Match List-I with List-II

List-I	List-II
Definite integral	Value
---	---
(A) $\int_{1}^{e} \frac{\log x}{x} dx$	(I) 4
(B) $\int_{-2}^{2} x^3(1 - x^2) dx$	(II) $\frac{1}{2}$
(C) $\int_{1}^{2} x \, dx$	(III) 0
(D) $\int_{-2}^{2} \vert x\vert dx$	(IV) $\frac{3}{2}$

Choose the correct answer from the options given below:

$\int_{1}^{e} \frac{\log x}{x}\, dx$

Let $t = \log x \implies dt = \frac{1}{x}\,dx$

When $x = 1$, $t = 0$ and when $x = e$, $t = 1$

$\int_{0}^{1} t\, dt = \left[\frac{t^2}{2}\right]_0^1$

$= \frac{1}{2} - 0$

$= \frac{1}{2}$

(A) → (II)

---

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

Let $f(x) = x^3(1 - x^2) = x^3 - x^5$

$f(-x) = (-x)^3 - (-x)^5 = -x^3 + x^5 = -f(x)$

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

$\int_{-2}^{2} x^3(1 - x^2)\, dx = 0$

(B) → (III)

---

$\int_{1}^{2} x\, dx$

$= \left[\frac{x^2}{2}\right]_1^2$

$= \frac{4}{2} - \frac{1}{2}$

$= \frac{3}{2}$

(C) → (IV)

---

$\int_{-2}^{2} |x|\, dx$

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

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

$= 2\left[\frac{x^2}{2}\right]_0^2$

$= 2\left(\frac{4}{2} - 0\right)$

$= 4$

(D) → (I)

---

List-I	List-II
(A)	(II) $\frac{1}{2}$
(B)	(III) $0$
(C)	(IV) $\frac{3}{2}$
(D)	(I) $4$