Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Definite integral | Value | ||
| (A) | \(\int_{1}^{e}\displaystyle\frac{\log x}{x}\,dx\) | (I) | 4 |
| (B) | \(\int_{-2}^{2}\displaystyle x^3(1 - x^2)\,dx\) | (II) | \(\displaystyle\frac{1}{2}\) |
| (C) | \(\int_{1}^{2}\displaystyle x\, dx\) | (III) | 0 |
| (D) | \(\int_{-2}^{2}\displaystyle |x|\, dx\) | (IV) | \(\displaystyle\frac{3}{2}\) |
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)
(A) \(\int_{1}^{e}\displaystyle\frac{\log x}{x}\,dx\)
This matches (II) \(\frac{1}{2}\). Let \(u = \log x\), so \(du = \frac{1}{x}\,dx\); the limits x = 1 and x = e become u = 0 and u = 1:
\( \displaystyle \int_{0}^{1} u\,du = \left[\frac{u^2}{2}\right]_{0}^{1} = \frac{1}{2} \)
(B) \(\int_{-2}^{2}\displaystyle x^3(1 - x^2)\,dx\)
This matches (III) 0. The integrand \(x^3 - x^5\) is an odd function, and the integral of an odd function over \([-a, a]\) is 0.
(C) \(\int_{1}^{2}\displaystyle x\, dx\)
This matches (IV) \(\frac{3}{2}\).
\( \displaystyle \int_{1}^{2} x\,dx = \left[\frac{x^2}{2}\right]_{1}^{2} = \frac{4 - 1}{2} = \frac{3}{2} \)
(D) \(\int_{-2}^{2}\displaystyle |x|\, dx\)
This matches (I) 4. \(|x|\) is an even function, so the integral is twice the integral over [0, 2], where \(|x| = x\):
\( \displaystyle 2\int_{0}^{2} x\,dx = 2\left[\frac{x^2}{2}\right]_{0}^{2} = 2 \times 2 = 4 \)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).