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.
Answer & explanation
Correct answer: option 1
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.