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:
Answer & explanation
Correct answer: option 4
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)