}xdx$$∫−11​∣x∣dx=∫−10​(−x)dx+∫01​xdx

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$$\int_{-1}^{0}(-x)dx = \left[-\frac{x^2}{2}\right]_{-1}^{0}$$∫−10​(−x)dx=[−2x2​]−10​

$$= -\frac{(0)^2}{2} - \left(-\frac{(-1)^2}{2}\right)$$=−2(0)2​−(−2(−1)2​)

$$= 0 + \frac{1}{2}$$=0+21​

$$= \frac{1}{2}$$=21​

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$$\int_{0}^{1}xdx = \left[\frac{x^2}{2}\right]_{0}^{1}$$∫01​xdx=[2x2​]01​

$$= \frac{(1)^2}{2} - \frac{(0)^2}{2}$$=2(1)2​−2(0)2​

$$= \frac{1}{2} - 0$$=21​−0

$$= \frac{1}{2}$$=21​

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$$\int_{-1}^{1}|x|dx = \frac{1}{2} + \frac{1}{2}$$∫−11​∣x∣dx=21​+21​

$$= 1$$=1

Therefore, $\displaystyle \int_{-1}^{1}|x|dx = 1$∫−11​∣x∣dx=1