\(\int_{-1}^{1}\displaystyle e^{|x|}\,dx =\)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2(e - 1)\)
Since \(e^{|-x|} = e^{|x|}\), the integrand is an even function, so
\( \displaystyle \int_{-1}^{1} e^{|x|}\,dx = 2\int_{0}^{1} e^{|x|}\,dx \)
For \(0 \le x \le 1\), |x| = x:
\( \displaystyle = 2\int_{0}^{1} e^{x}\,dx \)
\( \displaystyle = 2\left[e^{x}\right]_{0}^{1} \)
\(= 2(e - 1)\)
Why the other options are incorrect
Option (1) \(2\left(e^{-1} - 1\right)\) — Incorrect. This is negative, but \(e^{|x|} \gt 0\), so the integral must be positive.
Option (2) \(2(e + 1)\) — Incorrect. The lower limit gives \(e^{0} = 1\), which is subtracted, not added.
Option (3) \(e - 1\) — Incorrect. This is only the part from 0 to 1; the factor 2 for the part from −1 to 0 is missing.