The value of \(\int_{-1}^{1}\displaystyle e^{|x|}\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2(e-1)\)
Let \(f(x) = e^{|x|}\). Since \(|-x| = |x|\), \(f(-x) = f(x)\), so f is an even function. Hence
\( \displaystyle \int_{-1}^{1}e^{|x|}\,dx = 2\int_{0}^{1}e^{x}\,dx \)
\( \displaystyle = 2\left[e^{x}\right]_{0}^{1} \)
\( = 2(e-1)\)
Why the other options are incorrect
Option (1) \(e-1\) — Incorrect. This is only the part over \([0, 1]\); the equal part over \([-1, 0]\) is left out.
Option (2) 2e — Incorrect. It leaves out the value at the lower limit, \(e^{0} = 1\).
Option (4) 0 — Incorrect. The function is even, not odd, and \(e^{|x|} \gt 0\), so the integral is positive.