Match List-I with List-II

[.] denotes the greatest integer function.

List-I	List-II
(A) $\int_0^3 [x]dx$	(I) $\frac{1}{2}$
(B) $\int_0^1 [2x]dx$	(II) 1
(C) $\int_0^1 [3x]dx$	(III) $\frac{3}{2}$
(D) $\int_0^1 [4x]dx$	(IV) 3

Choose the correct answer from the options given below:

The greatest integer function $[x]$ (floor function) gives the largest integer less than or equal to $x$. The function $[x]$ stays constant between consecutive integers.

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For $\int_0^3 [x]dx$:

The interval is broken based on where $[x]$ changes value:

When $0 \leq x < 1$: $[x] = 0$

When $1 \leq x < 2$: $[x] = 1$

When $2 \leq x < 3$: $[x] = 2$

$\int_0^3 [x]dx = \int_0^1 0\,dx + \int_1^2 1\,dx + \int_2^3 2\,dx$

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

$= 0 + 1 + 2$

$= 3$

Therefore (A) matches with (IV)

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For $\int_0^1 [2x]dx$:

The function $[2x]$ changes when $2x$ crosses integers:

When $0 \leq x < \frac{1}{2}$: $2x \in [0,1)$, so $[2x] = 0$

When $\frac{1}{2} \leq x < 1$: $2x \in [1,2)$, so $[2x] = 1$

$\int_0^1 [2x]dx = \int_0^{1/2} 0\,dx + \int_{1/2}^1 1\,dx$

$= 0 + 1 \times (1-\frac{1}{2})$

$= \frac{1}{2}$

Therefore (B) matches with (I)

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For $\int_0^1 [3x]dx$:

The function $[3x]$ changes when $3x$ crosses integers:

When $0 \leq x < \frac{1}{3}$: $3x \in [0,1)$, so $[3x] = 0$

When $\frac{1}{3} \leq x < \frac{2}{3}$: $3x \in [1,2)$, so $[3x] = 1$

When $\frac{2}{3} \leq x < 1$: $3x \in [2,3)$, so $[3x] = 2$

$\int_0^1 [3x]dx = \int_0^{1/3} 0\,dx + \int_{1/3}^{2/3} 1\,dx + \int_{2/3}^1 2\,dx$

$= 0 + 1 \times (\frac{2}{3}-\frac{1}{3}) + 2 \times (1-\frac{2}{3})$

$= 0 + 1 \times \frac{1}{3} + 2 \times \frac{1}{3}$

$= \frac{1}{3} + \frac{2}{3}$

$= 1$

Therefore (C) matches with (II)

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For $\int_0^1 [4x]dx$:

The function $[4x]$ changes when $4x$ crosses integers:

When $0 \leq x < \frac{1}{4}$: $[4x] = 0$

When $\frac{1}{4} \leq x < \frac{1}{2}$: $[4x] = 1$

When $\frac{1}{2} \leq x < \frac{3}{4}$: $[4x] = 2$

When $\frac{3}{4} \leq x < 1$: $[4x] = 3$

$\int_0^1 [4x]dx = \int_0^{1/4} 0\,dx + \int_{1/4}^{1/2} 1\,dx + \int_{1/2}^{3/4} 2\,dx + \int_{3/4}^1 3\,dx$

$= 0 + 1(\frac{1}{4}) + 2(\frac{1}{4}) + 3(\frac{1}{4})$

$= \frac{1}{4} + \frac{2}{4} + \frac{3}{4}$

$= \frac{6}{4}$

$= \frac{3}{2}$

Therefore (D) matches with (III)

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Final matching:

(A) → (IV) = 3

(B) → (I) = $\frac{1}{2}$

(C) → (II) = 1

(D) → (III) = $\frac{3}{2}$