If \( \int_{0}^{40}\displaystyle\frac{dx}{2x+1}=\log_e k \), then the value of \( k \) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9
\( \displaystyle \int_0^{40}\frac{dx}{2x+1}=\frac{1}{2}\Big[\log_e(2x+1)\Big]_0^{40} \)
\( \displaystyle =\frac{1}{2}\left(\log_e 81-\log_e 1\right) \)
\( \displaystyle =\frac{1}{2}\log_e 81 \)
\( \displaystyle =\log_e 81^{1/2} \)
\( \displaystyle =\log_e 9 \)
So k = 9.