Match List-I with List-II

List-I	List-II
(A) The value of $\int_{0}^{4} \vert x\vert \, dx$ is	(I) 3
(B) The value of $\int_{-2}^{2} \vert x\vert \, dx$ is	(II) -1
(C) The value of $\int_{0}^{3} [x] \, dx$ is	(III) 8
(D) The value of $\int_{-1}^{1} [x] \, dx$ is	(IV) 4

Choose the correct answer from the options given below:

(A) $\int_{0}^{4} |x| \, dx$

Since $x \geq 0$ over $[0, 4]$, $|x| = x$.

$\int_{0}^{4} x \, dx = \left[\dfrac{x^2}{2}\right]_0^4$

$= \dfrac{16}{2} - 0$

$= 8 \longrightarrow \text{(III)}$

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(B) $\int_{-2}^{2} |x| \, dx$

$|x|$ is an even function, so:

$\int_{-a}^{a} |x|\,dx = 2\int_{0}^{a} x\,dx$

$= 2\int_{0}^{2} x \, dx$

$= 2\left[\dfrac{x^2}{2}\right]_0^2$

$= 2 \times 2$

$= 4 \longrightarrow \text{(IV)}$

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(C) $\int_{0}^{3} [x] \, dx$

Here $[x]$ is the greatest integer function (floor function) — it gives the largest integer $\leq x$.

Splitting at every integer where $[x]$ changes value:

$= \int_{0}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx$

$= 0 + 1(2-1) + 2(3-2)$

$= 0 + 1 + 2$

$= 3 \longrightarrow \text{(I)}$

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(D) $\int_{-1}^{1} [x] \, dx$

Splitting at integers where $[x]$ changes:

On $[-1, 0)$, $[x] = -1$

On $[0, 1)$, $[x] = 0$

$= \int_{-1}^{0} (-1) \, dx + \int_{0}^{1} 0 \, dx$

$= (-1)(0 - (-1)) + 0$

$= -1 + 0$

$= -1 \longrightarrow \text{(II)}$

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(A) $\rightarrow$ (III), (B) $\rightarrow$ (IV), (C) $\rightarrow$ (I), (D) $\rightarrow$ (II)