Discussion

\(\displaystyle\int_{-\log 3}^{+\log 3} e^{|x|}\, dx =\)

Answer: C

The integral is 4.

\(e^{|x|}\) is even, so

\[ \begin{aligned} \int_{-\log 3}^{\log 3} e^{|x|}dx &= 2\int_{0}^{\log 3} e^{x}dx \\ &= 2(e^{\log 3} - 1) = 2(3 - 1) = 4 \end{aligned} \]

3 is \(e^{\log 3}\) alone, 6 forgets to subtract \(e^{0} = 1\), and 0 wrongly treats the integrand as odd.

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