The value of \(\int_{1}^{5}\displaystyle 2^{x}\,dx\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{30}{\log_e 2}\)
An antiderivative of \(a^{x}\) is \(\displaystyle\frac{a^{x}}{\log_e a}\). So
\( \displaystyle \int_{1}^{5} 2^{x}\,dx = \left[\frac{2^{x}}{\log_e 2}\right]_{1}^{5} \)
\( \displaystyle = \frac{2^{5}-2^{1}}{\log_e 2} \)
\( \displaystyle = \frac{32-2}{\log_e 2} = \frac{30}{\log_e 2} \)
Why the other options are incorrect
Option (1) \(30\log_e 2\) — Incorrect. The factor \(\log_e 2\) divides; it does not multiply.
Option (2) \(\displaystyle\frac{31}{\log_e 2}\) — Incorrect. The lower limit gives \(2^{1} = 2\), not \(2^{0} = 1\).
Option (3) \(\displaystyle\frac{32}{\log_e 2}\) — Incorrect. This is only the value at the upper limit; the value at x = 1 must be subtracted.