Match List-I with List-II
| List-I | List-II |
| [ Integral ] | [ Value of the integral] |
| (A). \( \int_1^2\displaystyle e^{2\log x}\,dx \) | (I). \( \log 2 \) |
| (B). \( \int_{-1}^1\displaystyle|x|\,dx \) | (II). \( \displaystyle\frac{1}{2} \) |
| (C). \( \int_0^1\displaystyle\log e^{x}\,dx \) | (III). \( \displaystyle\frac{7}{3} \) |
| (D). \( \int_1^2\displaystyle\frac{1}{x}\,dx \) | (IV). 1 |
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) - (II), (D) - (I)
Each integrand simplifies with \( e^{\log a}=a \) and \( \log e^{x}=x \).
(A) \( \int_1^2\displaystyle e^{2\log x}\,dx \)
This matches (III) 7/3. Since \( e^{2\log x}=e^{\log x^2}=x^2 \):
\( \displaystyle \int_1^2x^2\,dx=\left[\frac{x^3}{3}\right]_1^2 \)
\( \displaystyle =\frac{8-1}{3}=\frac{7}{3} \)
(B) \( \int_{-1}^1\displaystyle|x|\,dx \)
This matches (IV) 1. |x| is even, so
\( \displaystyle \int_{-1}^1|x|\,dx=2\int_0^1x\,dx=1 \)
(C) \( \int_0^1\displaystyle\log e^{x}\,dx \)
This matches (II) 1/2. Since \( \log e^{x}=x \):
\( \displaystyle \int_0^1x\,dx=\frac{1}{2} \)
(D) \( \int_1^2\displaystyle\frac{1}{x}\,dx \)
This matches (I) log 2.
\( \displaystyle \int_1^2\frac{dx}{x}=\log 2-\log 1=\log 2 \)
So the matching is (A) - (III), (B) - (IV), (C) - (II), (D) - (I).