Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Definite Integration

Question:

Match List-I with List-II

List-I

List-II

(A) The value of $\int\limits_0^4|x|dx$ is

(I) 3

(B) The value of $\int\limits_{-2}^2|x|dx$ is

(II) -1

(C) The value of $\int\limits_0^3[x] dx$ is

(III) 8

(D) The value of $\int\limits_{-1}^1[x] dx$ is

(IV) 4

Choose the correct answer from the options given below:

Options:

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

(A)-(III), (B)-(II), (C)-(I), (D)-(IV)

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Correct Answer:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Explanation:

The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

List-I

List-II

(A) The value of $\int\limits_0^4|x|dx$ is

(III) 8

(B) The value of $\int\limits_{-2}^2|x|dx$ is

(IV) 4

(C) The value of $\int\limits_0^3[x] dx$ is

(I) 3

(D) The value of $\int\limits_{-1}^1[x] dx$ is

(II) -1

List-I and List-II Matching:


(A) $\displaystyle \int_0^4 |x|\, dx$

For $x \in [0, 4]$, $|x| = x$

$\Rightarrow \int_0^4 x\, dx = \left[\frac{x^2}{2}\right]_0^4 = \frac{16}{2} = 8$

⇒ (A) → (III)


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

Even function: $= 2 \int_0^2 x\, dx = 2 \cdot \left[\frac{x^2}{2}\right]_0^2 = 2 \cdot \frac{4}{2} = 4$

⇒ (B) → (IV)


(C) $\displaystyle \int_0^3 [x]\, dx$

$[x]$ is the greatest integer ≤ x

  • On $[0,1)$: $[x] = 0$ → Area = $0 \cdot 1 = 0$
  • On $[1,2)$: $[x] = 1$ → Area = $1 \cdot 1 = 1$
  • On $[2,3)$: $[x] = 2$ → Area = $2 \cdot 1 = 2$

Total = $0 + 1 + 2 = 3$

⇒ (C) → (I)


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

On $[-1, 0)$: $[x] = -1$ → Interval length = 1 → Area = $-1 \cdot 1 = -1$

On $[0,1)$: $[x] = 0$ → Area = $0$

Total = $-1 + 0 = -1$

⇒ (D) → (II)