Problem 24 - Calculus Early Transcendentals

Question

Evaluate the following integrals as they are written. $$\int_{0}^{\ln 2} \int_{e^{x}}^{2} d y d x$$

Step-by-Step Solution

Answer

The result of the double integral is $$2\ln 2 - 1$$.

Step 1: Evaluate the inner integral with respect to y

First, let's consider the inner integral. Treat x as a constant while integrating with respect to y. $$\int_{e^x}^{2} dy$$ As the integrand has no y dependence and is just constant "1", we simply compute the difference between the upper and lower limits. $$\int_{e^x}^{2} dy = 2 - e^x$$ Now, we have the result of the inner integral.

Step 2: Evaluate the outer integral with respect to x

Next, we will use the result from Step 1 to evaluate the outer integral with respect to x. $$\int_{0}^{\ln 2} (2 - e^x) dx$$ Now, integrate the expression with respect to x: $$\int_{0}^{\ln 2} (2 - e^x) dx = \left[ 2x - e^x \right]_{0}^{\ln 2}$$ Now, substitute the limits of integration and subtract: $$\left[ 2(\ln 2) - e^{\ln 2} \right] - \left[ 2(0) - e^{0} \right] = 2\ln 2 - 2 - (0 - 1) = 2\ln 2 - 1$$ So, the result of the double integral is: $$\int_{0}^{\ln 2} \int_{e^{x}}^{2} dy dx = 2\ln 2 - 1$$