Question

evaluate the integral ( int_{ln 8}^{ln 27} e^{\frac{x}{3}} dx ). ( int_{ln 8}^{ln 27} e^{\frac{x}{3}} dx = square ) (simplify your answer.)

Explanation:

Step1: Find the antiderivative of \( e^{\frac{x}{3}} \)

The integral of \( e^{ax} \) is \( \frac{1}{a}e^{ax} + C \). For \( a = \frac{1}{3} \), the antiderivative of \( e^{\frac{x}{3}} \) is \( 3e^{\frac{x}{3}} \).

Step2: Apply the Fundamental Theorem of Calculus

Evaluate \( 3e^{\frac{x}{3}} \) from \( \ln 8 \) to \( \ln 27 \).
First, at \( x = \ln 27 \): \( 3e^{\frac{\ln 27}{3}} = 3e^{\ln 27^{\frac{1}{3}}} = 3 \times 27^{\frac{1}{3}} = 3 \times 3 = 9 \)
Then, at \( x = \ln 8 \): \( 3e^{\frac{\ln 8}{3}} = 3e^{\ln 8^{\frac{1}{3}}} = 3 \times 8^{\frac{1}{3}} = 3 \times 2 = 6 \)
Subtract the lower limit from the upper limit: \( 9 - 6 = 3 \)

Answer:

\( 3 \)