Match List-I with List-II
| List-I | List-II | ||
| (A) | The value of \( I=\int_{0}^{1}\displaystyle\frac{1}{\sqrt{x}}\,dx \) is | (I) | 1 |
| (B) | The value of \( I=\int_{0}^{1}\displaystyle (5x^{2}+1)\,dx \) is | (II) | \( \displaystyle\frac{1}{2} \) |
| (C) | The value of \( I=\int_{0}^{\infty}\displaystyle e^{-x}\,dx \) is | (III) | \( \displaystyle\frac{8}{3} \) |
| (D) | The value of \( I=\int_{1}^{e}\displaystyle\frac{\log x}{x}\,dx \) is | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
(A) The value of \( I=\int_{0}^{1}\frac{1}{\sqrt{x}}\,dx \) is
This matches (IV) 2.
\( \displaystyle \int_0^1 x^{-1/2}\,dx=\Big[2\sqrt{x}\Big]_0^1=2 \)
(B) The value of \( I=\int_{0}^{1}(5x^{2}+1)\,dx \) is
This matches (III) \( \frac{8}{3} \).
\( \displaystyle \left[\frac{5x^3}{3}+x\right]_0^1=\frac{5}{3}+1=\frac{8}{3} \)
(C) The value of \( I=\int_{0}^{\infty}e^{-x}\,dx \) is
This matches (I) 1.
\( \displaystyle \int_0^\infty e^{-x}\,dx=\lim_{b\to\infty}\Big[-e^{-x}\Big]_0^b \)
\( \displaystyle =\lim_{b\to\infty}\left(1-e^{-b}\right) \)
\( \displaystyle =1 \)
(D) The value of \( I=\int_{1}^{e}\frac{\log x}{x}\,dx \) is
This matches (II) \( \frac{1}{2} \).
Put \( u=\log x \), \( du=\frac{dx}{x} \); the limits become 0 and 1:
\( \displaystyle \int_0^1 u\,du=\left[\frac{u^2}{2}\right]_0^1=\frac{1}{2} \)
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).