The value of \(\int_{0}^{\log_e 3}\displaystyle e^{2x}\,dx\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4
Since \(\frac{d}{dx}\left(e^{2x}\right) = 2e^{2x}\), an antiderivative of \(e^{2x}\) is \(\frac{1}{2}e^{2x}\). By the fundamental theorem of calculus,
\( \displaystyle \int_{0}^{\log_e 3}e^{2x}\,dx = \frac{1}{2}\Big[e^{2x}\Big]_{0}^{\log_e 3} \)
Using \(e^{2\log_e 3} = \left(e^{\log_e 3}\right)^{2} = 3^{2} = 9\) and \(e^{0} = 1\):
\( \displaystyle = \frac{1}{2}(9 - 1) \)
\( = 4 \)
Why the other options are incorrect
Option (1) 2 — Incorrect. This is \(e^{\log_e 3} - 1\), the value for \(e^{x}\), not \(e^{2x}\).
Option (3) \(\displaystyle\frac{9}{2}\) — Incorrect. This leaves out the lower-limit value \(\frac{1}{2}e^{0} = \frac{1}{2}\).
Option (4) 8 — Incorrect. This misses the factor \(\frac{1}{2}\) from the antiderivative.