Match List-I with List-II

List-I	List-II
Integral	Value
---	---
(A) $\int_{-1}^{1} (\vert x\vert + 1) dx$	(I) 0
(B) $\int_{-2}^{2} \vert x + 1\vert dx$	(II) 2
(C) $\int_{-1}^{1} 3\vert x^2\vert dx$	(III) 5
(D) $\int_{-1}^{1} x\vert x\vert dx$	(IV) 3

Choose the correct answer from the options given below:

(A) $\int_{-1}^{1} (|x| + 1)\, dx$

Since $|x|$ is an even function, $\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$

$= 2\int_{0}^{1} (x + 1)\, dx$

$= 2\left[\dfrac{x^2}{2} + x\right]_0^1$

$= 2\left(\dfrac{1}{2} + 1\right)$

$= 2 \times \dfrac{3}{2} = 3 \longrightarrow \text{(IV)}$

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(B) $\int_{-2}^{2} |x + 1|\, dx$

$|x + 1| = 0$ when $x = -1$, so splitting at $x = -1$:

For $x < -1$, $|x+1| = -(x+1)$

For $x \geq -1$, $|x+1| = (x+1)$

$= \int_{-2}^{-1} -(x+1)\, dx + \int_{-1}^{2} (x+1)\, dx$

First integral:

$= \left[-\dfrac{x^2}{2} - x\right]_{-2}^{-1}$

$= \left(-\dfrac{1}{2} + 1\right) - \left(-2 + 2\right) = \dfrac{1}{2}$

Second integral:

$= \left[\dfrac{x^2}{2} + x\right]_{-1}^{2}$

$= \left(2 + 2\right) - \left(\dfrac{1}{2} - 1\right) = 4 + \dfrac{1}{2} = \dfrac{9}{2}$

$= \dfrac{1}{2} + \dfrac{9}{2} = 5 \longrightarrow \text{(III)}$

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(C) $\int_{-1}^{1} 3|x^2|\, dx$

Since $x^2 \geq 0$ always, $|x^2| = x^2$. Also $x^2$ is an even function.

$= 3 \times 2\int_{0}^{1} x^2\, dx$

$= 6\left[\dfrac{x^3}{3}\right]_0^1$

$= 6 \times \dfrac{1}{3} = 2 \longrightarrow \text{(II)}$

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(D) $\int_{-1}^{1} x|x|\, dx$

$f(-x) = (-x)|-x| = -x|x| = -f(x)$

Since $f(x) = x|x|$ is an odd function, $\int_{-a}^{a} f(x)\, dx = 0$

$= 0 \longrightarrow \text{(I)}$

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(A) $\rightarrow$ (IV), (B) $\rightarrow$ (III), (C) $\rightarrow$ (II), (D) $\rightarrow$ (I)