Match List-I with List-II
Choose the correct answer from the options given below. |
(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) |
(A)-(II), (B)-(III), (C)-(IV), (D)-(I) |
The correct answer is Option (1) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(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. |