Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\int_{0}^{1}\displaystyle\frac{x^2}{1 + x^3}\,dx\) | (I) | 0 |
| (B) | \(\int_{0}^{\pi}\displaystyle 3\sin x\,dx\) | (II) | \(2\log_e\left(\displaystyle\frac{3}{2}\right)\) |
| (C) | \(\int_{-1}^{1}\displaystyle \sin^5 x\cos^6 x\,dx\) | (III) | 6 |
| (D) | \(\int_{2}^{3}\displaystyle\frac{4}{x^2 - 1}\,dx\) | (IV) | \(\displaystyle\frac{1}{3}\log_e 2\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) \(\int_{0}^{1}\frac{x^2}{1 + x^3}\,dx\)
This matches (IV) \(\frac{1}{3}\log_e 2\). Let \(t = 1 + x^3\), so \(dt = 3x^2\,dx\), and t goes from 1 to 2:
\( \displaystyle \int_{0}^{1}\frac{x^2}{1 + x^3}\,dx = \frac{1}{3}\int_{1}^{2}\frac{dt}{t} = \frac{1}{3}\log_e 2 \)
(B) \(\int_{0}^{\pi} 3\sin x\,dx\)
This matches (III) 6.
\( \displaystyle \int_{0}^{\pi} 3\sin x\,dx = 3\left[-\cos x\right]_{0}^{\pi} \)
\( \displaystyle = 3(1 + 1) = 6 \)
(C) \(\int_{-1}^{1}\sin^5 x\cos^6 x\,dx\)
This matches (I) 0. \(\sin^5 x\) is odd and \(\cos^6 x\) is even, so the integrand is an odd function, and its integral over \([-1, 1]\) is 0.
(D) \(\int_{2}^{3}\frac{4}{x^2 - 1}\,dx\)
This matches (II) \(2\log_e\left(\frac{3}{2}\right)\). By partial fractions:
\( \displaystyle \frac{4}{x^2 - 1} = \frac{2}{x - 1} - \frac{2}{x + 1} \)
\( \displaystyle \int_{2}^{3}\frac{4}{x^2 - 1}\,dx = 2\left[\log_e\left|\frac{x - 1}{x + 1}\right|\right]_{2}^{3} \)
\( \displaystyle = 2\left(\log_e\frac{1}{2} - \log_e\frac{1}{3}\right) = 2\log_e\left(\frac{3}{2}\right) \)
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).