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Home Questions 3SUETBNAGV
Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Definite Integration

Question:

Match List-I with List-II

List-I (Integral)

List-II (Value)

(A) $\int\limits_{-1}^1(|x|+1)dx$

(I) 0

(B) $\int\limits_{-2}^2|x+1|dx$

(II) 2

(C) $\int\limits_{-1}^13|x|^2dx$

(III) 5

(D) $\int\limits_{-1}^1x|x|dx$

(IV) 3

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

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

List-I (Integral)

List-II (Value)

(A) $\int\limits_{-1}^1(|x|+1)dx$

(IV) 3

(B) $\int\limits_{-2}^2|x+1|dx$

(III) 5

(C) $\int\limits_{-1}^13|x|^2dx$

(II) 2

(D) $\int\limits_{-1}^1x|x|dx$

(I) 0

(A) $\int_{-1}^{1}\left(|x|+1\right)\,dx = \int_{-1}^{1}|x|\,dx+\int_{-1}^{1}1\,dx = 2\int_{0}^{1}x\,dx+2 = 2\cdot\frac{1}{2}+2=3$ → (IV)

(B) $\int_{-2}^{2}|x+1|\,dx = \int_{-2}^{-1}-(x+1)\,dx+\int_{-1}^{2}(x+1)\,dx = \left[-\frac{x^{2}}{2}-x\right]_{-2}^{-1}+\left[\frac{x^{2}}{2}+x\right]_{-1}^{2} = \frac{1}{2}+\frac{9}{2}=5$ → (III)

(C) $\int_{-1}^{1}3|x|^{2}\,dx = 3\int_{-1}^{1}x^{2}\,dx = 3\left[\frac{x^{3}}{3}\right]_{-1}^{1} = 3\cdot\frac{2}{3}=2$ → (II)

(D) $x|x|$ is odd, hence $\int_{-1}^{1}x|x|\,dx=0$ → (I)

Final: (A)–(IV), (B)–(III), (C)–(II), (D)–(I)

CUET Questions