Problem 15 - Calculus Early Transcendentals

Question

Evaluate the following iterated integrals. $$\int_{1}^{\ln 5} \int_{0}^{\ln 3} e^{x+y} d x d y$$

Step-by-Step Solution

Answer

Question: Evaluate the iterated integral $$\int_{1}^{\ln 5} \int_{0}^{\ln 3} e^{x+y} d x d y$$. Answer: The evaluated iterated integral is $$2(5 - e)$$

Step 1: Integrate with respect to x

Integrate the inner integral with respect to x. $$\int_{0}^{\ln 3} e^{x+y} d x = e^y \int_{0}^{\ln 3} e^x d x$$ Now, integrate \(e^x\) with respect to x. $$e^y \int_{0}^{\ln 3} e^x d x = e^y \left[e^x\right]_{0}^{\ln 3} = e^y (e^{\ln 3} - e^0) = e^y (3 - 1) = 2e^y$$

Step 2: Integrate with respect to y

Now, integrate the outer integral $$\int_{1}^{\ln 5} 2e^y d y$$. Find the antiderivative of \(2e^y\) with respect to y. $$\int_{1}^{\ln 5} 2e^y d y = 2 \int_{1}^{\ln 5} e^y d y = 2 \left[e^y\right]_{1}^{\ln 5} = 2(e^{\ln 5} - e^1) = 2(5 - e)$$

Step 3: Final Solution

The evaluated iterated integral is: $$\int_{1}^{\ln 5} \int_{0}^{\ln 3} e^{x+y} d x d y = 2(5 - e)$$