Problem 62 - Calculus Early Transcendentals

Question

Reverse the order of integration in the following integrals. $$\int_{1}^{e} \int_{0}^{\ln x} f(x, y) d y d x$$

Step-by-Step Solution

Answer

In order to reverse the order of integration for the given integral $$\int_{1}^{e} \int_{0}^{\ln x} f(x, y) d y d x$$, we first identify the region in the xy-plane based on the original limits of integration. We then determine the new limits for x and y by integrating first with respect to x and then with respect to y. Finally, the integral with the reversed order of integration is $$\int_{0}^{1} \int_{e^y}^{e} f(x, y) d x d y$$.

Step 1: Identify the region

The given integral is: $$\int_{1}^{e} \int_{0}^{\ln x} f(x, y) d y d x$$ The outer integral has limits \(1 \leq x \leq e\). The inner integral has limits \(0 \leq y \leq \ln x\). Let's determine the region in the xy-plane based on these limits. 1. \(x\) varies from \(1\) to \(e\) 2. For a given \(x\), \(y\) varies from \(0\) to \(\ln x\) Since \(e^0 = 1\) and \(e^1=e\), we can identify that the region is bounded by the curves \(x=1\), \(x=e\), \(y=0\) and \(y=\ln x\).

Step 2: Determine the new limits of integration

Now, we keep in mind that we are reversing the order of integration, i.e., integrating first with respect to \(x\) and then with respect to \(y\). For a given value of \(y\), \(x\) varies from \(x=e^y\) to \(x=e\). To find the limits for \(y\), notice that the lowest point in the region is \((1, 0)\) and the highest point is \((e, 1)\), so \(y\) varies from \(0\) to \(1\). Therefore, the new limits of integration are: 1. \(0 \leq y \leq 1\) 2. \(e^y \leq x \leq e\)

Step 3: Rewrite the integral with the new order of integration

Now let's rewrite the given integral with the new limits of integration. The reversed integral is: $$\int_{0}^{1} \int_{e^y}^{e} f(x, y) d x d y$$