Match List-I with List-II
| List-I | List-II | ||
| (A) | \(\int_{0}^{1}\displaystyle\frac{2x}{1 + x^{2}}\,dx\) | (I) | \(\displaystyle\frac{1}{2}\) |
| (B) | \(\int_{0}^{\frac{\pi}{2}}\displaystyle\sin^{2}x\cos x\,dx\) | (II) | \(\displaystyle\frac{\pi}{4}\) |
| (C) | \(\int_{0}^{1}\displaystyle\frac{1}{1 + x^{2}}\,dx\) | (III) | \(\log_e 2\) |
| (D) | \(\int_{0}^{\frac{\pi}{4}}\displaystyle\tan x\sec^{2}x\,dx\) | (IV) | \(\displaystyle\frac{1}{3}\) |
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) \(\int_{0}^{1}\frac{2x}{1 + x^{2}}\,dx\)
This matches (III) \(\log_e 2\).
The numerator is the derivative of the denominator:
\( \displaystyle \Big[\log_e(1 + x^{2})\Big]_{0}^{1} = \log_e 2 - \log_e 1 = \log_e 2 \)
(B) \(\int_{0}^{\frac{\pi}{2}}\sin^{2}x\cos x\,dx\)
This matches (IV) \(\frac{1}{3}\).
With \(t = \sin x\), \(dt = \cos x\,dx\) and t runs from 0 to 1:
\( \displaystyle \int_{0}^{1}t^{2}\,dt = \frac{1}{3} \)
(C) \(\int_{0}^{1}\frac{1}{1 + x^{2}}\,dx\)
This matches (II) \(\frac{\pi}{4}\).
\( \displaystyle \Big[\tan^{-1}x\Big]_{0}^{1} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \)
(D) \(\int_{0}^{\frac{\pi}{4}}\tan x\sec^{2}x\,dx\)
This matches (I) \(\frac{1}{2}\).
With \(t = \tan x\), \(dt = \sec^{2}x\,dx\) and t runs from 0 to 1:
\( \displaystyle \int_{0}^{1}t\,dt = \frac{1}{2} \)