The value of \(\int_{0}^{1}\displaystyle\left(e^{x}+e^{-x}\right)dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(e-\displaystyle\frac{1}{e}\)
An antiderivative of \(e^{-x}\) is \(-e^{-x}\). So
\( \displaystyle \int_{0}^{1}\left(e^{x}+e^{-x}\right)dx = \left[e^{x}-e^{-x}\right]_{0}^{1} \)
\( \displaystyle = \left(e - \frac{1}{e}\right) - \left(1 - 1\right) \)
\( \displaystyle = e - \frac{1}{e} \)
Why the other options are incorrect
Option (2) \(e+\displaystyle\frac{1}{e}-2\) — Incorrect. It takes the antiderivative of \(e^{-x}\) as \(e^{-x}\) instead of \(-e^{-x}\).
Option (3) \(e+\displaystyle\frac{1}{e}\) — Incorrect. It has the same sign slip and also leaves out the value at the lower limit.
Option (4) \(e-1\) — Incorrect. This is only \(\int_{0}^{1}e^{x}\,dx\); the part \(\int_{0}^{1}e^{-x}\,dx = 1 - \frac{1}{e}\) is left out.