Question

evaluate the integral.
\\( \int_{\ln 1}^{\ln 3} e^x \\, dx \\)
\\( \int_{\ln 1}^{\ln 3} e^x \\, dx = \square \\)

Explanation:

Step1: Recall the antiderivative of \( e^x \)

The antiderivative of \( e^x \) with respect to \( x \) is \( e^x \) itself, since the derivative of \( e^x \) is \( e^x \). So, by the Fundamental Theorem of Calculus, the definite integral \( \int_{a}^{b} e^x dx = e^x \big|_{a}^{b} = e^b - e^a \).

Step2: Substitute the limits of integration

Here, the lower limit \( a = \ln 1 \) and the upper limit \( b = \ln 3 \). First, simplify \( \ln 1 \): we know that \( \ln 1 = 0 \) because \( \log_{e} 1 = 0 \) (since \( e^0 = 1 \)). Now, substitute into the formula \( e^b - e^a \):
\( e^{\ln 3} - e^{\ln 1} \)
We know that \( e^{\ln x} = x \) (this is the inverse property of the exponential and logarithmic functions). So, \( e^{\ln 3} = 3 \) and \( e^{\ln 1} = e^{0} = 1 \) (since we already found \( \ln 1 = 0 \)).

Step3: Calculate the result

Substitute these values back: \( 3 - 1 = 2 \).

Answer:

\( 2 \)