Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Definite integral | Value | ||
| (A) | \(\int_{0}^{1}\displaystyle\frac{dx}{1+x^{2}}\) | (I) | \(\displaystyle\frac{14}{3}\) |
| (B) | \(\int_{0}^{\frac{\pi}{4}}\displaystyle\sec^{2}x\,dx\) | (II) | 0 |
| (C) | \(\int_{-2}^{2}\displaystyle x|x|\,dx\) | (III) | \(\displaystyle\frac{\pi}{4}\) |
| (D) | \(\int_{1}^{4}\displaystyle\sqrt{x}\,dx\) | (IV) | 1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(A) \(\int_{0}^{1}\frac{dx}{1+x^{2}}\)
This matches (III) \(\frac{\pi}{4}\):
\( \displaystyle \left[\tan^{-1}x\right]_{0}^{1} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \)
(B) \(\int_{0}^{\pi/4}\sec^{2}x\,dx\)
This matches (IV) 1:
\( \displaystyle \left[\tan x\right]_{0}^{\pi/4} = 1 - 0 = 1 \)
(C) \(\int_{-2}^{2}x|x|\,dx\)
This matches (II) 0. Let \(f(x) = x|x|\); then \(f(-x) = (-x)|-x| = -x|x| = -f(x)\), so f is an odd function and its integral over \([-2, 2]\) is 0.
(D) \(\int_{1}^{4}\sqrt{x}\,dx\)
This matches (I) \(\frac{14}{3}\):
\( \displaystyle \left[\frac{2}{3}x^{3/2}\right]_{1}^{4} = \frac{2}{3}(8 - 1) = \frac{14}{3} \)