Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Definite integral | Value | ||
| (A) | \(\int_{0}^{3}\displaystyle\frac{dx}{9+x^{2}}\) | (I) | \(\pi\) |
| (B) | \(\int_{0}^{2}\displaystyle\frac{dx}{\sqrt{16-x^{2}}}\) | (II) | \(\displaystyle\frac{1}{2}\log_e\frac{3}{2}\) |
| (C) | \(\int_{0}^{2}\displaystyle\sqrt{4-x^{2}}\,dx\) | (III) | \(\displaystyle\frac{\pi}{12}\) |
| (D) | \(\int_{2}^{3}\displaystyle\frac{dx}{x^{2}-1}\) | (IV) | \(\displaystyle\frac{\pi}{6}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A) \(\int_{0}^{3}\frac{dx}{9+x^{2}}\)
This matches (III) \(\frac{\pi}{12}\), using \(\int\frac{dx}{a^{2}+x^{2}} = \frac{1}{a}\tan^{-1}\frac{x}{a}\) with \(a = 3\):
\( \displaystyle \left[\frac{1}{3}\tan^{-1}\frac{x}{3}\right]_{0}^{3} = \frac{1}{3}\cdot\frac{\pi}{4} = \frac{\pi}{12} \)
(B) \(\int_{0}^{2}\frac{dx}{\sqrt{16-x^{2}}}\)
This matches (IV) \(\frac{\pi}{6}\), using \(\int\frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin^{-1}\frac{x}{a}\) with \(a = 4\):
\( \displaystyle \left[\sin^{-1}\frac{x}{4}\right]_{0}^{2} = \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \)
(C) \(\int_{0}^{2}\sqrt{4-x^{2}}\,dx\)
This matches (I) \(\pi\), using \(\int\sqrt{a^{2}-x^{2}}\,dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\) with \(a = 2\):
\( \displaystyle \left[\frac{x}{2}\sqrt{4-x^{2}} + 2\sin^{-1}\frac{x}{2}\right]_{0}^{2} = 0 + 2\cdot\frac{\pi}{2} = \pi \)
(D) \(\int_{2}^{3}\frac{dx}{x^{2}-1}\)
This matches (II) \(\frac{1}{2}\log_e\frac{3}{2}\), using \(\int\frac{dx}{x^{2}-a^{2}} = \frac{1}{2a}\log_e\left|\frac{x-a}{x+a}\right|\) with \(a = 1\):
\( \displaystyle \frac{1}{2}\left[\log_e\frac{x-1}{x+1}\right]_{2}^{3} = \frac{1}{2}\left(\log_e\frac{1}{2} - \log_e\frac{1}{3}\right) = \frac{1}{2}\log_e\frac{3}{2} \)