Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Definite Integration

Question:

Match List-I with List-II

List-I Definite integral

List-II Value

(A) $\int\limits_1^e\frac{\log x}{x}dx$

(I) 4

(B) $\int\limits_{-2}^2x^3(1 – x^2)dx$

(II) $\frac{1}{2}$

(C) $\int\limits_1^2x\, dx$

(III) 0

(D) $\int\limits_{-2}^2|x| dx$

(IV) $\frac{3}{2}$

Choose the correct answer from the options given below.

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (1) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

List-I Definite integral

List-II Value

(A) $\int\limits_1^e\frac{\log x}{x}dx$

(II) $\frac{1}{2}$

(B) $\int\limits_{-2}^2x^3(1 – x^2)dx$

(III) 0

(C) $\int\limits_1^2x\, dx$

(IV) $\frac{3}{2}$

(D) $\int\limits_{-2}^2|x| dx$

(I) 4

(A) ∫ from 1 to e of $\frac{\log x}{x}\,dx$

Put $u=\log x,\; du=\frac{1}{x}dx$

$=\int_{0}^{1} u\,du = \frac{1}{2}$

(B) ∫ from −2 to 2 of $x^{3}(1-x^{2})\,dx$

$x^{3}$ is odd and $(1-x^{2})$ is even ⇒ product is odd ⇒ integral over symmetric limits is $0$

(C) ∫ from 1 to 2 of $x\,dx$

$=\frac{x^{2}}{2}\Big|_{1}^{2}=\frac{4-1}{2}=\frac{3}{2}$

(D) ∫ from −2 to 2 of $|x|\,dx$

$=2\int_{0}^{2} x\,dx=2\cdot \frac{2^{2}}{2}=4$

Thus, the correct matching is: A–II, B–III, C–IV, D–I.