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