Match List-I with List-II
| List-I | List-II | ||
| Integrals | Solution (where C is an arbitrary constant) | ||
| (A) | \( \int e^{2\log x}\,dx \) | (I) | \( \displaystyle\frac{x^{2}}{2}+C \) |
| (B) | \( \int e^{3\log x}\,dx \) | (II) | \( \displaystyle\frac{x^{3}}{3}+C \) |
| (C) | \( \int e^{-\log_{e}x}\,dx \) | (III) | \( \displaystyle\frac{x^{4}}{4}+C \) |
| (D) | \( \int e^{\log x}\,dx \) | (IV) | \( \log_{e}x+C \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
Each integrand simplifies with \( e^{k\log x}=x^{k} \).
(A) \( \int e^{2\log x}\,dx \)
This matches (II).
\( \displaystyle \int x^{2}\,dx=\frac{x^{3}}{3}+C \)
(B) \( \int e^{3\log x}\,dx \)
This matches (III).
\( \displaystyle \int x^{3}\,dx=\frac{x^{4}}{4}+C \)
(C) \( \int e^{-\log_{e}x}\,dx \)
This matches (IV).
\( \displaystyle \int x^{-1}\,dx=\log_{e}x+C \)
(D) \( \int e^{\log x}\,dx \)
This matches (I).
\( \displaystyle \int x\,dx=\frac{x^{2}}{2}+C \)
So the matching is (A) - (II), (B) - (III), (C) - (IV), (D) - (I).