Question

$\int_{-\log 3}^{+\log 3} e^{|x|} dx = $ ________.

• A. 3
• B. 6
• C. 4
• D. 8
• E. 0

Hint:

$e^{\log a} = a$.

Answer 1

The Correct Option is C

To solve the integral \(\int_{-\log 3}^{+\log 3} e^{|x|} dx\), we need to consider how the absolute value affects the expression. The absolute value function, \( |x| \), results in two different cases for \( x \):

• When \( x \geq 0 \), \( |x| = x \).
• When \( x < 0 \), \( |x| = -x \).

Given these cases, we can split the integral at \( x = 0 \). Therefore, the integral becomes:

\(\int_{-\log 3}^{+\log 3} e^{|x|} dx = \int_{-\log 3}^{0} e^{-x} dx + \int_{0}^{+\log 3} e^{x} dx\)

Let's evaluate each integral separately:

1. For the integral \(\int_{-\log 3}^{0} e^{-x} dx\):

\(\int e^{-x} dx = -e^{-x} + C\)

Evaluating from \(-\log 3\) to \(0\):

\(\left[-e^{-x}\right]_{-\log 3}^{0} = [-e^{0}] - [-e^{-\log 3}] = -1 + e^{\log 3}\) \(= -1 + 3\) \(= 2\)

1. For the integral \(\int_{0}^{+\log 3} e^{x} dx\):

\(\int e^{x} dx = e^{x} + C\)

Evaluating from \(0\) to \(\log 3\):

\(\left[e^{x}\right]_{0}^{\log 3} = [e^{\log 3}] - [e^{0}] = 3 - 1\) \(= 2\)

Adding the results of the two integrals:

\(2 + 2 = 4\)

The value of the integral \(\int_{-\log 3}^{+\log 3} e^{|x|} dx\) is \(4\). Thus, the correct answer is 4.